{"id":3974,"date":"2022-04-12T17:11:10","date_gmt":"2022-04-12T17:11:10","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=3974"},"modified":"2022-11-01T04:17:30","modified_gmt":"2022-11-01T04:17:30","slug":"double-integrals-over-nonrectangular-regions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/double-integrals-over-nonrectangular-regions\/","title":{"raw":"Double Integrals over Nonrectangular Regions","rendered":"Double Integrals over Nonrectangular Regions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul class=\"os-abstract\">\r\n \t<li><span class=\"os-abstract-content\">Recognize when a function of two variables is integrable over a general region.<\/span><\/li>\r\n \t<li><span class=\"os-abstract-content\">Evaluate a double integral by computing an iterated integral over a region bounded by two vertical lines and two functions of [latex]x[\/latex] or two horizontal lines and two functions of [latex]y[\/latex].<\/span><\/li>\r\n \t<li><span class=\"os-abstract-content\">Simplify the calculation of an iterated integral by changing the order of integration.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<section id=\"fs-id1167793238235\" data-depth=\"1\">\r\n<h2 data-type=\"title\">General Regions of Integration<\/h2>\r\n<p id=\"fs-id1167793871811\">An example of a general bounded region [latex]D[\/latex] on a plane is shown in\u00a0Figure 1. Since [latex]D[\/latex] is bounded on the plane, there must exist a rectangular region [latex]R[\/latex] on the same plane that encloses the region [latex]D[\/latex], that is, a rectangular region [latex]R[\/latex] exists such that [latex]D[\/latex] is a subset of [latex]R(D \\subseteq R)[\/latex].<\/p>\r\n\r\n<\/section>[caption id=\"attachment_1317\" align=\"aligncenter\" width=\"417\"]<img class=\"size-full wp-image-1317\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23152039\/5-2-1.jpeg\" alt=\"A rectangle R with a shape D inside of it. Inside D, there is a point labeled g(x, y) = f(x, y). Outside D but still inside R, there is a point labeled g(x, y) = 0.\" width=\"417\" height=\"310\" \/> Figure 1. For a region [latex]\\small{D}[\/latex] that is a subset of\u00a0[latex]\\small{R}[\/latex], we can define a function\u00a0[latex]\\small{g(x,y)}[\/latex] to equal\u00a0[latex]\\small{f(x,y)}[\/latex] at every point in\u00a0[latex]\\small{D}[\/latex] and\u00a0[latex]\\small{0}[\/latex] at every point of\u00a0[latex]\\small{R}[\/latex] not in\u00a0[latex]\\small{D}[\/latex].[\/caption]\r\n<p id=\"fs-id1167793267284\">Suppose [latex]z=f(x, y)[\/latex] is defined on a general planar bounded region [latex]D[\/latex] as in\u00a0Figure 1. In order to develop double integrals of [latex]f[\/latex] over [latex]D[\/latex], we extend the definition of the function to include all points on the rectangular region [latex]R[\/latex] and then use the concepts and tools from the preceding section. But how do we extend the definition of [latex]f[\/latex] to include all the points on [latex]R[\/latex]? We do this by defining a new function [latex]g(x, y)[\/latex] on [latex]R[\/latex] as follows:<\/p>\r\n[latex]\\hspace{8cm}\\large{g(x,y)=\\Bigg\\{\\begin{align}\r\n\r\n&amp;f(x,y)&amp;\\ &amp;\\text{if }(x,y)\\text{ is in }D \\\\\r\n\r\n&amp;0&amp;\\ &amp;\\text{if }(x,y)\\text{ is in }R\\text{ but not in }D\r\n\r\n\\end{align}}[\/latex]\r\n\r\nNote that we might have some technical difficulties if the boundary of [latex]D[\/latex] is complicated. So we assume the boundary to be a piecewise smooth and continuous simple closed curve. Also, since all the results developed in\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/double-integrals-over-rectangular-regions\/\">Double Integrals over Rectangular Regions<\/a>\u00a0used an integrable function [latex]f(x, y)[\/latex], we must be careful about [latex]g(x, y)[\/latex] and verify that [latex]g(x, y)[\/latex] is an integrable function over the rectangular region [latex]R[\/latex]. This happens as long as the region [latex]D[\/latex] is bounded by simple closed curves. For now we will concentrate on the descriptions of the regions rather than the function and extend our theory appropriately for integration.\r\n<p id=\"fs-id1167794060679\">We consider two types of planar bounded regions.<\/p>\r\n\r\n<div id=\"fs-id1167793857192\" class=\"ui-has-child-title\" data-type=\"note\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">DEFINITION<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167793884044\">A region [latex]D[\/latex] in the [latex](x, y)[\/latex]-plane is of\u00a0<span id=\"f783ca08-ca2f-4ebb-a559-d1b30c2f9b8c_term209\" data-type=\"term\">Type I<\/span>\u00a0if it lies between two vertical lines and the graphs of two continuous functions [latex]g_1(x)[\/latex] and [latex]g_2(x)[\/latex]. That is (Figure 2),<\/p>\r\n<p style=\"text-align: center;\">[latex]{D} = \\left \\{(x,y) \\mid a \\leq x \\leq \\ b,g_1 \\leq y \\leq g_2 \\ (x) \\right \\}[\/latex].<\/p>\r\n<p id=\"fs-id1167794337439\">A region [latex]D[\/latex] in the [latex]xy[\/latex] plane is of\u00a0<span id=\"f783ca08-ca2f-4ebb-a559-d1b30c2f9b8c_term210\" data-type=\"term\">Type II<\/span>\u00a0if it lies between two horizontal lines and the graphs of two continuous functions [latex]h_1(y)[\/latex] and [latex]h_2(y)[\/latex]. That is (Figure 3),<\/p>\r\n<p style=\"text-align: center;\">[latex]{D} = \\left \\{(x,y) \\mid c \\leq y \\leq \\ d,h_1 \\leq x \\leq h_2 \\ (y) \\right \\}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[caption id=\"attachment_1320\" align=\"alignnone\" width=\"963\"]<img class=\"size-full wp-image-1320\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23155027\/5-2-2.jpeg\" alt=\"The graphs showing a region marked D. In all instances, between a and b, there is a shape that is defined by two functions g1(x) and g2(x). In one instance, the two functions do not touch; in another instance, they touch at the end point a, and in the last instance they touch at both end points.\" width=\"963\" height=\"274\" \/> Figure 2.\u00a0A Type I region lies between two vertical lines and the graphs of two functions of\u00a0[latex]x[\/latex].[\/caption]\u00a0[caption id=\"attachment_1321\" align=\"aligncenter\" width=\"665\"]<img class=\"size-full wp-image-1321\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23155104\/5-2-3.jpeg\" alt=\"The graphs show a region marked D. In all instances, between c and d, there is a shape that is defined by two vertically oriented functions x = h1(y) and x = h2(y). In one instance, the two functions do not touch; in the other instance, they touch at the end point c.\" width=\"665\" height=\"310\" \/> Figure 3.\u00a0A Type II region lies between two horizontal lines and the graphs of two functions of [latex]y[\/latex].[\/caption]\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Describing a region as type I and also as type ii<\/h3>\r\nConsider the region in the first quadrant between the functions [latex]{y} = {\\sqrt{x}}[\/latex] and [latex]y=x^{3}[\/latex] (Figure 4). Describe the region first as Type I and then as Type II.\r\n\r\n[caption id=\"attachment_1323\" align=\"aligncenter\" width=\"503\"]<img class=\"size-full wp-image-1323\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23155305\/5-2-4.jpeg\" alt=\"The region D is drawn between two functions, namely, y = the square root of x and y = x3.\" width=\"503\" height=\"277\" \/> Figure 4. Region [latex]D[\/latex]\u00a0can be described as Type I or as Type II.[\/caption][reveal-answer q=\"375418035\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"375418035\"]\r\n<p id=\"fs-id1167793848514\">When describing a region as Type I, we need to identify the function that lies above the region and the function that lies below the region. Here, region [latex]D[\/latex] is bounded above by [latex]{y} = {\\sqrt{x}}[\/latex] and below by [latex]y=x^{3}[\/latex] in the interval for [latex]x[\/latex] in [latex][0,1][\/latex]. Hence, as Type I, [latex]D[\/latex] is described as the set [latex]{\\left \\{(x,y) \\mid 0 \\leq x \\leq 1,x^3 \\leq y \\leq \\sqrt{x} \\right \\}}[\/latex].<\/p>\r\n<p id=\"fs-id1167793524718\">However, when describing a region as Type II, we need to identify the function that lies on the left of the region and the function that lies on the right of the region. Here, the region [latex]D[\/latex] is bounded on the left by [latex]x=y^{2}[\/latex] and on the right by [latex]x = \\sqrt[3]{y}[\/latex] in the interval for [latex]y[\/latex] in [latex][0,1][\/latex]. Hence, as Type II, [latex]D[\/latex] is described as the set [latex]{\\left \\{(x,y) \\mid 0 \\leq y \\leq 1,y^2 \\leq x \\leq \\sqrt[3]{y} \\right \\}}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nConsider the region in the first quadrant between the functions [latex]y=2x[\/latex] and [latex]y=x^{2}[\/latex]. Describe the region first as Type I and then as Type II.\r\n\r\n[reveal-answer q=\"984622178\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"984622178\"]\r\n\r\nType I and Type II are expressed as [latex]\\{(x,y)|0\\leq{x}\\leq2,x^2\\leq{y}\\leq2x\\}[\/latex] and [latex]\\{(x,y)|0\\leq{y}\\leq4,\\frac12y\\leq{x}\\leq\\sqrt{y}\\}[\/latex], respectively.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2 data-type=\"title\">Double Integrals over Nonrectangular Regions<\/h2>\r\n<p id=\"fs-id1167794051685\">To develop the concept and tools for evaluation of a double integral over a general, nonrectangular region, we need to first understand the region and be able to express it as Type I or Type II or a combination of both. Without understanding the regions, we will not be able to decide the limits of integrations in double integrals. As a first step, let us look at the following theorem.<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Theorem: double integrals over nonrectangular regions<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167794200474\">Suppose [latex]g(x, y)[\/latex] is the extension to the rectangle [latex]R[\/latex] of the function [latex]f(x, y)[\/latex] defined on the regions [latex]D[\/latex] and [latex]R[\/latex] as shown in\u00a0Figure 1\u00a0inside [latex]R[\/latex]. Then [latex]g(x, y)[\/latex] is integrable and we define the double integral of [latex]f(x, y)[\/latex] over [latex]D[\/latex] by<\/p>\r\n<p style=\"text-align: center;\">[latex]\\underset{D}{\\displaystyle\\iint} {f(x,y)dA} = \\underset{R}{\\displaystyle\\iint} {g(x,y)dA}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793355329\">The right-hand side of this equation is what we have seen before, so this theorem is reasonable because [latex]R[\/latex] is a rectangle and [latex]\\underset{R}{\\displaystyle\\iint}{g(x,y)dA}[\/latex] has been discussed in the preceding section. Also, the equality works because the values of [latex]g(x, y)[\/latex] are [latex]0[\/latex] for any point [latex](x, y)[\/latex] that lies outside [latex]D[\/latex], and hence these points do not add anything to the integral. However, it is important that the rectangle [latex]R[\/latex] contains the region [latex]D[\/latex].<\/p>\r\n<p id=\"fs-id1167793568896\">As a matter of fact, if the region [latex]D[\/latex] is bounded by smooth curves on a plane and we are able to describe it as Type I or Type II or a mix of both, then we can use the following theorem and not have to find a rectangle [latex]R[\/latex] containing the region.<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Theorem: fubini's theorem (strong form)<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167793276048\">For a function [latex]f(x, y)[\/latex] that is continuous on a region [latex]D[\/latex] of Type I, we have<\/p>\r\n<p style=\"text-align: center;\">[latex]\\underset{D}{\\displaystyle\\iint}{f(x,y)dA} = \\underset{D}{\\displaystyle\\iint}{f(x,y)dy \\ dx}=\\displaystyle\\int_a^b\\left[\\displaystyle\\int_{g_1(x)}^{g_2(x)}f(x,y)dy\\right]dx.[\/latex]<\/p>\r\nSimilarly, for a function [latex]f(x, y)[\/latex] that is continuous on a region [latex]D[\/latex] of Type II, we have\r\n<p style=\"text-align: center;\">[latex]\\underset{D}{\\displaystyle\\iint}{f(x,y)dA} = \\underset{D}{\\displaystyle\\iint}{f(x,y)dx \\ dy}=\\displaystyle\\int_c^d\\left[\\displaystyle\\int_{h_1(y)}^{h_2(y)}f(x,y)dx\\right]dy.[\/latex]<\/p>\r\n\r\n<\/div>\r\nThe integral in each of these expressions is an iterated integral, similar to those we have seen before. Notice that, in the inner integral in the first expression, we integrate [latex]f(x, y)[\/latex] with [latex]x[\/latex] being held constant and the limits of integration being [latex]g_1(x)[\/latex] and [latex]g_2(x)[\/latex]. In the inner integral in the second expression, we integrate [latex]f(x, y)[\/latex] with [latex]y[\/latex] being held constant and the limits of integration are [latex]h_1(x)[\/latex] and\u00a0[latex]h_2(x)[\/latex].\r\n<div class=\"textbox exercises\">\r\n<h3>Example: evaluating an iterated integral over a type I region<\/h3>\r\nEvaluate the integral [latex]\\underset{D}{\\displaystyle\\iint}{x^2}{e^{xy}}{dA}[\/latex] where [latex]D[\/latex] is shown in\u00a0Figure 5 (see solution for image).\r\n\r\n[reveal-answer q=\"615274593\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"615274593\"]\r\n\r\nFirst construct the region [latex]D[\/latex] as a Type I region (Figure 5). Here [latex]{D} = {\\left \\{{(x,y)}{\\mid}{0} \\ {\\leq} \\ {x} \\ {\\leq} \\ {2,{\\frac{1}{2}x}} \\ {\\leq} \\ y \\ {\\leq} \\ {1} \\right \\}}[\/latex]. Then we have\r\n<p style=\"text-align: center;\">[latex]\\large{\\underset{D}{\\displaystyle\\iint}{x^2}{e^{xy}}{dA} = \\displaystyle\\int_{x=0}^{x=2}\\displaystyle\\int_{y=1\/2x}^{y=1}x^2e^{xy}dydx.}[\/latex]<\/p>\r\n\r\n[caption id=\"attachment_1324\" align=\"aligncenter\" width=\"251\"]<img class=\"size-full wp-image-1324\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23155437\/5-2-5.jpeg\" alt=\"A triangle marked D drawn with lines y = 1\/2 x and y = 1, with vertices (0, 0), (2, 1), and (0, 1). Here, there is a pair of red arrows reaching vertically from one edge to the other.\" width=\"251\" height=\"160\" \/> Figure 5. We can express region [latex]\\small{D}[\/latex] as a Type I region and integrate from [latex]\\small{y=\\frac{1}{2}x}[\/latex] to\u00a0[latex]\\small{y=1}[\/latex], between the lines\u00a0[latex]\\small{x=0}[\/latex] and\u00a0[latex]\\small{x=2}[\/latex].[\/caption]\r\nTherefore, we have\r\n[latex]\\begin{align}\\displaystyle\\int_{x=0}^{x=2}\\displaystyle\\int_{y=\\frac12x}^{y=1}x^2e^{xy}dydx &amp;=\\int_{x=0}^{x=2}[\\int_{y=\\frac12x}^{y=1}x^2e^{xy}dydx] &amp;\\text{Iterated for a Type I region.} \\\\&amp;=\\int_{x=0}^{x=2}[x^2\\frac{e^xy}x]\\Bigg|_{y=\\frac12x}^{y=1}dx &amp;\\text{Integrate with respect to }y\\text{ using }u\\text{-substitution} \\\\&amp; &amp;\\text{ with }u=xy\\text{ where }x\\text{ is held constant} \\\\&amp;=\\int_{x=0}^{x=2}[xe^x-xe^{\\frac{x^2}2}]dx &amp;\\text{Integrate with respect to }x\\text{ using }u\\text{-substitution with}u=\\frac12x^2 \\\\&amp;=[xe^x-e^x-e^{\\frac12x^2}]\\bigg|_{x=0}^{x=2}=2\\end{align}[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nIn\u00a0the Example: Evaluating an Iterated Integral over a Type I Region, we could have looked at the region in another way, such as [latex]{D} = {\\left \\{{(x,y)}{\\mid}{0} \\ {\\leq} \\ {y} \\ {\\leq} \\ {1,0} \\ {\\leq} \\ {x} \\ {\\leq} \\ {2y} \\right \\}}[\/latex] (Figure 6).\r\n\r\n[caption id=\"attachment_1340\" align=\"aligncenter\" width=\"229\"]<img class=\"size-full wp-image-1340\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25055920\/5-2-6.jpeg\" alt=\"A triangle marked D drawn with lines x = 2y and y = 1, with vertices (0, 0), (2, 1), and (0, 1). Here there is a pair of red arrows reaching horizontally from one edge to the other.\" width=\"229\" height=\"160\" \/> Figure 6.[\/caption]\r\n\r\nThis is a Type II region and the integral would then look like\r\n<p style=\"text-align: center;\">[latex]\\large{\\underset{D}{\\displaystyle\\iint}{x^2}{e^{xy}}{dA} = \\displaystyle\\int_{y=0}^{y=1}\\displaystyle\\int_{x=0}^{x=2u}{x^2}{e^{xy}}{dx} \\ {dy}}[\/latex]<\/p>\r\nHowever, if we integrate first with respect to [latex]x[\/latex], this integral is lengthy to compute because we have to use integration by parts twice.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: evaluating an iterated integral over a type II region<\/h3>\r\nEvaluate the integral [latex]\\underset{D}{\\displaystyle\\iint}{({3x^2}+{y^2})} \\ {dA}[\/latex] where\u00a0[latex]{=} \\ {\\left \\{{(x,y)}{\\mid}{-2} \\ {\\leq} \\ {y} \\ {\\leq} \\ {3,y^2-3} \\ {\\leq} \\ {x} \\ {\\leq} \\ {y+3} \\right \\}}[\/latex].\r\n\r\n[reveal-answer q=\"518274593\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"518274593\"]\r\n\r\nNotice that [latex]D[\/latex] can be seen as either a Type I or a Type II region, as shown in\u00a0Figure 7. However, in this case describing [latex]D[\/latex] as Type I is more complicated than describing it as Type II. Therefore, we use [latex]D[\/latex] as a Type II region for the integration.\r\n\r\n[caption id=\"attachment_1341\" align=\"aligncenter\" width=\"891\"]<img class=\"size-full wp-image-1341\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25060113\/5-2-7.jpeg\" alt=\"This figure consists of two figures labeled a and b. In figure a, a region is bounded by y = the square root of the quantity (x + 3), y = the negative of the square root of the quantity (x + 3), and y = x minus 3, which has points of intersection (6, 3), (1, negative 2), and (0, negative 3). There are vertical lines in the shape, and it is noted that this is a type I region: integrate first with respect to y. In figure b, a region is bounded by x = y2 minus 3 and x = y + 3, which has points of intersection (6, 3), (1, negative 2), and (0, negative 3). There are horizontal lines in the shape, and it is noted that this is a type II region: integrate first with respect to x.\" width=\"891\" height=\"372\" \/> Figure 7.\u00a0The region [latex]D[\/latex] in this example can be either (a) Type I or (b) Type II.[\/caption]Choosing this order of integration, we have[latex]\\begin{align}\\underset{D}{\\displaystyle\\iint}(3x^2+y^2)dA&amp;=\\displaystyle\\int_{y=-2}^{y=3}\\displaystyle\\int_{x=y^2-3}^{x=y+3}(3x^2+y^2)dxdy &amp;\\quad &amp;\\text{Interated integral, Type II region}.\\\\&amp;=\\displaystyle\\int_{y=-2}^{y=3}(x^2+xy^2\\Bigg|_{y^2-3}^{y+3}dy &amp;\\quad &amp;\\text{Integrate with respect to }x. \\\\&amp;=\\displaystyle\\int_{y=-2}^{y=3}\\left((y+3)^2+(y+3)y^2-(y^2-3)^3-(y^2-3)y^2\\right)dy \\\\&amp;=\\displaystyle\\int_{-2}^{3}(54+27y-12y^2+2y^3+8y^4-y^6)dy &amp;\\quad &amp;\\text{Integrate with respect to }y. \\\\&amp;=\\left[54y+\\frac{27y^2}2-4y^3+\\frac{y^4}2+\\frac{8y^5}5-\\frac{y^7}7\\right]\\bigg|_{-2}^3 \\\\&amp;=\\frac{2,375}7.\\end{align}[\/latex][\/hidden-answer]<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nSketch the region [latex]D[\/latex] and evaluate the iterated integral [latex]\\underset{D}{\\displaystyle\\iint}{xy} \\ {dy} \\ {dx}[\/latex] where [latex]D[\/latex] is the region bounded by the curves [latex]{y} = {\\cos{x}}[\/latex] and [latex]{y} = {\\sin{x}}[\/latex] in the interval [latex][{-3}{{\\pi}\/{4}},{{\\pi}\/{4}}][\/latex].\r\n\r\n[reveal-answer q=\"884561035\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"884561035\"]\r\n\r\n[latex]\\pi\/4[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793510065\">Recall from\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/double-integrals-over-rectangular-regions\/\">Double Integrals over Rectangular Regions<\/a>\u00a0the properties of double integrals. As we have seen from the examples here, all these properties are also valid for a function defined on a nonrectangular bounded region on a plane. In particular, property 3 states:<\/p>\r\n<p id=\"fs-id1167793287003\">If [latex]{R} = {S} \\ {\\cup} \\ {T}[\/latex] and [latex]{S} \\ {\\cap} \\ {T} = {\\emptyset}[\/latex] except at their boundaries, then<\/p>\r\n<p style=\"text-align: center;\">[latex]\\underset{R}{\\displaystyle\\iint}f(x,y)dA=\\underset{S}{\\displaystyle\\iint}f(x,y)dA=\\underset{T}{\\displaystyle\\iint}f(x,y)dA[\/latex].<\/p>\r\nSimilarly, we have the following property of double integrals over a nonrectangular bounded region on a plane.\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">theorem: decomposing regions into smaller regions<\/h3>\r\n\r\n<hr \/>\r\n\r\nSuppose the region [latex]D[\/latex] can be expressed as [latex]{D} = {D_1} \\ {\\cup} \\ {D_2}[\/latex] where [latex]D_1[\/latex] and [latex]D_2[\/latex] do not overlap except at their boundaries. Then\r\n<p style=\"text-align: center;\">[latex]\\underset{D}{\\displaystyle\\iint}f(x,y)dA=\\underset{D_1}{\\displaystyle\\iint}f(x,y)dA=\\underset{D_2}{\\displaystyle\\iint}f(x,y)dA[\/latex].<\/p>\r\n\r\n<\/div>\r\nThis theorem is particularly useful for nonrectangular regions because it allows us to split a region into a union of regions of Type I and Type II. Then we can compute the double integral on each piece in a convenient way, as in the next example.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: decomposing regions<\/h3>\r\nExpress the region [latex]D[\/latex] shown in\u00a0Figure 8\u00a0as a union of regions of Type I or Type II, and evaluate the integral\r\n<p style=\"text-align: center;\">[latex]\\underset{D}{\\displaystyle\\iint}(2x+5y)dA[\/latex].<\/p>\r\n\r\n\r\n[caption id=\"attachment_1343\" align=\"aligncenter\" width=\"342\"]<img class=\"size-full wp-image-1343\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25060228\/5-2-8.jpeg\" alt=\"A complicated shape enclosed by the lines y = (x + 2) squared, x = 16y minus y cubed, x = negative 2, and y = negative 4. This graph has intersection points (0, 4), (negative 2, 0), (0, negative 4), and (negative 2, negative 4).\" width=\"342\" height=\"347\" \/> Figure 8.\u00a0This region can be decomposed into a union of three regions of Type I or Type II.[\/caption]\r\n\r\n[reveal-answer q=\"589644316\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"589644316\"]\r\n\r\nThe region [latex]D[\/latex] is not easy to decompose into any one type; it is actually a combination of different types. So we can write it as a union of three regions [latex]D_1[\/latex], [latex]D_2[\/latex], and [latex]D_3[\/latex] where,\r\n\r\n[latex]{D_1} = {\\left \\{{(x,y)}{\\mid}{-2} \\ {\\leq} \\ {x} \\ {\\leq} \\ {0,0} \\ {\\leq} \\ {y} \\ {\\leq} \\ {(x+y)^2} \\right \\}}[\/latex],\r\n\r\n[latex]{D_2} = {\\left \\{{(x,y)}{\\mid}{0} \\ {\\leq} \\ {y} \\ {\\leq} \\ {4,0} \\ {\\leq} \\ {x} \\ {\\leq} \\ {(y-{\\frac{1}{16}y^3})} \\right \\}}[\/latex].\r\n\r\nThese regions are illustrated more clearly in\u00a0Figure 9.\r\n\r\n[caption id=\"attachment_1344\" align=\"aligncenter\" width=\"342\"]<img class=\"size-full wp-image-1344\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25060322\/5-2-9.jpeg\" alt=\"The same complicated shape enclosed by the lines y = (x + 2) squared, x = 16y minus y cubed, x = negative 2, and y = negative 4. This graph has intersection points (0, 4), (negative 2, 0), (0, negative 4), and (negative 2, negative 4). The area in the first quadrant is marked as D2 and a Type II region. The region in the second quadrant is marked as D1 and is a Type I region. The region in the third quadrant is marked as D3 and is a Type II region.\" width=\"342\" height=\"347\" \/> Figure 9.\u00a0Breaking the region into three subregions makes it easier to set up the integration.[\/caption]\r\n\r\nHere [latex]D_1[\/latex] is Type I\u00a0and [latex]D_2[\/latex] and [latex]D_3[\/latex] are both of Type II. Hence,\r\n\r\n[latex]\\begin{align}\r\n\r\n\\underset{D}{\\displaystyle\\iint}(2x+5y)dA&amp;=\\underset{D_1}{\\displaystyle\\iint}(2x+5y)dA+\\underset{D_2}{\\displaystyle\\iint}(2x+5y)dA+\\underset{D_3}{\\displaystyle\\iint}(2x+5y)dA \\\\\r\n\r\n&amp;=\\displaystyle\\int_{x=-2}^{x=0}\\displaystyle\\int_{y=0}^{y=(x+2)^2}(2x+5y)dydx+\\displaystyle\\int_{y=0}^{y=4}\\displaystyle\\int_{x=0}^{x=y-(1\/16)y^3}(2x+5y)dxdy+\\displaystyle\\int_{y=-4}^{y=0}\\displaystyle\\int_{x=-2}^{x=y-(1\/16)y^3}(2x+5y)dxdy \\\\\r\n\r\n&amp;=\\displaystyle\\int_{x=-2}^{x=0}\\left[\\frac12(2+x)^2(20+24x+5x^2)\\right] \\\\\r\n\r\n&amp;\\,\\,\\,+\\displaystyle\\int_{y=0}^{y=4}\\left[\\frac1{256}y^6-\\frac7{16}y^4+6y^2\\right] +\\displaystyle\\int_{y=-4}^{y=0}\\left[\\frac1{256}y^6-\\frac7{16}y^4+6y^2+10y-4\\right] \\\\\r\n\r\n&amp;=\\frac{40}3+\\frac{1,664}{35}-\\frac{1,696}{35}=\\frac{1,304}{105}.\r\n\r\n\\end{align}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nConsider the region bounded by the curves [latex]y = \\ln{x}[\/latex] and [latex]y = e^x[\/latex] in the interval [latex][1,2][\/latex]. Decompose the region into smaller regions of Type II.\r\n\r\n[reveal-answer q=\"710946832\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"710946832\"]\r\n\r\n[latex]\\{(x,y)|0\\leq{y}\\leq1,1\\leq{x}\\leq{e^y}\\}\\cup\\{(x,y)|1\\leq{y}\\leq{e},1\\leq{x}\\leq2\\}\\cup\\{(x,y)|e\\leq{y}\\leq{e}^2,\\ln{y}\\leq{x}\\leq2\\}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8197098&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=JCwVItEtOkg&amp;video_target=tpm-plugin-72owa3pa-JCwVItEtOkg\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP5.9_transcript.html\">transcript for \u201cCP 5.9\u201d here (opens in new window).<\/a><\/center>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nRedo\u00a0Example \"Decomposing Regions\" using a union of two Type II regions.\r\n\r\n[reveal-answer q=\"325791014\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"325791014\"]\r\n\r\nSame as in the example shown: [latex]\\{(x,y)|0\\leq{y}\\leq1,1\\leq{x}\\leq{e^y}\\}\\cup\\{(x,y)|1\\leq{y}\\leq{e},1\\leq{x}\\leq2\\}\\cup\\{(x,y)|e\\leq{y}\\leq{e}^2,\\ln{y}\\leq{x}\\leq2\\}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2 data-type=\"title\">Changing the Order of Integration<\/h2>\r\n<p id=\"fs-id1167793384878\">As we have already seen when we evaluate an iterated integral, sometimes one order of integration leads to a computation that is significantly simpler than the other order of integration. Sometimes the order of integration does not matter, but it is important to learn to recognize when a change in order will simplify our work.<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example: changing the order of integration<\/h3>\r\nReverse the order of integration in the iterated integral [latex]\\displaystyle\\int_{x=0}^{x=\\sqrt2}\\displaystyle\\int_{y=0}^{y=2-x^2}xe^{x^2}dydx[\/latex]. Then evaluate the new iterated integral.\r\n\r\n[reveal-answer q=\"557138862\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"557138862\"]\r\n\r\nThe region as presented is of Type I. To reverse the order of integration, we must first express the region as Type II. Refer to\u00a0Figure 10.\r\n\r\n[caption id=\"attachment_1346\" align=\"aligncenter\" width=\"560\"]<img class=\"size-full wp-image-1346\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25060441\/5-2-10.jpeg\" alt=\"This figure consists of two figures labeled Type I and Type II. In the Type I figure, a curve is given as y = 2 minus x squared, which forms a shape with the x and y axes. There is a vertical line with arrows on the end of it within this shape. In the Type II figure, a curve is given as x = the square root of the quantity (2 minus y), which forms a shape with the x and y axes. There is a horizontal line with arrows on the end of it within this shape.\" width=\"560\" height=\"278\" \/> Figure 10.\u00a0Converting a region from Type I to Type II.[\/caption]\r\n<p id=\"fs-id1167794122040\">We can see from the limits of integration that the region is bounded above by [latex]y=2-x^{2}[\/latex] and below by [latex]y=0[\/latex], where [latex]x[\/latex] is in the interval [latex][0,{\\sqrt{2}}][\/latex]. By reversing the order, we have the region bounded on the left by [latex]x=0[\/latex] and on the right by [latex]{x} = {{\\sqrt{2-y}}}[\/latex] where [latex]y[\/latex] is in the interval [latex][0,2][\/latex]. We solved [latex]y=2-x^2[\/latex] in terms of [latex]x[\/latex] to obtain [latex]{x} = {{\\sqrt{2-y}}}[\/latex].<\/p>\r\n<p id=\"fs-id1167793263187\">Hence<\/p>\r\n[latex]\\begin{align}\r\n\r\n\\displaystyle\\int_0^{\\sqrt2}\\displaystyle\\int_0^{2-x^2}xe^{x^2}dydx&amp;=\\displaystyle\\int_0^2\\displaystyle\\int_0^{\\sqrt{2-y}}xe^{x^2}dxdy \\,\\,\\,\\,\\,\\text{ Reverse the order of integration then use substitution} \\\\\r\n\r\n&amp;=\\displaystyle\\int_0^2\\left[\\frac12ex^2\\bigg|_0^{\\sqrt{2-y}}\\right]dy=\\displaystyle\\int_0^2\\frac12(e^{2-y}-1)dy=-\\frac12(e^{2-y}+y)\\bigg|_0^2 \\\\\r\n\r\n&amp;=\\frac12(e^2-3).\r\n\r\n\\end{align}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: evaluating an iterated integral by reversing the order of integration<\/h3>\r\n<p id=\"fs-id1167793521486\">Consider the iterated integral [latex]\\underset{R}{\\displaystyle\\iint}{f(x,y)}{dx} \\ {dy}[\/latex]\u00a0<span style=\"font-size: 1rem; text-align: initial;\">where [latex]{z} = {f(x,y)} = {x-2y}[\/latex]<\/span><span style=\"font-size: 1rem; text-align: initial;\">\u00a0over a triangular region [latex]R[\/latex]<\/span><span style=\"font-size: 1rem; text-align: initial;\">\u00a0that has sides on [latex]x=0,\\text{ }y=0[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">and the line [latex]x+y=1[\/latex].\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">Sketch the region, and then evaluate the iterated integral by<\/span><\/p>\r\n<p style=\"padding-left: 30px;\">a. integrating first with respect to [latex]y[\/latex] and then<\/p>\r\n<p style=\"padding-left: 30px;\">b. integrating first with respect to [latex]x[\/latex].<\/p>\r\n[reveal-answer q=\"913592004\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"913592004\"]\r\n\r\nA sketch of the region appears in\u00a0Figure 11.\r\n\r\n[caption id=\"attachment_1347\" align=\"aligncenter\" width=\"262\"]<img class=\"size-full wp-image-1347\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25060520\/5-2-11.jpeg\" alt=\"The line y = 1 minus x is drawn, and it is also marked as x = 1 minus y. There is a shaded region around x = 0 that comes from the y axis, which projects down to make a shaded region marked y = 0 from the x axis.\" width=\"262\" height=\"240\" \/> Figure 11.\u00a0A triangular region\u00a0<span style=\"font-size: 1rem; text-align: initial;\">[latex]R[\/latex]\u00a0<\/span>for integrating in two ways.[\/caption]\r\n<p id=\"fs-id1167793589762\">We can complete this integration in two different ways.<\/p>\r\n<p style=\"padding-left: 30px;\">a. One way to look at it is by first integrating [latex]y[\/latex] from [latex]y=0[\/latex] to [latex]y=1-x[\/latex] vertically and then integrating [latex]x[\/latex] from [latex]x=0[\/latex]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0to\u00a0[latex]x=1[\/latex]:<\/p>\r\n[latex]\\hspace{5cm}\\begin{align}\r\n\r\n\\underset{R}{\\displaystyle\\iint}f(x,y)dxdy&amp;=\\displaystyle\\int_{x=0}^{x=1}\\displaystyle\\int_{y=0}^{y=1-x}(x-2y)dydx=\\displaystyle\\int_{x=0}^{x=1}[xy-2y^2]_{y=0}^{y=1-x}dx \\\\\r\n\r\n&amp;=\\displaystyle\\int_{x=0}^{x=1}\\left[x(1-x)-(1-x)^2\\right]dx=\\displaystyle\\int_{x=0}^{x=1}[-1+3x-2x^2]dx \\\\\r\n\r\n&amp;=\\left[-x+\\frac32x^2-\\frac23x^3\\right]_{x=0}^{x=1}=-\\frac16\r\n\r\n\\end{align}[\/latex]\r\n<p style=\"padding-left: 30px;\">b. The other way to do this problem is by first integrating [latex]x[\/latex] from [latex]x=0[\/latex] to [latex]x=1-y[\/latex] horizontally and then integrating [latex]y[\/latex] from [latex]y=0[\/latex] to\u00a0[latex]y=1[\/latex]:<\/p>\r\n[latex]\\hspace{5cm}\\begin{align}\r\n\r\n\\underset{R}{\\displaystyle\\iint}f(x,y)dxdy&amp;=\\displaystyle\\int_{y=0}^{y=1}\\displaystyle\\int_{x=0}^{x=1-y}(x-2y)dxdy=\\displaystyle\\int_{y=0}^{y=1}[\\frac12x^2-2xy]_{x=0}^{x=1-y}dy \\\\\r\n\r\n&amp;=\\displaystyle\\int_{y=0}^{y=1}\\left[\\frac12(1-y)^2-2y(1-y)\\right]dy=\\displaystyle\\int_{y=0}^{y=1}[\\frac12-3y+\\frac52y^2]dy \\\\\r\n\r\n&amp;=\\left[\\frac12y-\\frac32y^2+\\frac56y^3\\right]_{y=0}^{y=1}=-\\frac16\r\n\r\n\\end{align}[\/latex]\r\n\r\n&nbsp;\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nEvaluate the iterated integral [latex]\\underset{R}{\\displaystyle\\iint}{(x^2+y^2)}{dA}[\/latex] over the region [latex]D[\/latex] in the first quadrant between the functions [latex]y=2x[\/latex] and [latex]y=x^{2}[\/latex]. Evaluate the iterated integral by integrating first with respect to [latex]y[\/latex] and then integrating first with resect to [latex]x[\/latex].\r\n\r\n[reveal-answer q=\"867530921\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"867530921\"]\r\n\r\n[latex]\\large{\\frac{216}{35}}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8197099&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=zcvSZfiNvqI&amp;video_target=tpm-plugin-zim0jvdd-zcvSZfiNvqI\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP5.11_transcript.html\">transcript for \u201cCP 5.11\u201d here (opens in new window).<\/a><\/center>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul class=\"os-abstract\">\n<li><span class=\"os-abstract-content\">Recognize when a function of two variables is integrable over a general region.<\/span><\/li>\n<li><span class=\"os-abstract-content\">Evaluate a double integral by computing an iterated integral over a region bounded by two vertical lines and two functions of [latex]x[\/latex] or two horizontal lines and two functions of [latex]y[\/latex].<\/span><\/li>\n<li><span class=\"os-abstract-content\">Simplify the calculation of an iterated integral by changing the order of integration.<\/span><\/li>\n<\/ul>\n<\/div>\n<section id=\"fs-id1167793238235\" data-depth=\"1\">\n<h2 data-type=\"title\">General Regions of Integration<\/h2>\n<p id=\"fs-id1167793871811\">An example of a general bounded region [latex]D[\/latex] on a plane is shown in\u00a0Figure 1. Since [latex]D[\/latex] is bounded on the plane, there must exist a rectangular region [latex]R[\/latex] on the same plane that encloses the region [latex]D[\/latex], that is, a rectangular region [latex]R[\/latex] exists such that [latex]D[\/latex] is a subset of [latex]R(D \\subseteq R)[\/latex].<\/p>\n<\/section>\n<div id=\"attachment_1317\" style=\"width: 427px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1317\" class=\"size-full wp-image-1317\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23152039\/5-2-1.jpeg\" alt=\"A rectangle R with a shape D inside of it. Inside D, there is a point labeled g(x, y) = f(x, y). Outside D but still inside R, there is a point labeled g(x, y) = 0.\" width=\"417\" height=\"310\" \/><\/p>\n<p id=\"caption-attachment-1317\" class=\"wp-caption-text\">Figure 1. For a region [latex]\\small{D}[\/latex] that is a subset of\u00a0[latex]\\small{R}[\/latex], we can define a function\u00a0[latex]\\small{g(x,y)}[\/latex] to equal\u00a0[latex]\\small{f(x,y)}[\/latex] at every point in\u00a0[latex]\\small{D}[\/latex] and\u00a0[latex]\\small{0}[\/latex] at every point of\u00a0[latex]\\small{R}[\/latex] not in\u00a0[latex]\\small{D}[\/latex].<\/p>\n<\/div>\n<p id=\"fs-id1167793267284\">Suppose [latex]z=f(x, y)[\/latex] is defined on a general planar bounded region [latex]D[\/latex] as in\u00a0Figure 1. In order to develop double integrals of [latex]f[\/latex] over [latex]D[\/latex], we extend the definition of the function to include all points on the rectangular region [latex]R[\/latex] and then use the concepts and tools from the preceding section. But how do we extend the definition of [latex]f[\/latex] to include all the points on [latex]R[\/latex]? We do this by defining a new function [latex]g(x, y)[\/latex] on [latex]R[\/latex] as follows:<\/p>\n<p>[latex]\\hspace{8cm}\\large{g(x,y)=\\Bigg\\{\\begin{align}    &f(x,y)&\\ &\\text{if }(x,y)\\text{ is in }D \\\\    &0&\\ &\\text{if }(x,y)\\text{ is in }R\\text{ but not in }D    \\end{align}}[\/latex]<\/p>\n<p>Note that we might have some technical difficulties if the boundary of [latex]D[\/latex] is complicated. So we assume the boundary to be a piecewise smooth and continuous simple closed curve. Also, since all the results developed in\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/double-integrals-over-rectangular-regions\/\">Double Integrals over Rectangular Regions<\/a>\u00a0used an integrable function [latex]f(x, y)[\/latex], we must be careful about [latex]g(x, y)[\/latex] and verify that [latex]g(x, y)[\/latex] is an integrable function over the rectangular region [latex]R[\/latex]. This happens as long as the region [latex]D[\/latex] is bounded by simple closed curves. For now we will concentrate on the descriptions of the regions rather than the function and extend our theory appropriately for integration.<\/p>\n<p id=\"fs-id1167794060679\">We consider two types of planar bounded regions.<\/p>\n<div id=\"fs-id1167793857192\" class=\"ui-has-child-title\" data-type=\"note\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">DEFINITION<\/h3>\n<hr \/>\n<p id=\"fs-id1167793884044\">A region [latex]D[\/latex] in the [latex](x, y)[\/latex]-plane is of\u00a0<span id=\"f783ca08-ca2f-4ebb-a559-d1b30c2f9b8c_term209\" data-type=\"term\">Type I<\/span>\u00a0if it lies between two vertical lines and the graphs of two continuous functions [latex]g_1(x)[\/latex] and [latex]g_2(x)[\/latex]. That is (Figure 2),<\/p>\n<p style=\"text-align: center;\">[latex]{D} = \\left \\{(x,y) \\mid a \\leq x \\leq \\ b,g_1 \\leq y \\leq g_2 \\ (x) \\right \\}[\/latex].<\/p>\n<p id=\"fs-id1167794337439\">A region [latex]D[\/latex] in the [latex]xy[\/latex] plane is of\u00a0<span id=\"f783ca08-ca2f-4ebb-a559-d1b30c2f9b8c_term210\" data-type=\"term\">Type II<\/span>\u00a0if it lies between two horizontal lines and the graphs of two continuous functions [latex]h_1(y)[\/latex] and [latex]h_2(y)[\/latex]. That is (Figure 3),<\/p>\n<p style=\"text-align: center;\">[latex]{D} = \\left \\{(x,y) \\mid c \\leq y \\leq \\ d,h_1 \\leq x \\leq h_2 \\ (y) \\right \\}[\/latex].<\/p>\n<\/div>\n<div id=\"attachment_1320\" style=\"width: 973px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1320\" class=\"size-full wp-image-1320\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23155027\/5-2-2.jpeg\" alt=\"The graphs showing a region marked D. In all instances, between a and b, there is a shape that is defined by two functions g1(x) and g2(x). In one instance, the two functions do not touch; in another instance, they touch at the end point a, and in the last instance they touch at both end points.\" width=\"963\" height=\"274\" \/><\/p>\n<p id=\"caption-attachment-1320\" class=\"wp-caption-text\">Figure 2.\u00a0A Type I region lies between two vertical lines and the graphs of two functions of\u00a0[latex]x[\/latex].<\/p>\n<\/div>\n<p>\u00a0<\/p>\n<div id=\"attachment_1321\" style=\"width: 675px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1321\" class=\"size-full wp-image-1321\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23155104\/5-2-3.jpeg\" alt=\"The graphs show a region marked D. In all instances, between c and d, there is a shape that is defined by two vertically oriented functions x = h1(y) and x = h2(y). In one instance, the two functions do not touch; in the other instance, they touch at the end point c.\" width=\"665\" height=\"310\" \/><\/p>\n<p id=\"caption-attachment-1321\" class=\"wp-caption-text\">Figure 3.\u00a0A Type II region lies between two horizontal lines and the graphs of two functions of [latex]y[\/latex].<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Describing a region as type I and also as type ii<\/h3>\n<p>Consider the region in the first quadrant between the functions [latex]{y} = {\\sqrt{x}}[\/latex] and [latex]y=x^{3}[\/latex] (Figure 4). Describe the region first as Type I and then as Type II.<\/p>\n<div id=\"attachment_1323\" style=\"width: 513px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1323\" class=\"size-full wp-image-1323\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23155305\/5-2-4.jpeg\" alt=\"The region D is drawn between two functions, namely, y = the square root of x and y = x3.\" width=\"503\" height=\"277\" \/><\/p>\n<p id=\"caption-attachment-1323\" class=\"wp-caption-text\">Figure 4. Region [latex]D[\/latex]\u00a0can be described as Type I or as Type II.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q375418035\">Show Solution<\/span><\/p>\n<div id=\"q375418035\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793848514\">When describing a region as Type I, we need to identify the function that lies above the region and the function that lies below the region. Here, region [latex]D[\/latex] is bounded above by [latex]{y} = {\\sqrt{x}}[\/latex] and below by [latex]y=x^{3}[\/latex] in the interval for [latex]x[\/latex] in [latex][0,1][\/latex]. Hence, as Type I, [latex]D[\/latex] is described as the set [latex]{\\left \\{(x,y) \\mid 0 \\leq x \\leq 1,x^3 \\leq y \\leq \\sqrt{x} \\right \\}}[\/latex].<\/p>\n<p id=\"fs-id1167793524718\">However, when describing a region as Type II, we need to identify the function that lies on the left of the region and the function that lies on the right of the region. Here, the region [latex]D[\/latex] is bounded on the left by [latex]x=y^{2}[\/latex] and on the right by [latex]x = \\sqrt[3]{y}[\/latex] in the interval for [latex]y[\/latex] in [latex][0,1][\/latex]. Hence, as Type II, [latex]D[\/latex] is described as the set [latex]{\\left \\{(x,y) \\mid 0 \\leq y \\leq 1,y^2 \\leq x \\leq \\sqrt[3]{y} \\right \\}}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Consider the region in the first quadrant between the functions [latex]y=2x[\/latex] and [latex]y=x^{2}[\/latex]. Describe the region first as Type I and then as Type II.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q984622178\">Show Solution<\/span><\/p>\n<div id=\"q984622178\" class=\"hidden-answer\" style=\"display: none\">\n<p>Type I and Type II are expressed as [latex]\\{(x,y)|0\\leq{x}\\leq2,x^2\\leq{y}\\leq2x\\}[\/latex] and [latex]\\{(x,y)|0\\leq{y}\\leq4,\\frac12y\\leq{x}\\leq\\sqrt{y}\\}[\/latex], respectively.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2 data-type=\"title\">Double Integrals over Nonrectangular Regions<\/h2>\n<p id=\"fs-id1167794051685\">To develop the concept and tools for evaluation of a double integral over a general, nonrectangular region, we need to first understand the region and be able to express it as Type I or Type II or a combination of both. Without understanding the regions, we will not be able to decide the limits of integrations in double integrals. As a first step, let us look at the following theorem.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Theorem: double integrals over nonrectangular regions<\/h3>\n<hr \/>\n<p id=\"fs-id1167794200474\">Suppose [latex]g(x, y)[\/latex] is the extension to the rectangle [latex]R[\/latex] of the function [latex]f(x, y)[\/latex] defined on the regions [latex]D[\/latex] and [latex]R[\/latex] as shown in\u00a0Figure 1\u00a0inside [latex]R[\/latex]. Then [latex]g(x, y)[\/latex] is integrable and we define the double integral of [latex]f(x, y)[\/latex] over [latex]D[\/latex] by<\/p>\n<p style=\"text-align: center;\">[latex]\\underset{D}{\\displaystyle\\iint} {f(x,y)dA} = \\underset{R}{\\displaystyle\\iint} {g(x,y)dA}[\/latex].<\/p>\n<\/div>\n<p id=\"fs-id1167793355329\">The right-hand side of this equation is what we have seen before, so this theorem is reasonable because [latex]R[\/latex] is a rectangle and [latex]\\underset{R}{\\displaystyle\\iint}{g(x,y)dA}[\/latex] has been discussed in the preceding section. Also, the equality works because the values of [latex]g(x, y)[\/latex] are [latex]0[\/latex] for any point [latex](x, y)[\/latex] that lies outside [latex]D[\/latex], and hence these points do not add anything to the integral. However, it is important that the rectangle [latex]R[\/latex] contains the region [latex]D[\/latex].<\/p>\n<p id=\"fs-id1167793568896\">As a matter of fact, if the region [latex]D[\/latex] is bounded by smooth curves on a plane and we are able to describe it as Type I or Type II or a mix of both, then we can use the following theorem and not have to find a rectangle [latex]R[\/latex] containing the region.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Theorem: fubini&#8217;s theorem (strong form)<\/h3>\n<hr \/>\n<p id=\"fs-id1167793276048\">For a function [latex]f(x, y)[\/latex] that is continuous on a region [latex]D[\/latex] of Type I, we have<\/p>\n<p style=\"text-align: center;\">[latex]\\underset{D}{\\displaystyle\\iint}{f(x,y)dA} = \\underset{D}{\\displaystyle\\iint}{f(x,y)dy \\ dx}=\\displaystyle\\int_a^b\\left[\\displaystyle\\int_{g_1(x)}^{g_2(x)}f(x,y)dy\\right]dx.[\/latex]<\/p>\n<p>Similarly, for a function [latex]f(x, y)[\/latex] that is continuous on a region [latex]D[\/latex] of Type II, we have<\/p>\n<p style=\"text-align: center;\">[latex]\\underset{D}{\\displaystyle\\iint}{f(x,y)dA} = \\underset{D}{\\displaystyle\\iint}{f(x,y)dx \\ dy}=\\displaystyle\\int_c^d\\left[\\displaystyle\\int_{h_1(y)}^{h_2(y)}f(x,y)dx\\right]dy.[\/latex]<\/p>\n<\/div>\n<p>The integral in each of these expressions is an iterated integral, similar to those we have seen before. Notice that, in the inner integral in the first expression, we integrate [latex]f(x, y)[\/latex] with [latex]x[\/latex] being held constant and the limits of integration being [latex]g_1(x)[\/latex] and [latex]g_2(x)[\/latex]. In the inner integral in the second expression, we integrate [latex]f(x, y)[\/latex] with [latex]y[\/latex] being held constant and the limits of integration are [latex]h_1(x)[\/latex] and\u00a0[latex]h_2(x)[\/latex].<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: evaluating an iterated integral over a type I region<\/h3>\n<p>Evaluate the integral [latex]\\underset{D}{\\displaystyle\\iint}{x^2}{e^{xy}}{dA}[\/latex] where [latex]D[\/latex] is shown in\u00a0Figure 5 (see solution for image).<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q615274593\">Show Solution<\/span><\/p>\n<div id=\"q615274593\" class=\"hidden-answer\" style=\"display: none\">\n<p>First construct the region [latex]D[\/latex] as a Type I region (Figure 5). Here [latex]{D} = {\\left \\{{(x,y)}{\\mid}{0} \\ {\\leq} \\ {x} \\ {\\leq} \\ {2,{\\frac{1}{2}x}} \\ {\\leq} \\ y \\ {\\leq} \\ {1} \\right \\}}[\/latex]. Then we have<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\underset{D}{\\displaystyle\\iint}{x^2}{e^{xy}}{dA} = \\displaystyle\\int_{x=0}^{x=2}\\displaystyle\\int_{y=1\/2x}^{y=1}x^2e^{xy}dydx.}[\/latex]<\/p>\n<div id=\"attachment_1324\" style=\"width: 261px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1324\" class=\"size-full wp-image-1324\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23155437\/5-2-5.jpeg\" alt=\"A triangle marked D drawn with lines y = 1\/2 x and y = 1, with vertices (0, 0), (2, 1), and (0, 1). Here, there is a pair of red arrows reaching vertically from one edge to the other.\" width=\"251\" height=\"160\" \/><\/p>\n<p id=\"caption-attachment-1324\" class=\"wp-caption-text\">Figure 5. We can express region [latex]\\small{D}[\/latex] as a Type I region and integrate from [latex]\\small{y=\\frac{1}{2}x}[\/latex] to\u00a0[latex]\\small{y=1}[\/latex], between the lines\u00a0[latex]\\small{x=0}[\/latex] and\u00a0[latex]\\small{x=2}[\/latex].<\/p>\n<\/div>\n<p>Therefore, we have<br \/>\n[latex]\\begin{align}\\displaystyle\\int_{x=0}^{x=2}\\displaystyle\\int_{y=\\frac12x}^{y=1}x^2e^{xy}dydx &=\\int_{x=0}^{x=2}[\\int_{y=\\frac12x}^{y=1}x^2e^{xy}dydx] &\\text{Iterated for a Type I region.} \\\\&=\\int_{x=0}^{x=2}[x^2\\frac{e^xy}x]\\Bigg|_{y=\\frac12x}^{y=1}dx &\\text{Integrate with respect to }y\\text{ using }u\\text{-substitution} \\\\& &\\text{ with }u=xy\\text{ where }x\\text{ is held constant} \\\\&=\\int_{x=0}^{x=2}[xe^x-xe^{\\frac{x^2}2}]dx &\\text{Integrate with respect to }x\\text{ using }u\\text{-substitution with}u=\\frac12x^2 \\\\&=[xe^x-e^x-e^{\\frac12x^2}]\\bigg|_{x=0}^{x=2}=2\\end{align}[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<p>In\u00a0the Example: Evaluating an Iterated Integral over a Type I Region, we could have looked at the region in another way, such as [latex]{D} = {\\left \\{{(x,y)}{\\mid}{0} \\ {\\leq} \\ {y} \\ {\\leq} \\ {1,0} \\ {\\leq} \\ {x} \\ {\\leq} \\ {2y} \\right \\}}[\/latex] (Figure 6).<\/p>\n<div id=\"attachment_1340\" style=\"width: 239px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1340\" class=\"size-full wp-image-1340\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25055920\/5-2-6.jpeg\" alt=\"A triangle marked D drawn with lines x = 2y and y = 1, with vertices (0, 0), (2, 1), and (0, 1). Here there is a pair of red arrows reaching horizontally from one edge to the other.\" width=\"229\" height=\"160\" \/><\/p>\n<p id=\"caption-attachment-1340\" class=\"wp-caption-text\">Figure 6.<\/p>\n<\/div>\n<p>This is a Type II region and the integral would then look like<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\underset{D}{\\displaystyle\\iint}{x^2}{e^{xy}}{dA} = \\displaystyle\\int_{y=0}^{y=1}\\displaystyle\\int_{x=0}^{x=2u}{x^2}{e^{xy}}{dx} \\ {dy}}[\/latex]<\/p>\n<p>However, if we integrate first with respect to [latex]x[\/latex], this integral is lengthy to compute because we have to use integration by parts twice.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: evaluating an iterated integral over a type II region<\/h3>\n<p>Evaluate the integral [latex]\\underset{D}{\\displaystyle\\iint}{({3x^2}+{y^2})} \\ {dA}[\/latex] where\u00a0[latex]{=} \\ {\\left \\{{(x,y)}{\\mid}{-2} \\ {\\leq} \\ {y} \\ {\\leq} \\ {3,y^2-3} \\ {\\leq} \\ {x} \\ {\\leq} \\ {y+3} \\right \\}}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q518274593\">Show Solution<\/span><\/p>\n<div id=\"q518274593\" class=\"hidden-answer\" style=\"display: none\">\n<p>Notice that [latex]D[\/latex] can be seen as either a Type I or a Type II region, as shown in\u00a0Figure 7. However, in this case describing [latex]D[\/latex] as Type I is more complicated than describing it as Type II. Therefore, we use [latex]D[\/latex] as a Type II region for the integration.<\/p>\n<div id=\"attachment_1341\" style=\"width: 901px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1341\" class=\"size-full wp-image-1341\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25060113\/5-2-7.jpeg\" alt=\"This figure consists of two figures labeled a and b. In figure a, a region is bounded by y = the square root of the quantity (x + 3), y = the negative of the square root of the quantity (x + 3), and y = x minus 3, which has points of intersection (6, 3), (1, negative 2), and (0, negative 3). There are vertical lines in the shape, and it is noted that this is a type I region: integrate first with respect to y. In figure b, a region is bounded by x = y2 minus 3 and x = y + 3, which has points of intersection (6, 3), (1, negative 2), and (0, negative 3). There are horizontal lines in the shape, and it is noted that this is a type II region: integrate first with respect to x.\" width=\"891\" height=\"372\" \/><\/p>\n<p id=\"caption-attachment-1341\" class=\"wp-caption-text\">Figure 7.\u00a0The region [latex]D[\/latex] in this example can be either (a) Type I or (b) Type II.<\/p>\n<\/div>\n<p>Choosing this order of integration, we have[latex]\\begin{align}\\underset{D}{\\displaystyle\\iint}(3x^2+y^2)dA&=\\displaystyle\\int_{y=-2}^{y=3}\\displaystyle\\int_{x=y^2-3}^{x=y+3}(3x^2+y^2)dxdy &\\quad &\\text{Interated integral, Type II region}.\\\\&=\\displaystyle\\int_{y=-2}^{y=3}(x^2+xy^2\\Bigg|_{y^2-3}^{y+3}dy &\\quad &\\text{Integrate with respect to }x. \\\\&=\\displaystyle\\int_{y=-2}^{y=3}\\left((y+3)^2+(y+3)y^2-(y^2-3)^3-(y^2-3)y^2\\right)dy \\\\&=\\displaystyle\\int_{-2}^{3}(54+27y-12y^2+2y^3+8y^4-y^6)dy &\\quad &\\text{Integrate with respect to }y. \\\\&=\\left[54y+\\frac{27y^2}2-4y^3+\\frac{y^4}2+\\frac{8y^5}5-\\frac{y^7}7\\right]\\bigg|_{-2}^3 \\\\&=\\frac{2,375}7.\\end{align}[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Sketch the region [latex]D[\/latex] and evaluate the iterated integral [latex]\\underset{D}{\\displaystyle\\iint}{xy} \\ {dy} \\ {dx}[\/latex] where [latex]D[\/latex] is the region bounded by the curves [latex]{y} = {\\cos{x}}[\/latex] and [latex]{y} = {\\sin{x}}[\/latex] in the interval [latex][{-3}{{\\pi}\/{4}},{{\\pi}\/{4}}][\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q884561035\">Show Solution<\/span><\/p>\n<div id=\"q884561035\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\pi\/4[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167793510065\">Recall from\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/double-integrals-over-rectangular-regions\/\">Double Integrals over Rectangular Regions<\/a>\u00a0the properties of double integrals. As we have seen from the examples here, all these properties are also valid for a function defined on a nonrectangular bounded region on a plane. In particular, property 3 states:<\/p>\n<p id=\"fs-id1167793287003\">If [latex]{R} = {S} \\ {\\cup} \\ {T}[\/latex] and [latex]{S} \\ {\\cap} \\ {T} = {\\emptyset}[\/latex] except at their boundaries, then<\/p>\n<p style=\"text-align: center;\">[latex]\\underset{R}{\\displaystyle\\iint}f(x,y)dA=\\underset{S}{\\displaystyle\\iint}f(x,y)dA=\\underset{T}{\\displaystyle\\iint}f(x,y)dA[\/latex].<\/p>\n<p>Similarly, we have the following property of double integrals over a nonrectangular bounded region on a plane.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">theorem: decomposing regions into smaller regions<\/h3>\n<hr \/>\n<p>Suppose the region [latex]D[\/latex] can be expressed as [latex]{D} = {D_1} \\ {\\cup} \\ {D_2}[\/latex] where [latex]D_1[\/latex] and [latex]D_2[\/latex] do not overlap except at their boundaries. Then<\/p>\n<p style=\"text-align: center;\">[latex]\\underset{D}{\\displaystyle\\iint}f(x,y)dA=\\underset{D_1}{\\displaystyle\\iint}f(x,y)dA=\\underset{D_2}{\\displaystyle\\iint}f(x,y)dA[\/latex].<\/p>\n<\/div>\n<p>This theorem is particularly useful for nonrectangular regions because it allows us to split a region into a union of regions of Type I and Type II. Then we can compute the double integral on each piece in a convenient way, as in the next example.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: decomposing regions<\/h3>\n<p>Express the region [latex]D[\/latex] shown in\u00a0Figure 8\u00a0as a union of regions of Type I or Type II, and evaluate the integral<\/p>\n<p style=\"text-align: center;\">[latex]\\underset{D}{\\displaystyle\\iint}(2x+5y)dA[\/latex].<\/p>\n<div id=\"attachment_1343\" style=\"width: 352px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1343\" class=\"size-full wp-image-1343\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25060228\/5-2-8.jpeg\" alt=\"A complicated shape enclosed by the lines y = (x + 2) squared, x = 16y minus y cubed, x = negative 2, and y = negative 4. This graph has intersection points (0, 4), (negative 2, 0), (0, negative 4), and (negative 2, negative 4).\" width=\"342\" height=\"347\" \/><\/p>\n<p id=\"caption-attachment-1343\" class=\"wp-caption-text\">Figure 8.\u00a0This region can be decomposed into a union of three regions of Type I or Type II.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q589644316\">Show Solution<\/span><\/p>\n<div id=\"q589644316\" class=\"hidden-answer\" style=\"display: none\">\n<p>The region [latex]D[\/latex] is not easy to decompose into any one type; it is actually a combination of different types. So we can write it as a union of three regions [latex]D_1[\/latex], [latex]D_2[\/latex], and [latex]D_3[\/latex] where,<\/p>\n<p>[latex]{D_1} = {\\left \\{{(x,y)}{\\mid}{-2} \\ {\\leq} \\ {x} \\ {\\leq} \\ {0,0} \\ {\\leq} \\ {y} \\ {\\leq} \\ {(x+y)^2} \\right \\}}[\/latex],<\/p>\n<p>[latex]{D_2} = {\\left \\{{(x,y)}{\\mid}{0} \\ {\\leq} \\ {y} \\ {\\leq} \\ {4,0} \\ {\\leq} \\ {x} \\ {\\leq} \\ {(y-{\\frac{1}{16}y^3})} \\right \\}}[\/latex].<\/p>\n<p>These regions are illustrated more clearly in\u00a0Figure 9.<\/p>\n<div id=\"attachment_1344\" style=\"width: 352px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1344\" class=\"size-full wp-image-1344\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25060322\/5-2-9.jpeg\" alt=\"The same complicated shape enclosed by the lines y = (x + 2) squared, x = 16y minus y cubed, x = negative 2, and y = negative 4. This graph has intersection points (0, 4), (negative 2, 0), (0, negative 4), and (negative 2, negative 4). The area in the first quadrant is marked as D2 and a Type II region. The region in the second quadrant is marked as D1 and is a Type I region. The region in the third quadrant is marked as D3 and is a Type II region.\" width=\"342\" height=\"347\" \/><\/p>\n<p id=\"caption-attachment-1344\" class=\"wp-caption-text\">Figure 9.\u00a0Breaking the region into three subregions makes it easier to set up the integration.<\/p>\n<\/div>\n<p>Here [latex]D_1[\/latex] is Type I\u00a0and [latex]D_2[\/latex] and [latex]D_3[\/latex] are both of Type II. Hence,<\/p>\n<p>[latex]\\begin{align}    \\underset{D}{\\displaystyle\\iint}(2x+5y)dA&=\\underset{D_1}{\\displaystyle\\iint}(2x+5y)dA+\\underset{D_2}{\\displaystyle\\iint}(2x+5y)dA+\\underset{D_3}{\\displaystyle\\iint}(2x+5y)dA \\\\    &=\\displaystyle\\int_{x=-2}^{x=0}\\displaystyle\\int_{y=0}^{y=(x+2)^2}(2x+5y)dydx+\\displaystyle\\int_{y=0}^{y=4}\\displaystyle\\int_{x=0}^{x=y-(1\/16)y^3}(2x+5y)dxdy+\\displaystyle\\int_{y=-4}^{y=0}\\displaystyle\\int_{x=-2}^{x=y-(1\/16)y^3}(2x+5y)dxdy \\\\    &=\\displaystyle\\int_{x=-2}^{x=0}\\left[\\frac12(2+x)^2(20+24x+5x^2)\\right] \\\\    &\\,\\,\\,+\\displaystyle\\int_{y=0}^{y=4}\\left[\\frac1{256}y^6-\\frac7{16}y^4+6y^2\\right] +\\displaystyle\\int_{y=-4}^{y=0}\\left[\\frac1{256}y^6-\\frac7{16}y^4+6y^2+10y-4\\right] \\\\    &=\\frac{40}3+\\frac{1,664}{35}-\\frac{1,696}{35}=\\frac{1,304}{105}.    \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Consider the region bounded by the curves [latex]y = \\ln{x}[\/latex] and [latex]y = e^x[\/latex] in the interval [latex][1,2][\/latex]. Decompose the region into smaller regions of Type II.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q710946832\">Show Solution<\/span><\/p>\n<div id=\"q710946832\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\{(x,y)|0\\leq{y}\\leq1,1\\leq{x}\\leq{e^y}\\}\\cup\\{(x,y)|1\\leq{y}\\leq{e},1\\leq{x}\\leq2\\}\\cup\\{(x,y)|e\\leq{y}\\leq{e}^2,\\ln{y}\\leq{x}\\leq2\\}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8197098&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=JCwVItEtOkg&amp;video_target=tpm-plugin-72owa3pa-JCwVItEtOkg\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP5.9_transcript.html\">transcript for \u201cCP 5.9\u201d here (opens in new window).<\/a><\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Redo\u00a0Example &#8220;Decomposing Regions&#8221; using a union of two Type II regions.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q325791014\">Show Solution<\/span><\/p>\n<div id=\"q325791014\" class=\"hidden-answer\" style=\"display: none\">\n<p>Same as in the example shown: [latex]\\{(x,y)|0\\leq{y}\\leq1,1\\leq{x}\\leq{e^y}\\}\\cup\\{(x,y)|1\\leq{y}\\leq{e},1\\leq{x}\\leq2\\}\\cup\\{(x,y)|e\\leq{y}\\leq{e}^2,\\ln{y}\\leq{x}\\leq2\\}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2 data-type=\"title\">Changing the Order of Integration<\/h2>\n<p id=\"fs-id1167793384878\">As we have already seen when we evaluate an iterated integral, sometimes one order of integration leads to a computation that is significantly simpler than the other order of integration. Sometimes the order of integration does not matter, but it is important to learn to recognize when a change in order will simplify our work.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: changing the order of integration<\/h3>\n<p>Reverse the order of integration in the iterated integral [latex]\\displaystyle\\int_{x=0}^{x=\\sqrt2}\\displaystyle\\int_{y=0}^{y=2-x^2}xe^{x^2}dydx[\/latex]. Then evaluate the new iterated integral.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q557138862\">Show Solution<\/span><\/p>\n<div id=\"q557138862\" class=\"hidden-answer\" style=\"display: none\">\n<p>The region as presented is of Type I. To reverse the order of integration, we must first express the region as Type II. Refer to\u00a0Figure 10.<\/p>\n<div id=\"attachment_1346\" style=\"width: 570px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1346\" class=\"size-full wp-image-1346\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25060441\/5-2-10.jpeg\" alt=\"This figure consists of two figures labeled Type I and Type II. In the Type I figure, a curve is given as y = 2 minus x squared, which forms a shape with the x and y axes. There is a vertical line with arrows on the end of it within this shape. In the Type II figure, a curve is given as x = the square root of the quantity (2 minus y), which forms a shape with the x and y axes. There is a horizontal line with arrows on the end of it within this shape.\" width=\"560\" height=\"278\" \/><\/p>\n<p id=\"caption-attachment-1346\" class=\"wp-caption-text\">Figure 10.\u00a0Converting a region from Type I to Type II.<\/p>\n<\/div>\n<p id=\"fs-id1167794122040\">We can see from the limits of integration that the region is bounded above by [latex]y=2-x^{2}[\/latex] and below by [latex]y=0[\/latex], where [latex]x[\/latex] is in the interval [latex][0,{\\sqrt{2}}][\/latex]. By reversing the order, we have the region bounded on the left by [latex]x=0[\/latex] and on the right by [latex]{x} = {{\\sqrt{2-y}}}[\/latex] where [latex]y[\/latex] is in the interval [latex][0,2][\/latex]. We solved [latex]y=2-x^2[\/latex] in terms of [latex]x[\/latex] to obtain [latex]{x} = {{\\sqrt{2-y}}}[\/latex].<\/p>\n<p id=\"fs-id1167793263187\">Hence<\/p>\n<p>[latex]\\begin{align}    \\displaystyle\\int_0^{\\sqrt2}\\displaystyle\\int_0^{2-x^2}xe^{x^2}dydx&=\\displaystyle\\int_0^2\\displaystyle\\int_0^{\\sqrt{2-y}}xe^{x^2}dxdy \\,\\,\\,\\,\\,\\text{ Reverse the order of integration then use substitution} \\\\    &=\\displaystyle\\int_0^2\\left[\\frac12ex^2\\bigg|_0^{\\sqrt{2-y}}\\right]dy=\\displaystyle\\int_0^2\\frac12(e^{2-y}-1)dy=-\\frac12(e^{2-y}+y)\\bigg|_0^2 \\\\    &=\\frac12(e^2-3).    \\end{align}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: evaluating an iterated integral by reversing the order of integration<\/h3>\n<p id=\"fs-id1167793521486\">Consider the iterated integral [latex]\\underset{R}{\\displaystyle\\iint}{f(x,y)}{dx} \\ {dy}[\/latex]\u00a0<span style=\"font-size: 1rem; text-align: initial;\">where [latex]{z} = {f(x,y)} = {x-2y}[\/latex]<\/span><span style=\"font-size: 1rem; text-align: initial;\">\u00a0over a triangular region [latex]R[\/latex]<\/span><span style=\"font-size: 1rem; text-align: initial;\">\u00a0that has sides on [latex]x=0,\\text{ }y=0[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">and the line [latex]x+y=1[\/latex].\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">Sketch the region, and then evaluate the iterated integral by<\/span><\/p>\n<p style=\"padding-left: 30px;\">a. integrating first with respect to [latex]y[\/latex] and then<\/p>\n<p style=\"padding-left: 30px;\">b. integrating first with respect to [latex]x[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q913592004\">Show Solution<\/span><\/p>\n<div id=\"q913592004\" class=\"hidden-answer\" style=\"display: none\">\n<p>A sketch of the region appears in\u00a0Figure 11.<\/p>\n<div id=\"attachment_1347\" style=\"width: 272px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1347\" class=\"size-full wp-image-1347\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25060520\/5-2-11.jpeg\" alt=\"The line y = 1 minus x is drawn, and it is also marked as x = 1 minus y. There is a shaded region around x = 0 that comes from the y axis, which projects down to make a shaded region marked y = 0 from the x axis.\" width=\"262\" height=\"240\" \/><\/p>\n<p id=\"caption-attachment-1347\" class=\"wp-caption-text\">Figure 11.\u00a0A triangular region\u00a0<span style=\"font-size: 1rem; text-align: initial;\">[latex]R[\/latex]\u00a0<\/span>for integrating in two ways.<\/p>\n<\/div>\n<p id=\"fs-id1167793589762\">We can complete this integration in two different ways.<\/p>\n<p style=\"padding-left: 30px;\">a. One way to look at it is by first integrating [latex]y[\/latex] from [latex]y=0[\/latex] to [latex]y=1-x[\/latex] vertically and then integrating [latex]x[\/latex] from [latex]x=0[\/latex]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0to\u00a0[latex]x=1[\/latex]:<\/p>\n<p>[latex]\\hspace{5cm}\\begin{align}    \\underset{R}{\\displaystyle\\iint}f(x,y)dxdy&=\\displaystyle\\int_{x=0}^{x=1}\\displaystyle\\int_{y=0}^{y=1-x}(x-2y)dydx=\\displaystyle\\int_{x=0}^{x=1}[xy-2y^2]_{y=0}^{y=1-x}dx \\\\    &=\\displaystyle\\int_{x=0}^{x=1}\\left[x(1-x)-(1-x)^2\\right]dx=\\displaystyle\\int_{x=0}^{x=1}[-1+3x-2x^2]dx \\\\    &=\\left[-x+\\frac32x^2-\\frac23x^3\\right]_{x=0}^{x=1}=-\\frac16    \\end{align}[\/latex]<\/p>\n<p style=\"padding-left: 30px;\">b. The other way to do this problem is by first integrating [latex]x[\/latex] from [latex]x=0[\/latex] to [latex]x=1-y[\/latex] horizontally and then integrating [latex]y[\/latex] from [latex]y=0[\/latex] to\u00a0[latex]y=1[\/latex]:<\/p>\n<p>[latex]\\hspace{5cm}\\begin{align}    \\underset{R}{\\displaystyle\\iint}f(x,y)dxdy&=\\displaystyle\\int_{y=0}^{y=1}\\displaystyle\\int_{x=0}^{x=1-y}(x-2y)dxdy=\\displaystyle\\int_{y=0}^{y=1}[\\frac12x^2-2xy]_{x=0}^{x=1-y}dy \\\\    &=\\displaystyle\\int_{y=0}^{y=1}\\left[\\frac12(1-y)^2-2y(1-y)\\right]dy=\\displaystyle\\int_{y=0}^{y=1}[\\frac12-3y+\\frac52y^2]dy \\\\    &=\\left[\\frac12y-\\frac32y^2+\\frac56y^3\\right]_{y=0}^{y=1}=-\\frac16    \\end{align}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Evaluate the iterated integral [latex]\\underset{R}{\\displaystyle\\iint}{(x^2+y^2)}{dA}[\/latex] over the region [latex]D[\/latex] in the first quadrant between the functions [latex]y=2x[\/latex] and [latex]y=x^{2}[\/latex]. Evaluate the iterated integral by integrating first with respect to [latex]y[\/latex] and then integrating first with resect to [latex]x[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q867530921\">Show Solution<\/span><\/p>\n<div id=\"q867530921\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\large{\\frac{216}{35}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8197099&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=zcvSZfiNvqI&amp;video_target=tpm-plugin-zim0jvdd-zcvSZfiNvqI\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP5.11_transcript.html\">transcript for \u201cCP 5.11\u201d here (opens in new window).<\/a><\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-3974\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>CP 5.9. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>CP 5.11. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 3. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":349141,"menu_order":8,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 3\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"CP 5.9\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"CP 5.11\",\"author\":\"Ryan 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