{"id":3964,"date":"2022-04-12T17:03:30","date_gmt":"2022-04-12T17:03:30","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=3964"},"modified":"2022-11-01T04:09:06","modified_gmt":"2022-11-01T04:09:06","slug":"double-integrals-over-rectangular-regions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/double-integrals-over-rectangular-regions\/","title":{"raw":"Double Integrals over Rectangular Regions","rendered":"Double Integrals over Rectangular Regions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li><span class=\"os-abstract-content\">Recognize when a function of two variables is integrable over a rectangular region.<\/span><\/li>\r\n \t<li><span class=\"os-abstract-content\">Recognize and use some of the properties of double integrals.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<section id=\"fs-id1167794098425\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Volumes and Double Integrals<\/h2>\r\nWe first begin with a review of the definition of the definite integral in terms of the limit of a Riemann Sum from single-variable calculus.\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Riemann SumS and the Definite Integral<\/h3>\r\nSuppose that [latex] y = f(x) [\/latex] is a continuous function and [latex] f(x) \\ge 0 [\/latex] on a closed interval [latex] [a,b] [\/latex].\u00a0 The area under the curve and above the [latex]y-[\/latex]axis can be approximated using rectangles, by subdividing the interval [latex] [a,b] [\/latex] into [latex] n [\/latex] subintervals of width [latex] \\Delta x = \\frac{b-a}{n} [\/latex].\u00a0 If [latex]x_i[\/latex] denotes a sample point within the [latex]i^{th} [\/latex] subinterval, then [latex]f(x_i^\\ast)[\/latex] gives the height of the [latex]i^{th} [\/latex] rectangle.\u00a0 Adding up each of these areas yields an approximation of the total area, called a <strong>Riemann Sum<\/strong> and denoted using the following:\r\n\r\n[latex]\\sum_{i=1}^n f(x_i^\\ast)\\Delta x [\/latex]\r\n\r\nThe definite integral of [latex] f(x) [\/latex] on [latex] [a,b] [\/latex] represents the exact area under the curve [latex] y = f(x) [\/latex], and is defined in terms as the following limit.\r\n\r\n[latex] \\int_a^b f(x)dx = \\lim_{n\\rightarrow\\infty}\\sum_{i=1}^n f(x_i^\\ast)\\Delta x [\/latex]\r\n\r\n<\/div>\r\n<p id=\"fs-id1167794128227\">To extend this notion into another dimension, we consider the space above a rectangular region [latex]R[\/latex]. Consider a continuous function [latex]{f(x,y)\\geq 0}[\/latex]\u00a0of two variables defined on the closed rectangle [latex]R[\/latex]:<\/p>\r\n\r\n<\/section>\r\n<p style=\"text-align: center;\">[latex]\\large{R = [a,b]\\times[c,d] = \\left \\{(x,y)\\in {\\mathbb{R}}^2 |a\\leq x\\leq b,c\\leq y\\leq d \\right \\}}[\/latex]<\/p>\r\nHere [latex]{{[a,b]}{\\times}{[c,d]}}[\/latex] denotes the Cartesian product of the two closed intervals [latex]{[a,b]}[\/latex] and [latex]{[c,d]}[\/latex]. It consists of rectangular pairs [latex](x, y)[\/latex] such that [latex]a\\leq x\\leq b[\/latex] and [latex]c\\leq y\\leq d[\/latex]. The graph of [latex]f[\/latex] represents a surface above the [latex]xy[\/latex]-plane with equation [latex]z=f(x, y)[\/latex] where [latex]z[\/latex] is the height of the surface at the point [latex](x, y)[\/latex]. Let [latex]S[\/latex] be the solid that lies above [latex]R[\/latex] and under the graph of [latex]f[\/latex] (Figure 1). The base of the solid is the rectangle [latex]R[\/latex] in the [latex]xy[\/latex]-plane. We want to find the volume [latex]V[\/latex] of the solid [latex]S[\/latex].\r\n\r\n[caption id=\"attachment_1299\" align=\"aligncenter\" width=\"487\"]<img class=\"size-full wp-image-1299\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23141334\/5-1-1.jpeg\" alt=\"In xyz space, there is a surface z = f(x, y). On the x axis, the lines denoting a and b are drawn; on the y axis the lines for c and d are drawn. When the surface is projected onto the xy plane, it forms a rectangle with corners (a, c), (a, d), (b, c), and (b, d).\" width=\"487\" height=\"434\" \/> Figure 1.\u00a0The graph of\u00a0[latex]f(x,y)[\/latex] over the rectangle\u00a0[latex]R[\/latex] in the\u00a0[latex]xy[\/latex]-plane is a curved surface.[\/caption]We divide the region [latex]R[\/latex] into small rectangles [latex]{R_{ij}}[\/latex], each with area [latex]{\\Delta{A}}[\/latex] and with sides [latex]{\\Delta{x}}[\/latex] and [latex]{\\Delta{y}}[\/latex] (Figure 2). We do this by dividing the interval [latex]{[a,b]}[\/latex] into [latex]m[\/latex] subintervals and dividing the interval [latex]{[c,d]}[\/latex] into [latex]n[\/latex] subintervals. Hence [latex]{{\\Delta{x}}={\\dfrac{b-a}{m}},{\\Delta{y}}={\\dfrac{d-c}{n}},}[\/latex] and\u00a0[latex]{{\\Delta{A}}={\\Delta{x}}{\\Delta{y}}}[\/latex].\r\n\r\n[caption id=\"attachment_1300\" align=\"aligncenter\" width=\"911\"]<img class=\"size-full wp-image-1300\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23141459\/5-1-2.jpeg\" alt=\"In the xy plane, there is a rectangle with corners (a, c), (a, d), (b, c), and (b, d). Between a and b on the x axis lines are drawn from a, x1, x2, \u2026, xi, \u2026, b with distance Delta x between each line; between c and d on the y axis lines are drawn from c, y1, y2, \u2026, yj, \u2026, d with distance Delta y between each line. Among the resulting subrectangles, the one in the second column and third row up has a point marked (x*23, y*23). The rectangle Rij is marked with upper right corner (xi, yj). Within this rectangle the point (x*ij, y*ij) is marked.\" width=\"911\" height=\"527\" \/> Figure 2. Rectangle\u00a0[latex]R[\/latex] is divided into small rectangles [latex]R_{ij}[\/latex], each with an area [latex]\\Delta{A}[\/latex].[\/caption]The volume of a thin rectangular box above [latex]{R_{ij}}[\/latex] is [latex]{f{({x{_{ij}^*}},{y{_{ij}^*}})}}{\\Delta{A}}[\/latex], where [latex]{({x{_{ij}^*}},{y{_{ij}^*}})}[\/latex] is an arbitrary sample point in each [latex]{R_{ij}}[\/latex] as shown in the following figure.\r\n\r\n&nbsp;\r\n\r\n[caption id=\"attachment_1301\" align=\"aligncenter\" width=\"487\"]<img class=\"size-full wp-image-1301\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23141611\/5-1-3.jpeg\" alt=\"In xyz space, there is a surface z = f(x, y). On the x axis, the lines denoting a and b are drawn; on the y axis the lines for c and d are drawn. When the surface is projected onto the xy plane, it forms a rectangle with corners (a, c), (a, d), (b, c), and (b, d). There are additional squares drawn to correspond to changes of Delta x and Delta y. On the surface, a square is marked and its projection onto the plane is marked as Rij. The average value for this small square is f(x*ij, y*ij).\" width=\"487\" height=\"474\" \/> Figure 3.\u00a0A thin rectangular box above\u00a0[latex]R_{ij}[\/latex] with height\u00a0[latex]f(x^{\\ast}_{ij},y^{\\ast}_{ij})[\/latex].[\/caption]\r\n<p id=\"fs-id1167794025753\">Using the same idea for all the subrectangles, we obtain an approximate volume of the solid [latex]S[\/latex] as [latex]{V}{\\approx} \\displaystyle\\sum_{i=1}^{m}\\sum_{j=1}^{n}\\ {f{({x{_{ij}^*}},{y{_{ij}^*}})}{\\Delta{A}}}[\/latex]. This sum is known as a\u00a0<strong><span id=\"d9da6d44-3ba3-4a90-adbc-dec26572e480_term205\" data-type=\"term\">double Riemann sum<\/span><\/strong>\u00a0and can be used to approximate the value of the volume of the solid. Here the double sum means that for each subrectangle we evaluate the function at the chosen point, multiply by the area of each rectangle, and then add all the results.<\/p>\r\n<p id=\"fs-id1167794139856\">As we have seen in the single-variable case, we obtain a better approximation to the actual volume if [latex]m[\/latex] and [latex]n[\/latex] become larger.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{{V}=\\displaystyle\\lim_{m,n\\to\\infty}\\displaystyle\\sum_{i=1}^{m}\\sum_{j=1}^{n}f{({x{_{ij}^*}},{y{_{ij}^*}}){\\Delta{A}}}}[\/latex] or [latex]\\large{V=\\displaystyle\\lim_{\\Delta x,\\Delta y\\to0}\\displaystyle\\sum_{i=1}^{m}\\sum_{j=1}^{n}f(x_{ij}^*,y_{ij}^*)\\Delta A}[\/latex].<\/p>\r\nNote that the sum approaches a limit in either case and the limit is the volume of the solid with the base [latex]R[\/latex]. Now we are ready to define the double integral.\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n\r\nThe\u00a0<span id=\"d9da6d44-3ba3-4a90-adbc-dec26572e480_term206\" data-type=\"term\">double integral<\/span>\u00a0of the function [latex]f(x, y)[\/latex] over the rectangular region [latex]R[\/latex] in the [latex]xy[\/latex]-plane is defined as\r\n<p style=\"text-align: center;\">[latex]\\large{\\underset{R}{\\displaystyle\\iint}f(x,y)dA=\\displaystyle\\lim_{m,n\\to\\infty}\\displaystyle\\sum_{i=1}^{m}\\sum_{j=1}^{n}f(x_i^*,y_j^*)\\Delta A}[\/latex].<\/p>\r\n\r\n<\/div>\r\nIf [latex]f(x, y)\\geq 0[\/latex], then the volume [latex]V[\/latex] of the solid [latex]S[\/latex], which lies above [latex]R[\/latex] in the [latex]xy[\/latex]-plane and under the graph of [latex]f[\/latex], is the double integral of the function [latex]f(x, y)[\/latex] over the rectangle [latex]R[\/latex]. If the function is ever negative, then the double integral can be considered a \u201csigned\u201d volume in a manner similar to the way we defined net signed area in\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/defining-and-evaluating-definite-integrals\/\">The Definite Integral<\/a>.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Setting up a double integral and approximating it by double sums<\/h3>\r\n<p id=\"fs-id1167793933224\">Consider the function [latex]{z = f(x,y) = 3x^2 - y}[\/latex] over the rectangular region [latex]{R = {[0,2]} \\times {[0,2]}}[\/latex] (Figure 4).<\/p>\r\n\r\n<ol id=\"fs-id1167793247300\" type=\"a\">\r\n \t<li>Set up a double integral for finding the value of the signed volume of the solid [latex]S[\/latex] that lies above [latex]R[\/latex] and \u201cunder\u201d the graph of [latex]f[\/latex].<\/li>\r\n \t<li>Divide [latex]R[\/latex] into four squares with [latex]m = n = 2[\/latex], and choose the sample point as the upper right corner point of each square [latex](1,1),(2,1),(1,2),[\/latex] and [latex](2,2)[\/latex] (Figure 4) to approximate the signed volume of the solid [latex]S[\/latex] that lies above [latex]R[\/latex] and \u201cunder\u201d the graph of [latex]f[\/latex].<\/li>\r\n \t<li>Divide [latex]R[\/latex] into four squares with [latex]m = n = 2[\/latex], and choose the sample point as the midpoint of each square: [latex](1\/2,1\/2),(3\/2,1\/2),(1\/2,3\/2),[\/latex] and [latex](3\/2,3\/2)[\/latex] to approximate the signed volume.<\/li>\r\n<\/ol>\r\n[caption id=\"attachment_1303\" align=\"aligncenter\" width=\"483\"]<img class=\"size-full wp-image-1303\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23141819\/5-1-4.jpeg\" alt=\"In xyz space, there is a surface z = f(x, y) = 3x2 minus y. The corners of the surface are given as (0, 0, 0), (2, 0, 12), (0, 2, negative 2), and (2, 2, 10). The surface is parabolic along the x axis.\" width=\"483\" height=\"252\" \/> Figure 4.\u00a0The function [latex]\\small{z=f(x,y)}[\/latex] graphed over the rectangular region [latex]\\small{R=[0,2]\\times[0,2]}[\/latex].[\/caption][reveal-answer q=\"193760238\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"193760238\"]\r\n<h2 data-type=\"solution-title\"><span class=\"os-title-label\">Solution<\/span><\/h2>\r\n<div class=\"os-solution-container\">\r\n<ol type=\"a\">\r\n \t<li style=\"list-style-type: none;\">\r\n<ol type=\"a\">\r\n \t<li style=\"list-style-type: none;\">\r\n<ol type=\"a\">\r\n \t<li style=\"text-align: center;\">As we can see, the function [latex]z = f(x,y) = 3x^2 - y[\/latex] is above the plane. To find the signed volume of [latex]S[\/latex], we need to divide the region [latex]R[\/latex] into small rectangles [latex]{R}_{ij}[\/latex] each with area [latex]{\\Delta{A}}[\/latex] and with sides [latex]{\\Delta{x}}[\/latex] and [latex]{\\Delta{y}}[\/latex], and choose [latex]{({x{_{ij}^*}},{y{_{ij}^*}})}[\/latex] as sample points in each [latex]{R}_{ij}[\/latex]. Hence, a double integral is set up as\r\n[latex]V = \\underset{R}{\\displaystyle\\iint}(3x^2-y)dA=\\displaystyle\\lim_{m,n\\to\\infty}\\displaystyle\\sum_{i=1}^{m}\\sum_{j=1}^{n}\\left[3(x_{ij}^*)^2-y_{ij}^*\\right]\\Delta A[\/latex].<span data-type=\"newline\">\r\n<\/span><\/li>\r\n \t<li>Approximating the signed volume using a Riemann sum with [latex]m = n = 2[\/latex] we have [latex]{\\Delta{A}} = {\\Delta{x}}{\\Delta{y}} = {1 {\\times} 1} = 1[\/latex]. Also, the sample points are [latex](1, 1), (2, 1), (1, 2),[\/latex] and [latex](2,2)[\/latex] as shown in the following figure.<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"CNX_Calc_Figure_15_01_005\" class=\"os-figure\">\r\n<div class=\"os-caption-container\">[caption id=\"attachment_1305\" align=\"aligncenter\" width=\"376\"]<img class=\"size-full wp-image-1305\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23142153\/5-1-4a.jpeg\" alt=\"On the xy plane, the points (1, 1), (1, 2), (2, 1), and (2, 2) are marked, and these form the upper right corners of four squares marked R11, R12, R21, and R22, respectively. Each square has area Delta A = 1.\" width=\"376\" height=\"484\" \/> Figure 5. Subrectangles for the rectangular region [latex]\\small{R=[0,2]\\times[0,2]}[\/latex].[\/caption]<\/div>\r\n<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>Hence,<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n[latex]\\hspace{3cm}\\begin{align}\r\n\r\nV&amp;=\\displaystyle\\sum_{i=1}^2\\sum_{j=1}^2f(x_{ij}^*,y_{ij}^*)\\Delta A \\\\\r\n\r\n&amp;=\\displaystyle\\sum_{i=1}^2(f(x_{i1}^*,y_{i1}^*)+ff(x_{i2}^*,y_{i2}^*))\\Delta A \\\\\r\n\r\n&amp;=(f(x_{11}^*,y_{11}^*)\\Delta A+(f(x_{21}^*,y_{21}^*)\\Delta A+(f(x_{12}^*,y_{12}^*)\\Delta A+(f(x_{22}^*,y_{22}^*)\\Delta A \\\\\r\n&amp;=f(1,1)(1)+f(2,1)(1)+f(1,2)(1)+f(2,2)(1) \\\\\r\n&amp;=(3-1)(1)+(12-1)(1)+(3-1)(1)+(12-2)(1) \\\\\r\n&amp;=2+11+1+10=24.\r\n\\end{align}[\/latex]<\/li>\r\n \t<li>Approximating the signed volume using a Riemann sum with [latex]m = n = 2[\/latex], we have[latex]{\\Delta{A}} = {\\Delta{x}}{\\Delta{y}} = {1 {\\times} 1} = 1[\/latex]. In this case the sample points are [latex](1\/2, 1\/2), (3\/2, 1\/2), (1\/2, 3\/2),[\/latex] and [latex](3\/2, 3\/2)[\/latex].<span data-type=\"newline\">\r\n<\/span>Hence,<\/li>\r\n<\/ol>\r\n[latex]\\hspace{3cm}\\begin{align}\r\nV&amp;=\\displaystyle\\sum_{i=1}^2\\sum_{j=1}^2f(x_{ij}^*,y_{ij}^*)\\Delta A \\\\&amp;=f(x_{11}^*,y_{11}^*)\\Delta A+f(x_{21}^*,y_{21}^*)\\Delta A+f(x_{12}^*,y_{12}^*)\\Delta A+f(x_{22}^*,y_{22}^*)\\Delta A \\\\\r\n&amp;=f(1\/2,1\/2)(1)+f(3\/2,1\/2)(1)+f(1\/2,3\/2)(1)+f(3\/2,3\/2)(1) \\\\\r\n&amp;=\\left(\\small{\\frac34}-\\small{\\frac14}\\right)(1)+\\left(\\small{\\frac{27}4}-\\small{\\frac12}\\right)(1)+\\left(\\small{\\frac34}-\\small{\\frac32}\\right)(1)+\\left(\\small{\\frac{27}4}-\\small{\\frac32}\\right)(1) \\\\\r\n&amp;=\\frac24+\\frac{25}4+\\left(-\\frac34\\right)+\\frac{21}4=\\frac{45}4.\r\n\\end{align}[\/latex]\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<h2 id=\"9\" data-type=\"commentary-title\"><span class=\"os-title-label\">Analysis<\/span><\/h2>\r\n<p id=\"fs-id1167794067077\">Notice that the approximate answers differ due to the choices of the sample points. In either case, we are introducing some error because we are using only a few sample points. Thus, we need to investigate how we can achieve an accurate answer.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it<\/h3>\r\n<p id=\"fs-id1167794324430\">Use the same function [latex]z = f(x,y) = 3x^2 - y[\/latex] over the rectangular region [latex]R = [0,2] \\times [0,2][\/latex].<\/p>\r\n<p id=\"fs-id1167793473779\">Divide [latex]R[\/latex] into the same four squares with [latex]m = n = 2[\/latex] and choose the sample points as the upper left corner point of each square [latex](0,1), (1,1), (0,2),[\/latex] and [latex](1,2)[\/latex] (Figure 5) to approximate the signed volume of the solid [latex]S[\/latex] that lies above [latex]R[\/latex] and \u201cunder\u201d the graph of [latex]f[\/latex].<\/p>\r\n[reveal-answer q=\"265173902\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"265173902\"]\r\n<p style=\"text-align: center;\">[latex]\\large{V=\\displaystyle\\sum_{i=1}^2\\sum_{j=1}^2f(x_{ij}^*,y_{ij}^*)\\Delta A=0}[\/latex]<\/p>\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=8197095&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=ZKNjm2OdGnU&amp;video_target=tpm-plugin-g6ue79lv-ZKNjm2OdGnU\" 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\/CP5.1_transcript.html\">transcript for \u201cCP 5.1\u201d here (opens in new window).<\/a><\/center>\r\n<p id=\"fs-id1167793263189\">Note that we developed the concept of double integral using a rectangular region [latex]R[\/latex]. This concept can be extended to any general region. However, when a region is not rectangular, the subrectangles may not all fit perfectly into [latex]R[\/latex], particularly if the base area is curved. We examine this situation in more detail in the next section, where we study regions that are not always rectangular and subrectangles may not fit perfectly in the region [latex]R[\/latex]. Also, the heights may not be exact if the surface [latex]z = f(x,y)[\/latex] is curved. However, the errors on the sides and the height where the pieces may not fit perfectly within the solid [latex]S[\/latex] approach 0 as [latex]m[\/latex] and [latex]n[\/latex] approach infinity. Also, the double integral of the function [latex]z = f(x,y)[\/latex] exists provided that the function [latex]f[\/latex] is not too discontinuous. If the function is bounded and continuous over [latex]R[\/latex] except on a finite number of smooth curves, then the double integral exists and we say that [latex]f[\/latex] is integrable over [latex]R[\/latex].<\/p>\r\n<p id=\"fs-id1167793456312\">Since [latex]{{\\Delta{A}}={\\Delta{x}}{\\Delta{y}}=\\Delta y\\Delta x}[\/latex], we can express [latex]dA[\/latex] as [latex]dx[\/latex] [latex]dy[\/latex] or [latex]dy[\/latex] [latex]dx[\/latex]. This means that, when we are using rectangular coordinates, the double integral over a region [latex]R[\/latex] denoted by [latex]\\underset{R}{\\displaystyle\\iint}f(x,y)dA[\/latex] can be written as<\/p>\r\n[latex]\\underset{R}{\\displaystyle\\iint}f(x,y)dx dy[\/latex] or\u00a0[latex]\\underset{R}{\\displaystyle\\iint}f(x,y)dy dx[\/latex].\r\n\r\n<section id=\"fs-id1167794098425\" data-depth=\"1\">\r\n<p id=\"fs-id1167793414884\">Now let\u2019s list some of the properties that can be helpful to compute double integrals.<\/p>\r\n\r\n<\/section><section id=\"fs-id1167793414890\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Properties of Double Integrals<\/h2>\r\n<p id=\"fs-id1167793414895\">The properties of double integrals are very helpful when computing them or otherwise working with them. We list here six properties of double integrals. Properties 1 and 2 are referred to as the linearity of the integral, property 3 is the additivity of the integral, property 4 is the monotonicity of the integral, and property 5 is used to find the bounds of the integral. Property 6 is used if [latex]f(x,y)[\/latex] is a product of two functions [latex]g(x)[\/latex] and\u00a0[latex]h(y)[\/latex].<\/p>\r\n\r\n<\/section>\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Theorem: properties of double integrals<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167793293479\">Assume that the functions [latex]f(x,y)[\/latex] and [latex]g(x,y)[\/latex] are integrable over the rectangular region [latex]R[\/latex]; [latex]S[\/latex]\u00a0and\u00a0[latex]T[\/latex]\u00a0are subregions of\u00a0[latex]R[\/latex]; and assume that\u00a0[latex]m[\/latex]\u00a0and\u00a0[latex]M[\/latex]\u00a0are real numbers.<\/p>\r\n\r\n<ol style=\"list-style-type: decimal;\">\r\n \t<li style=\"list-style-type: none\">\r\n<ol style=\"list-style-type: decimal;\">\r\n \t<li>The sum [latex]f(x,y) + g(x,y)[\/latex] is integrable and<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<p style=\"text-align: center;\">[latex]\\underset{R}{\\displaystyle\\iint}{[f(x,y)+g(x,y)]dA}=\\underset{R}{\\displaystyle\\iint}{f(x,y)dA}+\\underset{R}{\\displaystyle\\iint}{g(x,y)dA}[\/latex].<span data-type=\"newline\">\r\n<\/span><\/p>\r\n\r\n<ol style=\"list-style-type: decimal;\">\r\n \t<li style=\"list-style-type: none\">\r\n<ol style=\"list-style-type: decimal;\">\r\n \t<li>If [latex]c[\/latex] is a constant, then [latex]cf(x,y)[\/latex] is integrable and<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<p style=\"text-align: center;\">[latex]\\underset{R}{\\displaystyle\\iint}{cf(x,y)dA} = {c}\\underset{R}{\\displaystyle\\iint}{f(x,y)dA}[\/latex].<\/p>\r\n\r\n<ol style=\"list-style-type: decimal;\">\r\n \t<li style=\"list-style-type: none\">\r\n<ol style=\"list-style-type: decimal;\">\r\n \t<li>If [latex]R = S \\cup T[\/latex] and [latex]S \\cap T = \\emptyset [\/latex] except an overlap on the boundaries, then<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\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\n\r\n<ol style=\"list-style-type: decimal;\">\r\n \t<li style=\"list-style-type: none\">\r\n<ol style=\"list-style-type: decimal;\">\r\n \t<li>If [latex]f(x,y)\\geq g(x,y)[\/latex] for [latex](x,y)[\/latex] in [latex]R[\/latex], then<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<p style=\"text-align: center;\">[latex]\\underset{R}{\\displaystyle\\iint}{f(x,y)dA}\\geq \\underset{R}{\\displaystyle\\iint}{g(x,y)dA}[\/latex].<\/p>\r\n\r\n<ol style=\"list-style-type: decimal;\">\r\n \t<li style=\"list-style-type: none\">\r\n<ol style=\"list-style-type: decimal;\">\r\n \t<li>If [latex]m \\leq f(x,y)\\leq M[\/latex], then<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<p style=\"text-align: center;\">[latex]{m \\times A(R)} \\leq \\underset{R}{\\displaystyle\\iint}{f(x,y)dA} \\leq {M \\times A(R)}[\/latex].<\/p>\r\n\r\n<ol style=\"list-style-type: decimal;\">\r\n \t<li>In the case where [latex]f(x,y)[\/latex] can be factored as a product of a function [latex]g(x)[\/latex] of [latex]x[\/latex] only and a function [latex]h(y)[\/latex] of [latex]y[\/latex] only, then over the region [latex]{R} = {\\left \\{ (x,y)\\mid a \\leq x \\leq b,c \\leq y \\leq d \\right \\}}[\/latex], the double integral can be written as<\/li>\r\n<\/ol>\r\n<p style=\"text-align: center;\">[latex]\\underset{R}{\\displaystyle\\iint}{f(x,y)dA} = \\left(\\displaystyle\\int_a^bg(x)dx\\right)\\left(\\displaystyle\\int_c^dh(y)dy\\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n<section id=\"fs-id1167793414890\" data-depth=\"1\">\r\n<p id=\"fs-id1167793495486\">These properties are used in the evaluation of double integrals, as we will see later. We will become skilled in using these properties once we become familiar with the computational tools of double integrals. So let\u2019s get to that now.<\/p>\r\n\r\n<\/section><section id=\"fs-id1167793495494\" data-depth=\"1\"><\/section>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li><span class=\"os-abstract-content\">Recognize when a function of two variables is integrable over a rectangular region.<\/span><\/li>\n<li><span class=\"os-abstract-content\">Recognize and use some of the properties of double integrals.<\/span><\/li>\n<\/ul>\n<\/div>\n<section id=\"fs-id1167794098425\" data-depth=\"1\">\n<h2 data-type=\"title\">Volumes and Double Integrals<\/h2>\n<p>We first begin with a review of the definition of the definite integral in terms of the limit of a Riemann Sum from single-variable calculus.<\/p>\n<div class=\"textbox examples\">\n<h3>Recall: Riemann SumS and the Definite Integral<\/h3>\n<p>Suppose that [latex]y = f(x)[\/latex] is a continuous function and [latex]f(x) \\ge 0[\/latex] on a closed interval [latex][a,b][\/latex].\u00a0 The area under the curve and above the [latex]y-[\/latex]axis can be approximated using rectangles, by subdividing the interval [latex][a,b][\/latex] into [latex]n[\/latex] subintervals of width [latex]\\Delta x = \\frac{b-a}{n}[\/latex].\u00a0 If [latex]x_i[\/latex] denotes a sample point within the [latex]i^{th}[\/latex] subinterval, then [latex]f(x_i^\\ast)[\/latex] gives the height of the [latex]i^{th}[\/latex] rectangle.\u00a0 Adding up each of these areas yields an approximation of the total area, called a <strong>Riemann Sum<\/strong> and denoted using the following:<\/p>\n<p>[latex]\\sum_{i=1}^n f(x_i^\\ast)\\Delta x[\/latex]<\/p>\n<p>The definite integral of [latex]f(x)[\/latex] on [latex][a,b][\/latex] represents the exact area under the curve [latex]y = f(x)[\/latex], and is defined in terms as the following limit.<\/p>\n<p>[latex]\\int_a^b f(x)dx = \\lim_{n\\rightarrow\\infty}\\sum_{i=1}^n f(x_i^\\ast)\\Delta x[\/latex]<\/p>\n<\/div>\n<p id=\"fs-id1167794128227\">To extend this notion into another dimension, we consider the space above a rectangular region [latex]R[\/latex]. Consider a continuous function [latex]{f(x,y)\\geq 0}[\/latex]\u00a0of two variables defined on the closed rectangle [latex]R[\/latex]:<\/p>\n<\/section>\n<p style=\"text-align: center;\">[latex]\\large{R = [a,b]\\times[c,d] = \\left \\{(x,y)\\in {\\mathbb{R}}^2 |a\\leq x\\leq b,c\\leq y\\leq d \\right \\}}[\/latex]<\/p>\n<p>Here [latex]{{[a,b]}{\\times}{[c,d]}}[\/latex] denotes the Cartesian product of the two closed intervals [latex]{[a,b]}[\/latex] and [latex]{[c,d]}[\/latex]. It consists of rectangular pairs [latex](x, y)[\/latex] such that [latex]a\\leq x\\leq b[\/latex] and [latex]c\\leq y\\leq d[\/latex]. The graph of [latex]f[\/latex] represents a surface above the [latex]xy[\/latex]-plane with equation [latex]z=f(x, y)[\/latex] where [latex]z[\/latex] is the height of the surface at the point [latex](x, y)[\/latex]. Let [latex]S[\/latex] be the solid that lies above [latex]R[\/latex] and under the graph of [latex]f[\/latex] (Figure 1). The base of the solid is the rectangle [latex]R[\/latex] in the [latex]xy[\/latex]-plane. We want to find the volume [latex]V[\/latex] of the solid [latex]S[\/latex].<\/p>\n<div id=\"attachment_1299\" style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1299\" class=\"size-full wp-image-1299\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23141334\/5-1-1.jpeg\" alt=\"In xyz space, there is a surface z = f(x, y). On the x axis, the lines denoting a and b are drawn; on the y axis the lines for c and d are drawn. When the surface is projected onto the xy plane, it forms a rectangle with corners (a, c), (a, d), (b, c), and (b, d).\" width=\"487\" height=\"434\" \/><\/p>\n<p id=\"caption-attachment-1299\" class=\"wp-caption-text\">Figure 1.\u00a0The graph of\u00a0[latex]f(x,y)[\/latex] over the rectangle\u00a0[latex]R[\/latex] in the\u00a0[latex]xy[\/latex]-plane is a curved surface.<\/p>\n<\/div>\n<p>We divide the region [latex]R[\/latex] into small rectangles [latex]{R_{ij}}[\/latex], each with area [latex]{\\Delta{A}}[\/latex] and with sides [latex]{\\Delta{x}}[\/latex] and [latex]{\\Delta{y}}[\/latex] (Figure 2). We do this by dividing the interval [latex]{[a,b]}[\/latex] into [latex]m[\/latex] subintervals and dividing the interval [latex]{[c,d]}[\/latex] into [latex]n[\/latex] subintervals. Hence [latex]{{\\Delta{x}}={\\dfrac{b-a}{m}},{\\Delta{y}}={\\dfrac{d-c}{n}},}[\/latex] and\u00a0[latex]{{\\Delta{A}}={\\Delta{x}}{\\Delta{y}}}[\/latex].<\/p>\n<div id=\"attachment_1300\" style=\"width: 921px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1300\" class=\"size-full wp-image-1300\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23141459\/5-1-2.jpeg\" alt=\"In the xy plane, there is a rectangle with corners (a, c), (a, d), (b, c), and (b, d). Between a and b on the x axis lines are drawn from a, x1, x2, \u2026, xi, \u2026, b with distance Delta x between each line; between c and d on the y axis lines are drawn from c, y1, y2, \u2026, yj, \u2026, d with distance Delta y between each line. Among the resulting subrectangles, the one in the second column and third row up has a point marked (x*23, y*23). The rectangle Rij is marked with upper right corner (xi, yj). Within this rectangle the point (x*ij, y*ij) is marked.\" width=\"911\" height=\"527\" \/><\/p>\n<p id=\"caption-attachment-1300\" class=\"wp-caption-text\">Figure 2. Rectangle\u00a0[latex]R[\/latex] is divided into small rectangles [latex]R_{ij}[\/latex], each with an area [latex]\\Delta{A}[\/latex].<\/p>\n<\/div>\n<p>The volume of a thin rectangular box above [latex]{R_{ij}}[\/latex] is [latex]{f{({x{_{ij}^*}},{y{_{ij}^*}})}}{\\Delta{A}}[\/latex], where [latex]{({x{_{ij}^*}},{y{_{ij}^*}})}[\/latex] is an arbitrary sample point in each [latex]{R_{ij}}[\/latex] as shown in the following figure.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"attachment_1301\" style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1301\" class=\"size-full wp-image-1301\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23141611\/5-1-3.jpeg\" alt=\"In xyz space, there is a surface z = f(x, y). On the x axis, the lines denoting a and b are drawn; on the y axis the lines for c and d are drawn. When the surface is projected onto the xy plane, it forms a rectangle with corners (a, c), (a, d), (b, c), and (b, d). There are additional squares drawn to correspond to changes of Delta x and Delta y. On the surface, a square is marked and its projection onto the plane is marked as Rij. The average value for this small square is f(x*ij, y*ij).\" width=\"487\" height=\"474\" \/><\/p>\n<p id=\"caption-attachment-1301\" class=\"wp-caption-text\">Figure 3.\u00a0A thin rectangular box above\u00a0[latex]R_{ij}[\/latex] with height\u00a0[latex]f(x^{\\ast}_{ij},y^{\\ast}_{ij})[\/latex].<\/p>\n<\/div>\n<p id=\"fs-id1167794025753\">Using the same idea for all the subrectangles, we obtain an approximate volume of the solid [latex]S[\/latex] as [latex]{V}{\\approx} \\displaystyle\\sum_{i=1}^{m}\\sum_{j=1}^{n}\\ {f{({x{_{ij}^*}},{y{_{ij}^*}})}{\\Delta{A}}}[\/latex]. This sum is known as a\u00a0<strong><span id=\"d9da6d44-3ba3-4a90-adbc-dec26572e480_term205\" data-type=\"term\">double Riemann sum<\/span><\/strong>\u00a0and can be used to approximate the value of the volume of the solid. Here the double sum means that for each subrectangle we evaluate the function at the chosen point, multiply by the area of each rectangle, and then add all the results.<\/p>\n<p id=\"fs-id1167794139856\">As we have seen in the single-variable case, we obtain a better approximation to the actual volume if [latex]m[\/latex] and [latex]n[\/latex] become larger.<\/p>\n<p style=\"text-align: center;\">[latex]\\large{{V}=\\displaystyle\\lim_{m,n\\to\\infty}\\displaystyle\\sum_{i=1}^{m}\\sum_{j=1}^{n}f{({x{_{ij}^*}},{y{_{ij}^*}}){\\Delta{A}}}}[\/latex] or [latex]\\large{V=\\displaystyle\\lim_{\\Delta x,\\Delta y\\to0}\\displaystyle\\sum_{i=1}^{m}\\sum_{j=1}^{n}f(x_{ij}^*,y_{ij}^*)\\Delta A}[\/latex].<\/p>\n<p>Note that the sum approaches a limit in either case and the limit is the volume of the solid with the base [latex]R[\/latex]. Now we are ready to define the double integral.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<p>The\u00a0<span id=\"d9da6d44-3ba3-4a90-adbc-dec26572e480_term206\" data-type=\"term\">double integral<\/span>\u00a0of the function [latex]f(x, y)[\/latex] over the rectangular region [latex]R[\/latex] in the [latex]xy[\/latex]-plane is defined as<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\underset{R}{\\displaystyle\\iint}f(x,y)dA=\\displaystyle\\lim_{m,n\\to\\infty}\\displaystyle\\sum_{i=1}^{m}\\sum_{j=1}^{n}f(x_i^*,y_j^*)\\Delta A}[\/latex].<\/p>\n<\/div>\n<p>If [latex]f(x, y)\\geq 0[\/latex], then the volume [latex]V[\/latex] of the solid [latex]S[\/latex], which lies above [latex]R[\/latex] in the [latex]xy[\/latex]-plane and under the graph of [latex]f[\/latex], is the double integral of the function [latex]f(x, y)[\/latex] over the rectangle [latex]R[\/latex]. If the function is ever negative, then the double integral can be considered a \u201csigned\u201d volume in a manner similar to the way we defined net signed area in\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/defining-and-evaluating-definite-integrals\/\">The Definite Integral<\/a>.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Setting up a double integral and approximating it by double sums<\/h3>\n<p id=\"fs-id1167793933224\">Consider the function [latex]{z = f(x,y) = 3x^2 - y}[\/latex] over the rectangular region [latex]{R = {[0,2]} \\times {[0,2]}}[\/latex] (Figure 4).<\/p>\n<ol id=\"fs-id1167793247300\" type=\"a\">\n<li>Set up a double integral for finding the value of the signed volume of the solid [latex]S[\/latex] that lies above [latex]R[\/latex] and \u201cunder\u201d the graph of [latex]f[\/latex].<\/li>\n<li>Divide [latex]R[\/latex] into four squares with [latex]m = n = 2[\/latex], and choose the sample point as the upper right corner point of each square [latex](1,1),(2,1),(1,2),[\/latex] and [latex](2,2)[\/latex] (Figure 4) to approximate the signed volume of the solid [latex]S[\/latex] that lies above [latex]R[\/latex] and \u201cunder\u201d the graph of [latex]f[\/latex].<\/li>\n<li>Divide [latex]R[\/latex] into four squares with [latex]m = n = 2[\/latex], and choose the sample point as the midpoint of each square: [latex](1\/2,1\/2),(3\/2,1\/2),(1\/2,3\/2),[\/latex] and [latex](3\/2,3\/2)[\/latex] to approximate the signed volume.<\/li>\n<\/ol>\n<div id=\"attachment_1303\" style=\"width: 493px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1303\" class=\"size-full wp-image-1303\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23141819\/5-1-4.jpeg\" alt=\"In xyz space, there is a surface z = f(x, y) = 3x2 minus y. The corners of the surface are given as (0, 0, 0), (2, 0, 12), (0, 2, negative 2), and (2, 2, 10). The surface is parabolic along the x axis.\" width=\"483\" height=\"252\" \/><\/p>\n<p id=\"caption-attachment-1303\" class=\"wp-caption-text\">Figure 4.\u00a0The function [latex]\\small{z=f(x,y)}[\/latex] graphed over the rectangular region [latex]\\small{R=[0,2]\\times[0,2]}[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q193760238\">Show Solution<\/span><\/p>\n<div id=\"q193760238\" class=\"hidden-answer\" style=\"display: none\">\n<h2 data-type=\"solution-title\"><span class=\"os-title-label\">Solution<\/span><\/h2>\n<div class=\"os-solution-container\">\n<ol type=\"a\">\n<li style=\"list-style-type: none;\">\n<ol type=\"a\">\n<li style=\"list-style-type: none;\">\n<ol type=\"a\">\n<li style=\"text-align: center;\">As we can see, the function [latex]z = f(x,y) = 3x^2 - y[\/latex] is above the plane. To find the signed volume of [latex]S[\/latex], we need to divide the region [latex]R[\/latex] into small rectangles [latex]{R}_{ij}[\/latex] each with area [latex]{\\Delta{A}}[\/latex] and with sides [latex]{\\Delta{x}}[\/latex] and [latex]{\\Delta{y}}[\/latex], and choose [latex]{({x{_{ij}^*}},{y{_{ij}^*}})}[\/latex] as sample points in each [latex]{R}_{ij}[\/latex]. Hence, a double integral is set up as<br \/>\n[latex]V = \\underset{R}{\\displaystyle\\iint}(3x^2-y)dA=\\displaystyle\\lim_{m,n\\to\\infty}\\displaystyle\\sum_{i=1}^{m}\\sum_{j=1}^{n}\\left[3(x_{ij}^*)^2-y_{ij}^*\\right]\\Delta A[\/latex].<span data-type=\"newline\"><br \/>\n<\/span><\/li>\n<li>Approximating the signed volume using a Riemann sum with [latex]m = n = 2[\/latex] we have [latex]{\\Delta{A}} = {\\Delta{x}}{\\Delta{y}} = {1 {\\times} 1} = 1[\/latex]. Also, the sample points are [latex](1, 1), (2, 1), (1, 2),[\/latex] and [latex](2,2)[\/latex] as shown in the following figure.<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"CNX_Calc_Figure_15_01_005\" class=\"os-figure\">\n<div class=\"os-caption-container\">\n<div id=\"attachment_1305\" style=\"width: 386px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1305\" class=\"size-full wp-image-1305\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23142153\/5-1-4a.jpeg\" alt=\"On the xy plane, the points (1, 1), (1, 2), (2, 1), and (2, 2) are marked, and these form the upper right corners of four squares marked R11, R12, R21, and R22, respectively. Each square has area Delta A = 1.\" width=\"376\" height=\"484\" \/><\/p>\n<p id=\"caption-attachment-1305\" class=\"wp-caption-text\">Figure 5. Subrectangles for the rectangular region [latex]\\small{R=[0,2]\\times[0,2]}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span>Hence,<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p>[latex]\\hspace{3cm}\\begin{align}    V&=\\displaystyle\\sum_{i=1}^2\\sum_{j=1}^2f(x_{ij}^*,y_{ij}^*)\\Delta A \\\\    &=\\displaystyle\\sum_{i=1}^2(f(x_{i1}^*,y_{i1}^*)+ff(x_{i2}^*,y_{i2}^*))\\Delta A \\\\    &=(f(x_{11}^*,y_{11}^*)\\Delta A+(f(x_{21}^*,y_{21}^*)\\Delta A+(f(x_{12}^*,y_{12}^*)\\Delta A+(f(x_{22}^*,y_{22}^*)\\Delta A \\\\  &=f(1,1)(1)+f(2,1)(1)+f(1,2)(1)+f(2,2)(1) \\\\  &=(3-1)(1)+(12-1)(1)+(3-1)(1)+(12-2)(1) \\\\  &=2+11+1+10=24.  \\end{align}[\/latex]<\/li>\n<li>Approximating the signed volume using a Riemann sum with [latex]m = n = 2[\/latex], we have[latex]{\\Delta{A}} = {\\Delta{x}}{\\Delta{y}} = {1 {\\times} 1} = 1[\/latex]. In this case the sample points are [latex](1\/2, 1\/2), (3\/2, 1\/2), (1\/2, 3\/2),[\/latex] and [latex](3\/2, 3\/2)[\/latex].<span data-type=\"newline\"><br \/>\n<\/span>Hence,<\/li>\n<\/ol>\n<p>[latex]\\hspace{3cm}\\begin{align}  V&=\\displaystyle\\sum_{i=1}^2\\sum_{j=1}^2f(x_{ij}^*,y_{ij}^*)\\Delta A \\\\&=f(x_{11}^*,y_{11}^*)\\Delta A+f(x_{21}^*,y_{21}^*)\\Delta A+f(x_{12}^*,y_{12}^*)\\Delta A+f(x_{22}^*,y_{22}^*)\\Delta A \\\\  &=f(1\/2,1\/2)(1)+f(3\/2,1\/2)(1)+f(1\/2,3\/2)(1)+f(3\/2,3\/2)(1) \\\\  &=\\left(\\small{\\frac34}-\\small{\\frac14}\\right)(1)+\\left(\\small{\\frac{27}4}-\\small{\\frac12}\\right)(1)+\\left(\\small{\\frac34}-\\small{\\frac32}\\right)(1)+\\left(\\small{\\frac{27}4}-\\small{\\frac32}\\right)(1) \\\\  &=\\frac24+\\frac{25}4+\\left(-\\frac34\\right)+\\frac{21}4=\\frac{45}4.  \\end{align}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<h2 id=\"9\" data-type=\"commentary-title\"><span class=\"os-title-label\">Analysis<\/span><\/h2>\n<p id=\"fs-id1167794067077\">Notice that the approximate answers differ due to the choices of the sample points. In either case, we are introducing some error because we are using only a few sample points. Thus, we need to investigate how we can achieve an accurate answer.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try it<\/h3>\n<p id=\"fs-id1167794324430\">Use the same function [latex]z = f(x,y) = 3x^2 - y[\/latex] over the rectangular region [latex]R = [0,2] \\times [0,2][\/latex].<\/p>\n<p id=\"fs-id1167793473779\">Divide [latex]R[\/latex] into the same four squares with [latex]m = n = 2[\/latex] and choose the sample points as the upper left corner point of each square [latex](0,1), (1,1), (0,2),[\/latex] and [latex](1,2)[\/latex] (Figure 5) to approximate the signed volume of the solid [latex]S[\/latex] that lies above [latex]R[\/latex] and \u201cunder\u201d the graph of [latex]f[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q265173902\">Show Solution<\/span><\/p>\n<div id=\"q265173902\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\large{V=\\displaystyle\\sum_{i=1}^2\\sum_{j=1}^2f(x_{ij}^*,y_{ij}^*)\\Delta A=0}[\/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=8197095&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=ZKNjm2OdGnU&amp;video_target=tpm-plugin-g6ue79lv-ZKNjm2OdGnU\" 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\/CP5.1_transcript.html\">transcript for \u201cCP 5.1\u201d here (opens in new window).<\/a><\/div>\n<p id=\"fs-id1167793263189\">Note that we developed the concept of double integral using a rectangular region [latex]R[\/latex]. This concept can be extended to any general region. However, when a region is not rectangular, the subrectangles may not all fit perfectly into [latex]R[\/latex], particularly if the base area is curved. We examine this situation in more detail in the next section, where we study regions that are not always rectangular and subrectangles may not fit perfectly in the region [latex]R[\/latex]. Also, the heights may not be exact if the surface [latex]z = f(x,y)[\/latex] is curved. However, the errors on the sides and the height where the pieces may not fit perfectly within the solid [latex]S[\/latex] approach 0 as [latex]m[\/latex] and [latex]n[\/latex] approach infinity. Also, the double integral of the function [latex]z = f(x,y)[\/latex] exists provided that the function [latex]f[\/latex] is not too discontinuous. If the function is bounded and continuous over [latex]R[\/latex] except on a finite number of smooth curves, then the double integral exists and we say that [latex]f[\/latex] is integrable over [latex]R[\/latex].<\/p>\n<p id=\"fs-id1167793456312\">Since [latex]{{\\Delta{A}}={\\Delta{x}}{\\Delta{y}}=\\Delta y\\Delta x}[\/latex], we can express [latex]dA[\/latex] as [latex]dx[\/latex] [latex]dy[\/latex] or [latex]dy[\/latex] [latex]dx[\/latex]. This means that, when we are using rectangular coordinates, the double integral over a region [latex]R[\/latex] denoted by [latex]\\underset{R}{\\displaystyle\\iint}f(x,y)dA[\/latex] can be written as<\/p>\n<p>[latex]\\underset{R}{\\displaystyle\\iint}f(x,y)dx dy[\/latex] or\u00a0[latex]\\underset{R}{\\displaystyle\\iint}f(x,y)dy dx[\/latex].<\/p>\n<section id=\"fs-id1167794098425\" data-depth=\"1\">\n<p id=\"fs-id1167793414884\">Now let\u2019s list some of the properties that can be helpful to compute double integrals.<\/p>\n<\/section>\n<section id=\"fs-id1167793414890\" data-depth=\"1\">\n<h2 data-type=\"title\">Properties of Double Integrals<\/h2>\n<p id=\"fs-id1167793414895\">The properties of double integrals are very helpful when computing them or otherwise working with them. We list here six properties of double integrals. Properties 1 and 2 are referred to as the linearity of the integral, property 3 is the additivity of the integral, property 4 is the monotonicity of the integral, and property 5 is used to find the bounds of the integral. Property 6 is used if [latex]f(x,y)[\/latex] is a product of two functions [latex]g(x)[\/latex] and\u00a0[latex]h(y)[\/latex].<\/p>\n<\/section>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Theorem: properties of double integrals<\/h3>\n<hr \/>\n<p id=\"fs-id1167793293479\">Assume that the functions [latex]f(x,y)[\/latex] and [latex]g(x,y)[\/latex] are integrable over the rectangular region [latex]R[\/latex]; [latex]S[\/latex]\u00a0and\u00a0[latex]T[\/latex]\u00a0are subregions of\u00a0[latex]R[\/latex]; and assume that\u00a0[latex]m[\/latex]\u00a0and\u00a0[latex]M[\/latex]\u00a0are real numbers.<\/p>\n<ol style=\"list-style-type: decimal;\">\n<li style=\"list-style-type: none\">\n<ol style=\"list-style-type: decimal;\">\n<li>The sum [latex]f(x,y) + g(x,y)[\/latex] is integrable and<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p style=\"text-align: center;\">[latex]\\underset{R}{\\displaystyle\\iint}{[f(x,y)+g(x,y)]dA}=\\underset{R}{\\displaystyle\\iint}{f(x,y)dA}+\\underset{R}{\\displaystyle\\iint}{g(x,y)dA}[\/latex].<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<ol style=\"list-style-type: decimal;\">\n<li style=\"list-style-type: none\">\n<ol style=\"list-style-type: decimal;\">\n<li>If [latex]c[\/latex] is a constant, then [latex]cf(x,y)[\/latex] is integrable and<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p style=\"text-align: center;\">[latex]\\underset{R}{\\displaystyle\\iint}{cf(x,y)dA} = {c}\\underset{R}{\\displaystyle\\iint}{f(x,y)dA}[\/latex].<\/p>\n<ol style=\"list-style-type: decimal;\">\n<li style=\"list-style-type: none\">\n<ol style=\"list-style-type: decimal;\">\n<li>If [latex]R = S \\cup T[\/latex] and [latex]S \\cap T = \\emptyset[\/latex] except an overlap on the boundaries, then<\/li>\n<\/ol>\n<\/li>\n<\/ol>\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<ol style=\"list-style-type: decimal;\">\n<li style=\"list-style-type: none\">\n<ol style=\"list-style-type: decimal;\">\n<li>If [latex]f(x,y)\\geq g(x,y)[\/latex] for [latex](x,y)[\/latex] in [latex]R[\/latex], then<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p style=\"text-align: center;\">[latex]\\underset{R}{\\displaystyle\\iint}{f(x,y)dA}\\geq \\underset{R}{\\displaystyle\\iint}{g(x,y)dA}[\/latex].<\/p>\n<ol style=\"list-style-type: decimal;\">\n<li style=\"list-style-type: none\">\n<ol style=\"list-style-type: decimal;\">\n<li>If [latex]m \\leq f(x,y)\\leq M[\/latex], then<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p style=\"text-align: center;\">[latex]{m \\times A(R)} \\leq \\underset{R}{\\displaystyle\\iint}{f(x,y)dA} \\leq {M \\times A(R)}[\/latex].<\/p>\n<ol style=\"list-style-type: decimal;\">\n<li>In the case where [latex]f(x,y)[\/latex] can be factored as a product of a function [latex]g(x)[\/latex] of [latex]x[\/latex] only and a function [latex]h(y)[\/latex] of [latex]y[\/latex] only, then over the region [latex]{R} = {\\left \\{ (x,y)\\mid a \\leq x \\leq b,c \\leq y \\leq d \\right \\}}[\/latex], the double integral can be written as<\/li>\n<\/ol>\n<p style=\"text-align: center;\">[latex]\\underset{R}{\\displaystyle\\iint}{f(x,y)dA} = \\left(\\displaystyle\\int_a^bg(x)dx\\right)\\left(\\displaystyle\\int_c^dh(y)dy\\right)[\/latex].<\/p>\n<\/div>\n<section id=\"fs-id1167793414890\" data-depth=\"1\">\n<p id=\"fs-id1167793495486\">These properties are used in the evaluation of double integrals, as we will see later. We will become skilled in using these properties once we become familiar with the computational tools of double integrals. So let\u2019s get to that now.<\/p>\n<\/section>\n<section id=\"fs-id1167793495494\" data-depth=\"1\"><\/section>\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-3964\">\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 5.1. <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":3,"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 5.1\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-3964","chapter","type-chapter","status-publish","hentry"],"part":23,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3964","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/349141"}],"version-history":[{"count":18,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3964\/revisions"}],"predecessor-version":[{"id":6450,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3964\/revisions\/6450"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/23"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3964\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=3964"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=3964"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=3964"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=3964"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}