{"id":689,"date":"2025-10-09T15:30:35","date_gmt":"2025-10-09T15:30:35","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=689"},"modified":"2025-10-29T16:19:47","modified_gmt":"2025-10-29T16:19:47","slug":"double-integrals-in-polar-coordinates-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/double-integrals-in-polar-coordinates-learn-it-1\/","title":{"raw":"Double Integrals in Polar Coordinates: Learn It 1","rendered":"Double Integrals in Polar Coordinates: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Use polar coordinates to evaluate double integrals<\/li>\r\n \t<li>Use double integrals in polar coordinates to calculate areas and volumes<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2 data-type=\"title\">Polar Rectangular Regions of Integration<\/h2>\r\nDouble integrals are sometimes much easier to evaluate if we change rectangular coordinates to polar coordinates. However, before we describe how to make this change, we need to establish the concept of a double integral in a polar rectangular region.\r\n<p class=\"whitespace-normal break-words\">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 or constant [latex]y[\/latex]-values.<\/p>\r\n<p class=\"whitespace-normal break-words\">In polar coordinates, the shape we work with is a <strong>polar rectangle<\/strong>, whose sides have constant [latex]r[\/latex]-values or constant [latex]\\theta[\/latex]-values. We can describe a polar rectangle as shown in Figure 1(a):<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]R = {(r,\\theta)\\mid a \\leq r \\leq b, \\alpha \\leq \\theta \\leq \\beta }[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">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 = \\frac{b-a}{m}[\/latex]. Similarly, we divide the interval [latex][\\alpha, \\beta][\/latex] into [latex]n[\/latex] subintervals [latex][\\theta_{j-1},\\theta_{j}][\/latex] of width [latex]\\Delta\\theta = \\frac{\\beta-\\alpha}{n}[\/latex]. 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 class=\"whitespace-normal break-words\">As before, we need to find the area [latex]\\Delta A[\/latex] of the polar subrectangle [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]. The polar rectangle [latex]R_{ij}[\/latex] looks like a trapezoid with parallel sides [latex]r_{i-1}\\Delta\\theta[\/latex] and [latex]r_{i}\\Delta\\theta[\/latex], and 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]\r\n\\Delta A = \\frac{1}{2} \\Delta r \\left( r_{i-1} \\Delta \\theta + r_i \\Delta \\theta \\right).\r\n[\/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] (Figure 2) is:\r\n<p style=\"text-align: center;\">[latex]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:<\/p>\r\n<p style=\"text-align: center;\">[latex]\r\n\\displaystyle\\sum_{i=1}^{m} \\displaystyle\\sum_{j=1}^{n} f(r_{ij}^*, \\theta_{ij}^*) \\, r_{ij}^* \\, \\Delta r \\, \\Delta \\theta.\r\n[\/latex]<\/p>\r\nWe 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]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\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>double integral in polar coordinates<\/h3>\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<\/section>\r\n<p class=\"whitespace-normal break-words\">Just as with rectangular regions, the double integral over a polar rectangular region can be expressed as an iterated integral in polar coordinates:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{R}{\\displaystyle\\iint}f(r,\\theta)dA = \\underset{R}{\\displaystyle\\iint}f(r,\\theta)rdrd\\theta = \\displaystyle\\int_{\\theta=\\alpha}^{\\theta=\\beta}\\displaystyle\\int_{r=a}^{r=b}f(r,\\theta)rdrd\\theta[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Notice that the expression for [latex]dA[\/latex] is replaced by [latex]rdrd\\theta[\/latex] when working in polar coordinates. The extra [latex]r[\/latex] factor is crucial and comes from the geometry of polar rectangles.<\/p>\r\n<p class=\"whitespace-normal break-words\">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[\/latex], [latex]y = r\\sin\\theta[\/latex], and [latex]dA = rdrd\\theta[\/latex] changes it to<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{R}{\\displaystyle\\iint}f(x,y)dA=\\underset{R}{\\displaystyle\\iint}f(r\\cos\\theta,r\\sin\\theta)rdrd\\theta[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Note that all the properties we listed earlier for the double integral in rectangular coordinates hold true for the double integral in polar coordinates as well.<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">Sketch the polar rectangular region\u00a0[latex]{R} = {\\left \\{{(r,{\\theta})}{\\mid}{1} \\ {\\leq} \\ {r} \\ {\\leq} \\ {3,0} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {\\pi} \\right \\}}[\/latex].[reveal-answer q=\"137726982\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"137726982\"]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.[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]<\/section>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.\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">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].[reveal-answer q=\"862144637\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"862144637\"]First we sketch a figure similar to\u00a0Figure 3\u00a0but with outer radius 2. From the figure we can see that we have[latex]\\begin{align}\\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.} \\\\&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. \\\\&amp;=\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi}7\\cos\\theta{d}\\theta=7\\sin\\theta\\bigg|_{\\theta=0}^{\\theta=\\pi}=0.\\end{align}[\/latex][\/hidden-answer]<\/section><section class=\"textbox example\" aria-label=\"Example\">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.[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<\/section><section class=\"textbox example\" aria-label=\"Example\">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].[reveal-answer q=\"984622178\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"984622178\"]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).[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]<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Use polar coordinates to evaluate double integrals<\/li>\n<li>Use double integrals in polar coordinates to calculate areas and volumes<\/li>\n<\/ul>\n<\/section>\n<h2 data-type=\"title\">Polar Rectangular Regions of Integration<\/h2>\n<p>Double integrals are sometimes much easier to evaluate if we change rectangular coordinates to polar coordinates. However, before we describe how to make this change, we need to establish the concept of a double integral in a polar rectangular region.<\/p>\n<p class=\"whitespace-normal break-words\">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 or constant [latex]y[\/latex]-values.<\/p>\n<p class=\"whitespace-normal break-words\">In polar coordinates, the shape we work with is a <strong>polar rectangle<\/strong>, whose sides have constant [latex]r[\/latex]-values or constant [latex]\\theta[\/latex]-values. We can describe a polar rectangle as shown in Figure 1(a):<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]R = {(r,\\theta)\\mid a \\leq r \\leq b, \\alpha \\leq \\theta \\leq \\beta }[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">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 = \\frac{b-a}{m}[\/latex]. Similarly, we divide the interval [latex][\\alpha, \\beta][\/latex] into [latex]n[\/latex] subintervals [latex][\\theta_{j-1},\\theta_{j}][\/latex] of width [latex]\\Delta\\theta = \\frac{\\beta-\\alpha}{n}[\/latex]. 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<figure id=\"attachment_1357\" aria-describedby=\"caption-attachment-1357\" style=\"width: 970px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" 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\" \/><figcaption 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.<\/figcaption><\/figure>\n<p class=\"whitespace-normal break-words\">As before, we need to find the area [latex]\\Delta A[\/latex] of the polar subrectangle [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]. The polar rectangle [latex]R_{ij}[\/latex] looks like a trapezoid with parallel sides [latex]r_{i-1}\\Delta\\theta[\/latex] and [latex]r_{i}\\Delta\\theta[\/latex], and 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]\\Delta A = \\frac{1}{2} \\Delta r \\left( r_{i-1} \\Delta \\theta + r_i \\Delta \\theta \\right).[\/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] (Figure 2) is:<\/p>\n<p style=\"text-align: center;\">[latex]f{({{r}^{*}_{ij}},{{\\theta}^{*}_{ij}})}{\\Delta}{A} = {f}{({{r}^{*}_{ij}},{{\\theta}^{*}_{ij}})}{{r}^{*}_{ij}}{\\Delta}{r}{\\Delta}{\\theta}[\/latex].<\/p>\n<figure id=\"attachment_1359\" aria-describedby=\"caption-attachment-1359\" style=\"width: 273px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" 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\" \/><figcaption 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]<\/figcaption><\/figure>\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:<\/p>\n<p style=\"text-align: center;\">[latex]\\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>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]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<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>double integral in polar coordinates<\/h3>\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<\/section>\n<p class=\"whitespace-normal break-words\">Just as with rectangular regions, the double integral over a polar rectangular region can be expressed as an iterated integral in polar coordinates:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{R}{\\displaystyle\\iint}f(r,\\theta)dA = \\underset{R}{\\displaystyle\\iint}f(r,\\theta)rdrd\\theta = \\displaystyle\\int_{\\theta=\\alpha}^{\\theta=\\beta}\\displaystyle\\int_{r=a}^{r=b}f(r,\\theta)rdrd\\theta[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Notice that the expression for [latex]dA[\/latex] is replaced by [latex]rdrd\\theta[\/latex] when working in polar coordinates. The extra [latex]r[\/latex] factor is crucial and comes from the geometry of polar rectangles.<\/p>\n<p class=\"whitespace-normal break-words\">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[\/latex], [latex]y = r\\sin\\theta[\/latex], and [latex]dA = rdrd\\theta[\/latex] changes it to<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{R}{\\displaystyle\\iint}f(x,y)dA=\\underset{R}{\\displaystyle\\iint}f(r\\cos\\theta,r\\sin\\theta)rdrd\\theta[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Note that all the properties we listed earlier for the double integral in rectangular coordinates hold true for the double integral in polar coordinates as well.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q137726982\">Show Solution<\/button><\/p>\n<div id=\"q137726982\" class=\"hidden-answer\" style=\"display: none\">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<figure id=\"attachment_1360\" aria-describedby=\"caption-attachment-1360\" style=\"width: 267px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" 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\" \/><figcaption id=\"caption-attachment-1360\" class=\"wp-caption-text\">Figure 3.\u00a0The polar region [latex]R[\/latex] lies between two semicircles.<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/section>\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<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q862144637\">Show Solution<\/button><\/p>\n<div id=\"q862144637\" class=\"hidden-answer\" style=\"display: none\">First we sketch a figure similar to\u00a0Figure 3\u00a0but with outer radius 2. From the figure we can see that we have[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]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q100254682\">Show Solution<\/button><\/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<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q984622178\">Show Solution<\/button><\/p>\n<div id=\"q984622178\" class=\"hidden-answer\" style=\"display: none\">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<figure id=\"attachment_1361\" aria-describedby=\"caption-attachment-1361\" style=\"width: 293px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" 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\" \/><figcaption id=\"caption-attachment-1361\" class=\"wp-caption-text\">Figure 4.\u00a0The annular region of integration\u00a0[latex]R[\/latex].<\/figcaption><\/figure>\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<\/section>\n","protected":false},"author":15,"menu_order":19,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-689","chapter","type-chapter","status-publish","hentry"],"part":432,"module-header":"learn_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/689","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":13,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/689\/revisions"}],"predecessor-version":[{"id":1874,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/689\/revisions\/1874"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/432"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/689\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=689"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=689"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=689"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=689"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}