{"id":3981,"date":"2022-04-12T17:24:17","date_gmt":"2022-04-12T17:24:17","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=3981"},"modified":"2022-11-01T04:22:07","modified_gmt":"2022-11-01T04:22:07","slug":"double-integrals-in-polar-coordinates","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/double-integrals-in-polar-coordinates\/","title":{"raw":"Double Integrals in Polar Coordinates","rendered":"Double Integrals in Polar Coordinates"},"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 the format of a double integral over a polar rectangular region.<\/span><\/li>\r\n \t<li><span class=\"os-abstract-content\">Evaluate a double integral in polar coordinates by using an iterated integral.<\/span><\/li>\r\n \t<li><span class=\"os-abstract-content\">Recognize the format of a double integral over a general polar region.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2 data-type=\"title\">Polar Rectangular Regions of Integration<\/h2>\r\n<p id=\"fs-id1167793292241\">When we defined the double integral for a continuous function in rectangular coordinates\u2014say, [latex]g[\/latex] over a region [latex]R[\/latex] in the [latex]xy[\/latex]-plane\u2014we divided [latex]R[\/latex] into subrectangles with sides parallel to the coordinate axes. These sides have either constant [latex]x[\/latex]-values and\/or constant [latex]y[\/latex]-values. In polar coordinates, the shape we work with is a\u00a0<span id=\"82852cef-a933-495b-9be8-6b3aa0676026_term217\" data-type=\"term\">polar rectangle<\/span>, whose sides have constant [latex]r[\/latex]-values and\/or constant [latex]\\theta[\/latex]-values. This means we can describe a polar rectangle as in\u00a0Figure 1(a), with [latex]{R} = {\\left \\{{(r,{\\theta})}{\\mid}{a} \\ {\\leq} \\ {r} \\ {\\leq} \\ {b,{\\alpha}} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {\\beta} \\right \\}}[\/latex].<\/p>\r\n<p id=\"fs-id1167793371538\">In this section, we are looking to integrate over polar rectangles. Consider a function [latex]{f}{(r,{\\theta})}[\/latex] over a polar rectangle [latex]R[\/latex]. We divide the interval [latex][a,b][\/latex] into [latex]m[\/latex] subintervals [latex]{[{r_{i-1}},{r_i}]}[\/latex] of length [latex]{\\Delta}{r} = {(b-a)}{\/m}[\/latex] and divide the interval [latex]{[{\\alpha}, {\\beta}]}[\/latex] into [latex]n[\/latex] subintervals [latex]{[{{\\theta}_{j-1}},{{\\theta}_{j}}]}[\/latex] of width [latex]{\\Delta}{\\theta} = {({\\beta}-{\\alpha})}{\/n}[\/latex]. This means that the circles [latex]{r} = {r_{i}}[\/latex] and rays [latex]{\\theta} = {{\\theta}_{j}}[\/latex] for [latex]{1} \\ {\\leq} \\ {i} \\ {\\leq} \\ {m}[\/latex] and [latex]{1} \\ {\\leq} \\ {j} \\ {\\leq} \\ {n}[\/latex] divide the polar rectangle [latex]R[\/latex] into smaller polar subrectangles [latex]{R}_{ij}[\/latex] (Figure 1(b)).<\/p>\r\n\r\n[caption id=\"attachment_1357\" align=\"aligncenter\" width=\"970\"]<img class=\"size-full wp-image-1357\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25142814\/5-3-1.jpeg\" alt=\"This figure consists of three figures labeled a, b, and c. In figure a, a sector of an annulus is shown in the polar coordinate plane with radii a and b and angles alpha and beta from the theta = 0 axis. In figure b, this sector of an annulus is cut up into subsectors in a manner similar to the way in which previous spaces were cut up into subrectangles. In figure c, one of these subsectors is shown with angle Delta theta, distance between inner and outer radii Delta r, and area Delta A = r* sub theta Delta r Delta theta, where the center point is given as (r* sub i j, theta* sub i j).\" width=\"970\" height=\"339\" \/> Figure 1.\u00a0(a) A polar rectangle\u00a0[latex]R[\/latex]\u00a0(b) divided into subrectangles [latex]R_{ij}[\/latex] (c) Close-up of a subrectangle.[\/caption]\r\n<p id=\"fs-id1167793940970\">As before, we need to find the area [latex]{\\Delta}{A}[\/latex] of the polar subrectangle [latex]{R}_{ij}[\/latex] and the \u201cpolar\u201d volume of the thin box above [latex]{R}_{ij}[\/latex]. Recall that, in a circle of radius [latex]r[\/latex], the length [latex]s[\/latex] of an arc subtended by a central angle of [latex]{\\theta}[\/latex] radians is [latex]{s} = {r}{\\theta}[\/latex]. Notice that the polar rectangle [latex]{R}_{ij}[\/latex] looks a lot like a trapezoid with parallel sides [latex]{r}_{i-1}{\\Delta}{\\theta}[\/latex] and [latex]{r}_{i}{\\Delta}{\\theta}[\/latex]and with a width [latex]{\\Delta}{r}[\/latex]. Hence the area of the polar subrectangle [latex]{R}_{ij}[\/latex] is<\/p>\r\n<p style=\"text-align: center;\">[latex]\\Large{\\Delta}{A} = {\\dfrac{1}{2}}{\\Delta}{r}{({r_{i-1}}{\\Delta}{\\theta}+{r_1}{\\Delta}{\\theta})}[\/latex].<\/p>\r\nSimplifying and letting [latex]{{r}^{*}_{ij}} = {\\dfrac{1}{2}}{({r_{i-1}}+{r_i})}[\/latex], we have [latex]{\\Delta}{A} = {{r}^{*}_{ij}}{\\Delta}{r}{\\Delta}{\\theta}[\/latex]. Therefore, the polar volume of the thin box above [latex]{R}_{ij}[\/latex]\u00a0(Figure 2) is\r\n<p style=\"text-align: center;\">[latex]\\Large{f}{({{r}^{*}_{ij}},{{\\theta}^{*}_{ij}})}{\\Delta}{A} = {f}{({{r}^{*}_{ij}},{{\\theta}^{*}_{ij}})}{{r}^{*}_{ij}}{\\Delta}{r}{\\Delta}{\\theta}[\/latex].<\/p>\r\n\r\n[caption id=\"attachment_1359\" align=\"aligncenter\" width=\"273\"]<img class=\"size-full wp-image-1359\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25142903\/5-3-2.jpeg\" alt=\"In x y z space, there is a surface f (r, theta). On the x y plane, a series of subsectors of annuli are drawn as in the previous figure with radius between annuli Delta r and angle between subsectors Delta theta. A subsector from the surface f(r, theta) is projected down onto one of these subsectors. This subsector has center point marked (r* sub i j, theta* sub i j).\" width=\"273\" height=\"307\" \/> Figure 2.\u00a0Finding the volume of the thin box above polar rectangle\u00a0[latex]\\small{R_{ij}}[\/latex][\/caption]\r\n<p id=\"fs-id1167794096760\">Using the same idea for all the subrectangles and summing the volumes of the rectangular boxes, we obtain a double Riemann sum as<\/p>\r\n<p style=\"text-align: center;\">[latex]\\Large{\\displaystyle\\sum_{i=1}^{m}\\displaystyle\\sum_{j=1}^{n}}{f}{({{r}^{*}_{ij}},{{\\theta}^{*}_{ij}})}{{r}^{*}_{ij}}{\\Delta}{r}{\\Delta}{\\theta}[\/latex].<\/p>\r\nAs we have seen before, we obtain a better approximation to the polar volume of the solid above the region [latex]R[\/latex] when we let [latex]m[\/latex] and [latex]n[\/latex] become larger. Hence, we define the polar volume as the limit of the double Riemann sum,\r\n<p style=\"text-align: center;\">[latex]\\Large{V} = {\\displaystyle\\lim_{m,n \\rightarrow \\infty}} \\ {\\displaystyle\\sum_{i=1}^{m}\\displaystyle\\sum_{j=1}^{n}}{f}{({{r}^{*}_{ij}},{{\\theta}^{*}_{ij}})}{{r}^{*}_{ij}}{\\Delta}{r}{\\Delta}{\\theta}[\/latex].<\/p>\r\nThis becomes the expression for 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 double integral of the function [latex]{f}{(r,{\\theta})}[\/latex] over the polar rectangular region [latex]R[\/latex] in the [latex]{r}{\\theta}[\/latex]-plane is defined as\r\n<p style=\"text-align: center;\">[latex]\\large{\\underset{R}{\\displaystyle\\iint}f(r,\\theta)dA = \\displaystyle\\lim_{m,n \\rightarrow \\infty}\\displaystyle\\sum_{i=1}^{m}\\displaystyle\\sum_{j=1}^{n}f(r^*_{ij},\\theta^*_{ij})\\Delta{A} = \\displaystyle\\lim_{m,n \\rightarrow \\infty}\\displaystyle\\sum_{i=1}^{m}\\displaystyle\\sum_{j=1}^{n}f(r^{*}_{ij},\\theta^{*}_{ij})r^{*}_{ij}\\Delta{r}\\Delta\\theta}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<p id=\"fs-id1167794293302\">Again, just as in\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/double-integrals-over-rectangular-regions\/\">Double Integrals over Rectangular Regions<\/a>, the double integral over a polar rectangular region can be expressed as an iterated integral in polar coordinates. Hence,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\Large{\\underset{R}{\\displaystyle\\iint}f(r,\\theta)dA = \\underset{R}{\\displaystyle\\iint}f(r,\\theta)r \\ dr \\ d\\theta = \\displaystyle\\int_{\\theta=\\alpha}^{\\theta=\\beta}\\displaystyle\\int_{r=a}^{r=b}f(r,\\theta)r \\ dr \\ d\\theta}[\/latex].<\/p>\r\nNotice that the expression for [latex]dA[\/latex] is replaced by [latex]{r} \\ {dr} \\ {d{\\theta}}[\/latex] when working in polar coordinates. Another way to look at the polar double integral is to change the double integral in rectangular coordinates by substitution. When the function [latex]f[\/latex] is given in terms of [latex]x[\/latex] and [latex]y[\/latex], using [latex]x=r\\cos\\theta,y = r\\sin\\theta[\/latex] and [latex]{dA} = {{r} \\ {dr} \\ {d{\\theta}}}[\/latex] changes it to\r\n<p style=\"text-align: center;\">[latex]\\Large{\\underset{R}{\\displaystyle\\iint}f(x,y)dA=\\underset{R}{\\displaystyle\\iint}f(r\\cos\\theta,r\\sin\\theta)r \\ dr \\ d\\theta}[\/latex].<\/p>\r\nNote that all the properties listed in\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/double-integrals-over-rectangular-regions\/\" data-page-slug=\"5-1-double-integrals-over-rectangular-regions\" data-page-uuid=\"d9da6d44-3ba3-4a90-adbc-dec26572e480\" data-page-fragment=\"page_d9da6d44-3ba3-4a90-adbc-dec26572e480\">Double Integrals over Rectangular Regions<\/a>\u00a0for the double integral in rectangular coordinates hold true for the double integral in polar coordinates as well, so we can use them without hesitation.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: sketching a polar rectangular region<\/h3>\r\nSketch the polar rectangular region\u00a0[latex]{R} = {\\left \\{{(r,{\\theta})}{\\mid}{1} \\ {\\leq} \\ {r} \\ {\\leq} \\ {3,0} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {\\pi} \\right \\}}[\/latex].\r\n\r\n[reveal-answer q=\"137726982\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"137726982\"]\r\n\r\nAs we can see from\u00a0Figure 3, [latex]r=1[\/latex] and [latex]r=3[\/latex] are circles of radius [latex]1[\/latex] and [latex]3[\/latex] and [latex]{0} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {\\pi}[\/latex] covers the entire top half of the plane. Hence the region [latex]R[\/latex] looks like a semicircular band.\r\n\r\n[caption id=\"attachment_1360\" align=\"aligncenter\" width=\"267\"]<img class=\"size-full wp-image-1360\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25142956\/5-3-3.jpeg\" alt=\"Half an annulus R is drawn with inner radius 1 and outer radius 3. That is, the inner semicircle is given by x squared + y squared = 1, whereas the outer semicircle is given by x squared + y squared = 9.\" width=\"267\" height=\"234\" \/> Figure 3.\u00a0The polar region [latex]R[\/latex] lies between two semicircles.[\/caption][\/hidden-answer]<\/div>\r\nNow that we have sketched a polar rectangular region, let us demonstrate how to evaluate a double integral over this region by using polar coordinates.\r\n<div class=\"textbox exercises\">\r\n<h3>example: evaluating a double integral over a polar rectangular region<\/h3>\r\nEvaluate the integral\u00a0[latex]\\underset{R}{\\displaystyle\\iint}{3x}{dA}[\/latex] over the region\u00a0[latex]{R} = {\\left \\{{(r,{\\theta})}{\\mid}{1} \\ {\\leq} \\ {r} \\ {\\leq} \\ {2,0} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {\\pi} \\right \\}}[\/latex].\r\n\r\n[reveal-answer q=\"862144637\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"862144637\"]\r\n\r\nFirst we sketch a figure similar to\u00a0Figure 3\u00a0but with outer radius 2. From the figure we can see that we have\r\n\r\n[latex]\\begin{align}\r\n\r\n\\underset{R}{\\displaystyle\\iint}3x \\ dA &amp;= \\displaystyle\\int_{\\theta=0}^{\\theta=\\pi}\\displaystyle\\int_{r=1}^{r=2}3r\\cos\\theta{r} \\ dr \\ d\\theta &amp;\\quad &amp;\\text{Use an iterated integral with correct limits of integration.} \\\\\r\n\r\n&amp;=\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi}\\cos\\theta\\left[r^3\\bigg|_{r=1}^{r=2}\\right]d\\theta &amp;\\quad &amp;\\text{Integrate first with respect to }r. \\\\\r\n\r\n&amp;=\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi}7\\cos\\theta{d}\\theta=7\\sin\\theta\\bigg|_{\\theta=0}^{\\theta=\\pi}=0.\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\nSketch the region\u00a0[latex]{R} = {\\left \\{{(r,{\\theta})}{\\mid}{1} \\ {\\leq} \\ {r} \\ {\\leq} \\ {2,-{\\frac{\\pi}{2}}} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {\\frac{\\pi}{2}} \\right \\}}[\/latex], and evaluate\u00a0[latex]\\underset{R}{\\displaystyle\\iint}{x} \\ {dA}[\/latex].\r\n\r\n[reveal-answer q=\"733260947\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"733260947\"]\r\n\r\n[latex]\\frac{14}3[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Evaluating a double integral by converting from rectangular coordinates<\/h3>\r\nEvaluate the integral [latex]\\underset{R}{\\displaystyle\\iint}{(1-{x^2}-{y^2})}{dA}[\/latex] where [latex]R[\/latex] is the unit circle on the [latex]xy[\/latex]-plane.\r\n\r\n[reveal-answer q=\"100254682\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"100254682\"]\r\n<p id=\"fs-id1167794218699\">The region [latex]R[\/latex] is a unit circle, so we can describe it as [latex]{R} = {\\left \\{{(r,{\\theta})}{\\mid}{0} \\ {\\leq} \\ {r} \\ {\\leq} \\ {1,0} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {2{\\pi}} \\right \\}}[\/latex].<\/p>\r\n<p id=\"fs-id1167793543369\">Using the conversion [latex]x=r\\cos\\theta,y=r\\sin\\theta[\/latex], and [latex]{dA} = {{r} \\ {dr} \\ {d{\\theta}}}[\/latex], we have<\/p>\r\n[latex]\\hspace{4cm}\\begin{align}\r\n\r\n\\underset{R}{\\displaystyle\\iint}(1-x^2-y^2)dA&amp;=\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^1(1-r^2)r \\ dr \\ d\\theta =\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^1(r - r^3)dr \\ d\\theta \\\\\r\n\r\n&amp;=\\displaystyle\\int_0^{2\\pi}\\left[\\frac{r^2}2-\\frac{r^4}4\\right]_0^1{d}\\theta=\\displaystyle\\int_0^{2\\pi}\\frac14d\\theta=\\frac{\\pi}2.\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 a double integral by converting from rectangular coordinates<\/h3>\r\nEvaluate the integral\u00a0[latex]\\underset{R}{\\displaystyle\\iint}{(x+y)}{dA}[\/latex] where\u00a0[latex]{R} = {\\left \\{{(x,y)}{\\mid}{1} \\ {\\leq} \\ {{x^2}+{y^2}} \\ {\\leq} \\ {4,x} \\ {\\leq} \\ {0} \\right \\}}[\/latex].\r\n\r\n&nbsp;\r\n\r\n[reveal-answer q=\"984622178\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"984622178\"]\r\n\r\nWe can see that [latex]R[\/latex] is an annular region that can be converted to polar coordinates and described as [latex]{R} = {\\left \\{{(r,{\\theta})}{\\mid}{1} \\ {\\leq} \\ {r} \\ {\\leq} \\ {2,{\\frac{\\pi}{2}}} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {\\frac{3{\\pi}}{2}} \\right \\}}[\/latex] (see the following graph).\r\n\r\n[caption id=\"attachment_1361\" align=\"aligncenter\" width=\"293\"]<img class=\"size-full wp-image-1361\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25143041\/5-3-4.jpeg\" alt=\"Two semicircles are drawn in the second and third quadrants, with equations x squared + y squared = 1 and x squared + y squared = 2.\" width=\"293\" height=\"272\" \/> Figure 4.\u00a0The annular region of integration\u00a0[latex]R[\/latex].[\/caption]Hence, using the conversion[latex]x =r\\cos\\theta,y =r\\sin\\theta[\/latex], and [latex]{dA} = {{r} \\ {dr} \\ {d{\\theta}}}[\/latex], we have\r\n[latex]\\hspace{5cm}\\begin{align}\\underset{R}{\\displaystyle\\iint}(x+y)dA&amp;=\\displaystyle\\int_{\\theta=\\pi\/2}^{\\theta=3\\pi\/2}\\displaystyle\\int_{r=1}^{r=2}(r\\cos\\theta+r\\sin\\theta)r \\ dr \\ d\\theta \\\\&amp;=\\left(\\displaystyle\\int_{r=1}^{r=2}r^2dr\\right)\\left(\\displaystyle\\int_{\\pi\/2}^{3\\pi\/2}(\\cos\\theta+\\sin\\theta)d\\theta\\right) \\\\&amp;=\\left[\\frac{r^3}3\\right]_1^2[\\sin\\theta-\\cos\\theta]\\bigg|_{\\pi\/2}^{3\\pi\/2} \\\\&amp;=-\\frac{14}3.\\end{align}[\/latex]\r\n[\/hidden-answer]<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nEvaluate the integral [latex]\\underset{R}{\\displaystyle\\iint}{(4-{x^2}-{y^2})}{dA}[\/latex] where [latex]R[\/latex] is the circle of radius 2 on the [latex]xy[\/latex]-plane.\r\n\r\n[reveal-answer q=\"784312663\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"784312663\"]\r\n\r\n[latex]8\\pi[\/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=8197102&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=UKNpIyyDQCQ&amp;video_target=tpm-plugin-0qwtbnvq-UKNpIyyDQCQ\" 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\/CP5.18_transcript.html\">transcript for \u201cCP 5.18<span style=\"font-size: 1em;\">\u201d here (opens in new window).<\/span><\/a><\/center>\r\n<h2 data-type=\"title\">General Polar Regions of Integration<\/h2>\r\n<p id=\"fs-id1167793928915\">To evaluate the double integral of a continuous function by iterated integrals over general polar regions, we consider two types of regions, analogous to Type I and Type II as discussed for rectangular coordinates in\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/introduction-to-double-integrals-over-general-regions\/\" data-page-slug=\"5-2-double-integrals-over-general-regions\" data-page-uuid=\"f783ca08-ca2f-4ebb-a559-d1b30c2f9b8c\" data-page-fragment=\"page_f783ca08-ca2f-4ebb-a559-d1b30c2f9b8c\">Double Integrals over General Regions<\/a>. It is more common to write polar equations as [latex]{r} = {f}{(\\theta)}[\/latex] than [latex]{\\theta} = {f}{(r)}[\/latex], so we describe a general polar region as [latex]{R} = {\\left \\{{(r,{\\theta})}{\\mid}{\\alpha} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {{\\beta},{h_1}}{(\\theta)} \\ {\\leq} \\ {r} \\ {\\leq} \\ {{h_2}{(\\theta)}} \\right \\}}[\/latex] (see the following figure).<\/p>\r\n\r\n[caption id=\"attachment_1362\" align=\"aligncenter\" width=\"354\"]<img class=\"size-full wp-image-1362\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25143121\/5-3-5.jpeg\" alt=\"A region D is shown in polar coordinates with edges given by theta = alpha, theta = beta, r = h2(theta), and r = h1(theta).\" width=\"354\" height=\"333\" \/> Figure 5.\u00a0A general polar region between [latex]\\alpha &lt; 0&lt;\\beta[\/latex] and\u00a0[latex]h_{1}(\\theta) &lt; r&lt;h_{2}(\\theta)[\/latex].[\/caption]\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">theorem: double integrals over a general polar regions<\/h3>\r\n\r\n<hr \/>\r\n\r\nIf [latex]{f}{(r,{\\theta})}[\/latex] is continuous on a general polar region [latex]D[\/latex] as described above, then\r\n<p style=\"text-align: center;\">[latex]\\large{\\underset{R}{\\displaystyle\\iint}f(r,\\theta)r \\ dr \\ d\\theta=\\displaystyle\\int_{\\theta=\\alpha}^{\\theta=\\beta}\\displaystyle\\int_{r=h_1(\\theta)}^{r=h_2(\\theta)}f(r,\\theta)r \\ dr \\ d\\theta}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: evaluating a double integral over a general polar region<\/h3>\r\nEvaluate the integral [latex]\\underset{D}{\\displaystyle\\iint}{r^2}{\\sin\\theta}{r} \\ {dr} \\ {d{\\theta}}[\/latex] where [latex]D[\/latex] is the region bounded by the polar axis and the upper half of the cardioid [latex]{r} = {1} + {\\cos\\theta}[\/latex].\r\n\r\n[reveal-answer q=\"425517802\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"425517802\"]\r\n\r\nWe can describe the region [latex]D[\/latex] as [latex]\\{(r,\\theta)\\mid0\\leq\\theta\\leq\\pi,0\\leq{r}\\leq1+\\cos\\theta\\}[\/latex] as shown in the following figure.\r\n\r\n[caption id=\"attachment_1363\" align=\"aligncenter\" width=\"425\"]<img class=\"size-full wp-image-1363\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25143147\/5-3-6.jpeg\" alt=\"A region D is given as the top half of a cardioid with equation r = 1 + cos theta.\" width=\"425\" height=\"360\" \/> Figure 6.\u00a0The region [latex]D[\/latex] is the top half of a cardioid.[\/caption]Hence, we have<center>\r\n[latex]\r\n\\begin{array}{ccc}\r\n\\underset{D}{\\displaystyle\\iint}r^2\\sin\\theta{r}dr \\ d\\theta \\hfill &amp;= \\hfill \\displaystyle\\int_{\\theta=0}^{\\theta=\\pi}\\displaystyle\\int_{r=0}^{r=1+\\cos\\theta}(r^2\\sin\\theta)r dr d\\theta \\\\\r\n\\hfill &amp;= \\hfill \\frac{1}{4}\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi}[r^4]_{r=0}^{r=1+\\cos\\theta}\\sin\\theta d\\theta \\\\\r\n\\hfill &amp;= \\hfill \\frac{1}{4}\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi}(1+\\cos\\theta)^4\\sin\\theta d\\theta \\\\\r\n\\hfill &amp;= \\hfill -\\frac{1}{4}[\\frac{(1+\\cos\\theta)^5}{5}]_0^{\\pi}=\\frac{8}{5}.\r\n\\end{array}\r\n[\/latex]<\/center>[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nEvaluate the integral\u00a0[latex]\\underset{D}{\\displaystyle\\iint}r^2\\sin^2{2}\\theta{r} \\ dr \\ d\\theta[\/latex] where\u00a0[latex]{D} = {\\left \\{{(r,{\\theta})}{\\mid}{-{\\frac{\\pi}{4}}} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {\\frac{\\pi}{4}} \\ {\\leq} \\ {r} \\ {\\leq} \\ {{2}{\\sqrt{\\cos2{\\theta}}}} \\right \\}}[\/latex].\r\n\r\n[reveal-answer q=\"098544213\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"098544213\"]\r\n\r\n[latex]\\pi\/4[\/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=8197103&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=BXLTeH6Mbsg&amp;video_target=tpm-plugin-1yq4ji0v-BXLTeH6Mbsg\" 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\/CP5.19_transcript.html\">transcript for \u201cCP 5.19\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 the format of a double integral over a polar rectangular region.<\/span><\/li>\n<li><span class=\"os-abstract-content\">Evaluate a double integral in polar coordinates by using an iterated integral.<\/span><\/li>\n<li><span class=\"os-abstract-content\">Recognize the format of a double integral over a general polar region.<\/span><\/li>\n<\/ul>\n<\/div>\n<h2 data-type=\"title\">Polar Rectangular Regions of Integration<\/h2>\n<p id=\"fs-id1167793292241\">When we defined the double integral for a continuous function in rectangular coordinates\u2014say, [latex]g[\/latex] over a region [latex]R[\/latex] in the [latex]xy[\/latex]-plane\u2014we divided [latex]R[\/latex] into subrectangles with sides parallel to the coordinate axes. These sides have either constant [latex]x[\/latex]-values and\/or constant [latex]y[\/latex]-values. In polar coordinates, the shape we work with is a\u00a0<span id=\"82852cef-a933-495b-9be8-6b3aa0676026_term217\" data-type=\"term\">polar rectangle<\/span>, whose sides have constant [latex]r[\/latex]-values and\/or constant [latex]\\theta[\/latex]-values. This means we can describe a polar rectangle as in\u00a0Figure 1(a), with [latex]{R} = {\\left \\{{(r,{\\theta})}{\\mid}{a} \\ {\\leq} \\ {r} \\ {\\leq} \\ {b,{\\alpha}} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {\\beta} \\right \\}}[\/latex].<\/p>\n<p id=\"fs-id1167793371538\">In this section, we are looking to integrate over polar rectangles. Consider a function [latex]{f}{(r,{\\theta})}[\/latex] over a polar rectangle [latex]R[\/latex]. We divide the interval [latex][a,b][\/latex] into [latex]m[\/latex] subintervals [latex]{[{r_{i-1}},{r_i}]}[\/latex] of length [latex]{\\Delta}{r} = {(b-a)}{\/m}[\/latex] and divide the interval [latex]{[{\\alpha}, {\\beta}]}[\/latex] into [latex]n[\/latex] subintervals [latex]{[{{\\theta}_{j-1}},{{\\theta}_{j}}]}[\/latex] of width [latex]{\\Delta}{\\theta} = {({\\beta}-{\\alpha})}{\/n}[\/latex]. This means that the circles [latex]{r} = {r_{i}}[\/latex] and rays [latex]{\\theta} = {{\\theta}_{j}}[\/latex] for [latex]{1} \\ {\\leq} \\ {i} \\ {\\leq} \\ {m}[\/latex] and [latex]{1} \\ {\\leq} \\ {j} \\ {\\leq} \\ {n}[\/latex] divide the polar rectangle [latex]R[\/latex] into smaller polar subrectangles [latex]{R}_{ij}[\/latex] (Figure 1(b)).<\/p>\n<div id=\"attachment_1357\" style=\"width: 980px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1357\" class=\"size-full wp-image-1357\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25142814\/5-3-1.jpeg\" alt=\"This figure consists of three figures labeled a, b, and c. In figure a, a sector of an annulus is shown in the polar coordinate plane with radii a and b and angles alpha and beta from the theta = 0 axis. In figure b, this sector of an annulus is cut up into subsectors in a manner similar to the way in which previous spaces were cut up into subrectangles. In figure c, one of these subsectors is shown with angle Delta theta, distance between inner and outer radii Delta r, and area Delta A = r* sub theta Delta r Delta theta, where the center point is given as (r* sub i j, theta* sub i j).\" width=\"970\" height=\"339\" \/><\/p>\n<p id=\"caption-attachment-1357\" class=\"wp-caption-text\">Figure 1.\u00a0(a) A polar rectangle\u00a0[latex]R[\/latex]\u00a0(b) divided into subrectangles [latex]R_{ij}[\/latex] (c) Close-up of a subrectangle.<\/p>\n<\/div>\n<p id=\"fs-id1167793940970\">As before, we need to find the area [latex]{\\Delta}{A}[\/latex] of the polar subrectangle [latex]{R}_{ij}[\/latex] and the \u201cpolar\u201d volume of the thin box above [latex]{R}_{ij}[\/latex]. Recall that, in a circle of radius [latex]r[\/latex], the length [latex]s[\/latex] of an arc subtended by a central angle of [latex]{\\theta}[\/latex] radians is [latex]{s} = {r}{\\theta}[\/latex]. Notice that the polar rectangle [latex]{R}_{ij}[\/latex] looks a lot like a trapezoid with parallel sides [latex]{r}_{i-1}{\\Delta}{\\theta}[\/latex] and [latex]{r}_{i}{\\Delta}{\\theta}[\/latex]and with a width [latex]{\\Delta}{r}[\/latex]. Hence the area of the polar subrectangle [latex]{R}_{ij}[\/latex] is<\/p>\n<p style=\"text-align: center;\">[latex]\\Large{\\Delta}{A} = {\\dfrac{1}{2}}{\\Delta}{r}{({r_{i-1}}{\\Delta}{\\theta}+{r_1}{\\Delta}{\\theta})}[\/latex].<\/p>\n<p>Simplifying and letting [latex]{{r}^{*}_{ij}} = {\\dfrac{1}{2}}{({r_{i-1}}+{r_i})}[\/latex], we have [latex]{\\Delta}{A} = {{r}^{*}_{ij}}{\\Delta}{r}{\\Delta}{\\theta}[\/latex]. Therefore, the polar volume of the thin box above [latex]{R}_{ij}[\/latex]\u00a0(Figure 2) is<\/p>\n<p style=\"text-align: center;\">[latex]\\Large{f}{({{r}^{*}_{ij}},{{\\theta}^{*}_{ij}})}{\\Delta}{A} = {f}{({{r}^{*}_{ij}},{{\\theta}^{*}_{ij}})}{{r}^{*}_{ij}}{\\Delta}{r}{\\Delta}{\\theta}[\/latex].<\/p>\n<div id=\"attachment_1359\" style=\"width: 283px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1359\" class=\"size-full wp-image-1359\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25142903\/5-3-2.jpeg\" alt=\"In x y z space, there is a surface f (r, theta). On the x y plane, a series of subsectors of annuli are drawn as in the previous figure with radius between annuli Delta r and angle between subsectors Delta theta. A subsector from the surface f(r, theta) is projected down onto one of these subsectors. This subsector has center point marked (r* sub i j, theta* sub i j).\" width=\"273\" height=\"307\" \/><\/p>\n<p id=\"caption-attachment-1359\" class=\"wp-caption-text\">Figure 2.\u00a0Finding the volume of the thin box above polar rectangle\u00a0[latex]\\small{R_{ij}}[\/latex]<\/p>\n<\/div>\n<p id=\"fs-id1167794096760\">Using the same idea for all the subrectangles and summing the volumes of the rectangular boxes, we obtain a double Riemann sum as<\/p>\n<p style=\"text-align: center;\">[latex]\\Large{\\displaystyle\\sum_{i=1}^{m}\\displaystyle\\sum_{j=1}^{n}}{f}{({{r}^{*}_{ij}},{{\\theta}^{*}_{ij}})}{{r}^{*}_{ij}}{\\Delta}{r}{\\Delta}{\\theta}[\/latex].<\/p>\n<p>As we have seen before, we obtain a better approximation to the polar volume of the solid above the region [latex]R[\/latex] when we let [latex]m[\/latex] and [latex]n[\/latex] become larger. Hence, we define the polar volume as the limit of the double Riemann sum,<\/p>\n<p style=\"text-align: center;\">[latex]\\Large{V} = {\\displaystyle\\lim_{m,n \\rightarrow \\infty}} \\ {\\displaystyle\\sum_{i=1}^{m}\\displaystyle\\sum_{j=1}^{n}}{f}{({{r}^{*}_{ij}},{{\\theta}^{*}_{ij}})}{{r}^{*}_{ij}}{\\Delta}{r}{\\Delta}{\\theta}[\/latex].<\/p>\n<p>This becomes the expression for the double integral.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">definition<\/h3>\n<hr \/>\n<p>The double integral of the function [latex]{f}{(r,{\\theta})}[\/latex] over the polar rectangular region [latex]R[\/latex] in the [latex]{r}{\\theta}[\/latex]-plane is defined as<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\underset{R}{\\displaystyle\\iint}f(r,\\theta)dA = \\displaystyle\\lim_{m,n \\rightarrow \\infty}\\displaystyle\\sum_{i=1}^{m}\\displaystyle\\sum_{j=1}^{n}f(r^*_{ij},\\theta^*_{ij})\\Delta{A} = \\displaystyle\\lim_{m,n \\rightarrow \\infty}\\displaystyle\\sum_{i=1}^{m}\\displaystyle\\sum_{j=1}^{n}f(r^{*}_{ij},\\theta^{*}_{ij})r^{*}_{ij}\\Delta{r}\\Delta\\theta}[\/latex].<\/p>\n<\/div>\n<p id=\"fs-id1167794293302\">Again, just as in\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/double-integrals-over-rectangular-regions\/\">Double Integrals over Rectangular Regions<\/a>, the double integral over a polar rectangular region can be expressed as an iterated integral in polar coordinates. Hence,<\/p>\n<p style=\"text-align: center;\">[latex]\\Large{\\underset{R}{\\displaystyle\\iint}f(r,\\theta)dA = \\underset{R}{\\displaystyle\\iint}f(r,\\theta)r \\ dr \\ d\\theta = \\displaystyle\\int_{\\theta=\\alpha}^{\\theta=\\beta}\\displaystyle\\int_{r=a}^{r=b}f(r,\\theta)r \\ dr \\ d\\theta}[\/latex].<\/p>\n<p>Notice that the expression for [latex]dA[\/latex] is replaced by [latex]{r} \\ {dr} \\ {d{\\theta}}[\/latex] when working in polar coordinates. Another way to look at the polar double integral is to change the double integral in rectangular coordinates by substitution. When the function [latex]f[\/latex] is given in terms of [latex]x[\/latex] and [latex]y[\/latex], using [latex]x=r\\cos\\theta,y = r\\sin\\theta[\/latex] and [latex]{dA} = {{r} \\ {dr} \\ {d{\\theta}}}[\/latex] changes it to<\/p>\n<p style=\"text-align: center;\">[latex]\\Large{\\underset{R}{\\displaystyle\\iint}f(x,y)dA=\\underset{R}{\\displaystyle\\iint}f(r\\cos\\theta,r\\sin\\theta)r \\ dr \\ d\\theta}[\/latex].<\/p>\n<p>Note that all the properties listed in\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/double-integrals-over-rectangular-regions\/\" data-page-slug=\"5-1-double-integrals-over-rectangular-regions\" data-page-uuid=\"d9da6d44-3ba3-4a90-adbc-dec26572e480\" data-page-fragment=\"page_d9da6d44-3ba3-4a90-adbc-dec26572e480\">Double Integrals over Rectangular Regions<\/a>\u00a0for the double integral in rectangular coordinates hold true for the double integral in polar coordinates as well, so we can use them without hesitation.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: sketching a polar rectangular region<\/h3>\n<p>Sketch the polar rectangular region\u00a0[latex]{R} = {\\left \\{{(r,{\\theta})}{\\mid}{1} \\ {\\leq} \\ {r} \\ {\\leq} \\ {3,0} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {\\pi} \\right \\}}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q137726982\">Show Solution<\/span><\/p>\n<div id=\"q137726982\" class=\"hidden-answer\" style=\"display: none\">\n<p>As we can see from\u00a0Figure 3, [latex]r=1[\/latex] and [latex]r=3[\/latex] are circles of radius [latex]1[\/latex] and [latex]3[\/latex] and [latex]{0} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {\\pi}[\/latex] covers the entire top half of the plane. Hence the region [latex]R[\/latex] looks like a semicircular band.<\/p>\n<div id=\"attachment_1360\" style=\"width: 277px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1360\" class=\"size-full wp-image-1360\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25142956\/5-3-3.jpeg\" alt=\"Half an annulus R is drawn with inner radius 1 and outer radius 3. That is, the inner semicircle is given by x squared + y squared = 1, whereas the outer semicircle is given by x squared + y squared = 9.\" width=\"267\" height=\"234\" \/><\/p>\n<p id=\"caption-attachment-1360\" class=\"wp-caption-text\">Figure 3.\u00a0The polar region [latex]R[\/latex] lies between two semicircles.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Now that we have sketched a polar rectangular region, let us demonstrate how to evaluate a double integral over this region by using polar coordinates.<\/p>\n<div class=\"textbox exercises\">\n<h3>example: evaluating a double integral over a polar rectangular region<\/h3>\n<p>Evaluate the integral\u00a0[latex]\\underset{R}{\\displaystyle\\iint}{3x}{dA}[\/latex] over the region\u00a0[latex]{R} = {\\left \\{{(r,{\\theta})}{\\mid}{1} \\ {\\leq} \\ {r} \\ {\\leq} \\ {2,0} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {\\pi} \\right \\}}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q862144637\">Show Solution<\/span><\/p>\n<div id=\"q862144637\" class=\"hidden-answer\" style=\"display: none\">\n<p>First we sketch a figure similar to\u00a0Figure 3\u00a0but with outer radius 2. From the figure we can see that we have<\/p>\n<p>[latex]\\begin{align}    \\underset{R}{\\displaystyle\\iint}3x \\ dA &= \\displaystyle\\int_{\\theta=0}^{\\theta=\\pi}\\displaystyle\\int_{r=1}^{r=2}3r\\cos\\theta{r} \\ dr \\ d\\theta &\\quad &\\text{Use an iterated integral with correct limits of integration.} \\\\    &=\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi}\\cos\\theta\\left[r^3\\bigg|_{r=1}^{r=2}\\right]d\\theta &\\quad &\\text{Integrate first with respect to }r. \\\\    &=\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi}7\\cos\\theta{d}\\theta=7\\sin\\theta\\bigg|_{\\theta=0}^{\\theta=\\pi}=0.    \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Sketch the region\u00a0[latex]{R} = {\\left \\{{(r,{\\theta})}{\\mid}{1} \\ {\\leq} \\ {r} \\ {\\leq} \\ {2,-{\\frac{\\pi}{2}}} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {\\frac{\\pi}{2}} \\right \\}}[\/latex], and evaluate\u00a0[latex]\\underset{R}{\\displaystyle\\iint}{x} \\ {dA}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q733260947\">Show Solution<\/span><\/p>\n<div id=\"q733260947\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{14}3[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Evaluating a double integral by converting from rectangular coordinates<\/h3>\n<p>Evaluate the integral [latex]\\underset{R}{\\displaystyle\\iint}{(1-{x^2}-{y^2})}{dA}[\/latex] where [latex]R[\/latex] is the unit circle on the [latex]xy[\/latex]-plane.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q100254682\">Show Solution<\/span><\/p>\n<div id=\"q100254682\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794218699\">The region [latex]R[\/latex] is a unit circle, so we can describe it as [latex]{R} = {\\left \\{{(r,{\\theta})}{\\mid}{0} \\ {\\leq} \\ {r} \\ {\\leq} \\ {1,0} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {2{\\pi}} \\right \\}}[\/latex].<\/p>\n<p id=\"fs-id1167793543369\">Using the conversion [latex]x=r\\cos\\theta,y=r\\sin\\theta[\/latex], and [latex]{dA} = {{r} \\ {dr} \\ {d{\\theta}}}[\/latex], we have<\/p>\n<p>[latex]\\hspace{4cm}\\begin{align}    \\underset{R}{\\displaystyle\\iint}(1-x^2-y^2)dA&=\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^1(1-r^2)r \\ dr \\ d\\theta =\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^1(r - r^3)dr \\ d\\theta \\\\    &=\\displaystyle\\int_0^{2\\pi}\\left[\\frac{r^2}2-\\frac{r^4}4\\right]_0^1{d}\\theta=\\displaystyle\\int_0^{2\\pi}\\frac14d\\theta=\\frac{\\pi}2.    \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Evaluating a double integral by converting from rectangular coordinates<\/h3>\n<p>Evaluate the integral\u00a0[latex]\\underset{R}{\\displaystyle\\iint}{(x+y)}{dA}[\/latex] where\u00a0[latex]{R} = {\\left \\{{(x,y)}{\\mid}{1} \\ {\\leq} \\ {{x^2}+{y^2}} \\ {\\leq} \\ {4,x} \\ {\\leq} \\ {0} \\right \\}}[\/latex].<\/p>\n<p>&nbsp;<\/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>We can see that [latex]R[\/latex] is an annular region that can be converted to polar coordinates and described as [latex]{R} = {\\left \\{{(r,{\\theta})}{\\mid}{1} \\ {\\leq} \\ {r} \\ {\\leq} \\ {2,{\\frac{\\pi}{2}}} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {\\frac{3{\\pi}}{2}} \\right \\}}[\/latex] (see the following graph).<\/p>\n<div id=\"attachment_1361\" style=\"width: 303px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1361\" class=\"size-full wp-image-1361\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25143041\/5-3-4.jpeg\" alt=\"Two semicircles are drawn in the second and third quadrants, with equations x squared + y squared = 1 and x squared + y squared = 2.\" width=\"293\" height=\"272\" \/><\/p>\n<p id=\"caption-attachment-1361\" class=\"wp-caption-text\">Figure 4.\u00a0The annular region of integration\u00a0[latex]R[\/latex].<\/p>\n<\/div>\n<p>Hence, using the conversion[latex]x =r\\cos\\theta,y =r\\sin\\theta[\/latex], and [latex]{dA} = {{r} \\ {dr} \\ {d{\\theta}}}[\/latex], we have<br \/>\n[latex]\\hspace{5cm}\\begin{align}\\underset{R}{\\displaystyle\\iint}(x+y)dA&=\\displaystyle\\int_{\\theta=\\pi\/2}^{\\theta=3\\pi\/2}\\displaystyle\\int_{r=1}^{r=2}(r\\cos\\theta+r\\sin\\theta)r \\ dr \\ d\\theta \\\\&=\\left(\\displaystyle\\int_{r=1}^{r=2}r^2dr\\right)\\left(\\displaystyle\\int_{\\pi\/2}^{3\\pi\/2}(\\cos\\theta+\\sin\\theta)d\\theta\\right) \\\\&=\\left[\\frac{r^3}3\\right]_1^2[\\sin\\theta-\\cos\\theta]\\bigg|_{\\pi\/2}^{3\\pi\/2} \\\\&=-\\frac{14}3.\\end{align}[\/latex]\n<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Evaluate the integral [latex]\\underset{R}{\\displaystyle\\iint}{(4-{x^2}-{y^2})}{dA}[\/latex] where [latex]R[\/latex] is the circle of radius 2 on the [latex]xy[\/latex]-plane.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q784312663\">Show Solution<\/span><\/p>\n<div id=\"q784312663\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]8\\pi[\/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=8197102&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=UKNpIyyDQCQ&amp;video_target=tpm-plugin-0qwtbnvq-UKNpIyyDQCQ\" 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\/CP5.18_transcript.html\">transcript for \u201cCP 5.18<span style=\"font-size: 1em;\">\u201d here (opens in new window).<\/span><\/a><\/div>\n<h2 data-type=\"title\">General Polar Regions of Integration<\/h2>\n<p id=\"fs-id1167793928915\">To evaluate the double integral of a continuous function by iterated integrals over general polar regions, we consider two types of regions, analogous to Type I and Type II as discussed for rectangular coordinates in\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/introduction-to-double-integrals-over-general-regions\/\" data-page-slug=\"5-2-double-integrals-over-general-regions\" data-page-uuid=\"f783ca08-ca2f-4ebb-a559-d1b30c2f9b8c\" data-page-fragment=\"page_f783ca08-ca2f-4ebb-a559-d1b30c2f9b8c\">Double Integrals over General Regions<\/a>. It is more common to write polar equations as [latex]{r} = {f}{(\\theta)}[\/latex] than [latex]{\\theta} = {f}{(r)}[\/latex], so we describe a general polar region as [latex]{R} = {\\left \\{{(r,{\\theta})}{\\mid}{\\alpha} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {{\\beta},{h_1}}{(\\theta)} \\ {\\leq} \\ {r} \\ {\\leq} \\ {{h_2}{(\\theta)}} \\right \\}}[\/latex] (see the following figure).<\/p>\n<div id=\"attachment_1362\" style=\"width: 364px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1362\" class=\"size-full wp-image-1362\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25143121\/5-3-5.jpeg\" alt=\"A region D is shown in polar coordinates with edges given by theta = alpha, theta = beta, r = h2(theta), and r = h1(theta).\" width=\"354\" height=\"333\" \/><\/p>\n<p id=\"caption-attachment-1362\" class=\"wp-caption-text\">Figure 5.\u00a0A general polar region between [latex]\\alpha &lt; 0&lt;\\beta[\/latex] and\u00a0[latex]h_{1}(\\theta) &lt; r&lt;h_{2}(\\theta)[\/latex].<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">theorem: double integrals over a general polar regions<\/h3>\n<hr \/>\n<p>If [latex]{f}{(r,{\\theta})}[\/latex] is continuous on a general polar region [latex]D[\/latex] as described above, then<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\underset{R}{\\displaystyle\\iint}f(r,\\theta)r \\ dr \\ d\\theta=\\displaystyle\\int_{\\theta=\\alpha}^{\\theta=\\beta}\\displaystyle\\int_{r=h_1(\\theta)}^{r=h_2(\\theta)}f(r,\\theta)r \\ dr \\ d\\theta}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: evaluating a double integral over a general polar region<\/h3>\n<p>Evaluate the integral [latex]\\underset{D}{\\displaystyle\\iint}{r^2}{\\sin\\theta}{r} \\ {dr} \\ {d{\\theta}}[\/latex] where [latex]D[\/latex] is the region bounded by the polar axis and the upper half of the cardioid [latex]{r} = {1} + {\\cos\\theta}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q425517802\">Show Solution<\/span><\/p>\n<div id=\"q425517802\" class=\"hidden-answer\" style=\"display: none\">\n<p>We can describe the region [latex]D[\/latex] as [latex]\\{(r,\\theta)\\mid0\\leq\\theta\\leq\\pi,0\\leq{r}\\leq1+\\cos\\theta\\}[\/latex] as shown in the following figure.<\/p>\n<div id=\"attachment_1363\" style=\"width: 435px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1363\" class=\"size-full wp-image-1363\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25143147\/5-3-6.jpeg\" alt=\"A region D is given as the top half of a cardioid with equation r = 1 + cos theta.\" width=\"425\" height=\"360\" \/><\/p>\n<p id=\"caption-attachment-1363\" class=\"wp-caption-text\">Figure 6.\u00a0The region [latex]D[\/latex] is the top half of a cardioid.<\/p>\n<\/div>\n<p>Hence, we have<\/p>\n<div style=\"text-align: center;\">\n[latex]\\begin{array}{ccc}  \\underset{D}{\\displaystyle\\iint}r^2\\sin\\theta{r}dr \\ d\\theta \\hfill &= \\hfill \\displaystyle\\int_{\\theta=0}^{\\theta=\\pi}\\displaystyle\\int_{r=0}^{r=1+\\cos\\theta}(r^2\\sin\\theta)r dr d\\theta \\\\  \\hfill &= \\hfill \\frac{1}{4}\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi}[r^4]_{r=0}^{r=1+\\cos\\theta}\\sin\\theta d\\theta \\\\  \\hfill &= \\hfill \\frac{1}{4}\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi}(1+\\cos\\theta)^4\\sin\\theta d\\theta \\\\  \\hfill &= \\hfill -\\frac{1}{4}[\\frac{(1+\\cos\\theta)^5}{5}]_0^{\\pi}=\\frac{8}{5}.  \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Evaluate the integral\u00a0[latex]\\underset{D}{\\displaystyle\\iint}r^2\\sin^2{2}\\theta{r} \\ dr \\ d\\theta[\/latex] where\u00a0[latex]{D} = {\\left \\{{(r,{\\theta})}{\\mid}{-{\\frac{\\pi}{4}}} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {\\frac{\\pi}{4}} \\ {\\leq} \\ {r} \\ {\\leq} \\ {{2}{\\sqrt{\\cos2{\\theta}}}} \\right \\}}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q098544213\">Show Solution<\/span><\/p>\n<div id=\"q098544213\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\pi\/4[\/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=8197103&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=BXLTeH6Mbsg&amp;video_target=tpm-plugin-1yq4ji0v-BXLTeH6Mbsg\" 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\/CP5.19_transcript.html\">transcript for \u201cCP 5.19\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-3981\">\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 5.18. <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 5.19. <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":13,"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 5.18\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"CP 5.19\",\"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-3981","chapter","type-chapter","status-publish","hentry"],"part":23,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3981","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":22,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3981\/revisions"}],"predecessor-version":[{"id":6453,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3981\/revisions\/6453"}],"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\/3981\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=3981"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=3981"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=3981"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=3981"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}