{"id":691,"date":"2025-10-09T15:30:30","date_gmt":"2025-10-09T15:30:30","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=691"},"modified":"2025-11-11T18:12:41","modified_gmt":"2025-11-11T18:12:41","slug":"double-integrals-in-polar-coordinates-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/double-integrals-in-polar-coordinates-fresh-take\/","title":{"raw":"Double Integrals in Polar Coordinates: Fresh Take","rendered":"Double Integrals in Polar Coordinates: Fresh Take"},"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\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Some regions and functions are naturally suited to polar coordinates rather than rectangular coordinates. Converting to polar coordinates can transform a difficult integral into a manageable one, especially for circular or radial regions.<\/p>\r\n<p class=\"whitespace-normal break-words\">A polar rectangular region is bounded by constant [latex]r[\/latex]-values and constant [latex]\\theta[\/latex]-values:<\/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\">This describes a sector of an annulus (ring) with inner radius [latex]a[\/latex], outer radius [latex]b[\/latex], and angular span from [latex]\\alpha[\/latex] to [latex]\\beta[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Critical Formula<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">When converting to polar coordinates, the area element changes:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]dA = r\\,dr\\,d\\theta[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">The extra [latex]r[\/latex] factor is essential\u2014it accounts for how polar \"rectangles\" get wider as you move away from the origin. Never forget this [latex]r[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Double Integral in Polar Form<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{R}{\\displaystyle\\iint}f(r,\\theta)dA = \\displaystyle\\int_{\\theta=\\alpha}^{\\theta=\\beta}\\displaystyle\\int_{r=a}^{r=b}f(r,\\theta)r\\,dr\\,d\\theta[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Converting from Rectangular<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">When [latex]f[\/latex] is given in [latex]x[\/latex] and [latex]y[\/latex], use the conversions:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]x = r\\cos\\theta[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]y = r\\sin\\theta[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]dA = r,dr,d\\theta[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">For example, [latex]x^2 + y^2[\/latex] becomes simply [latex]r^2[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Identify if polar is appropriate<\/strong>: Look for circles, sectors, or expressions involving [latex]x^2 + y^2[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Determine bounds<\/strong>: Find the [latex]r[\/latex]-range (radii) and [latex]\\theta[\/latex]-range (angles) that describe your region<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Convert the integrand<\/strong>: Replace [latex]x[\/latex] and [latex]y[\/latex] with [latex]r\\cos\\theta[\/latex] and [latex]r\\sin\\theta[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Don't forget the [latex]r[\/latex]<\/strong>: Include it in [latex]r\\,dr\\,d\\theta[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Integrate<\/strong>: Usually integrate with respect to [latex]r[\/latex] first, then [latex]\\theta[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\">Use polar coordinates when:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">The region is circular, semicircular, or a sector<\/li>\r\n \t<li class=\"whitespace-normal break-words\">The integrand contains [latex]x^2 + y^2[\/latex] (which becomes [latex]r^2[\/latex])<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Boundaries are easier to express as [latex]r = \\text{constant}[\/latex] or [latex]\\theta = \\text{constant}[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">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].[reveal-answer q=\"733260947\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"733260947\"][latex]\\frac{14}3[\/latex][\/hidden-answer]<\/section><section class=\"textbox example\" aria-label=\"Example\">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.[reveal-answer q=\"784312663\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"784312663\"][latex]8\\pi[\/latex][\/hidden-answer]<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">\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\" data-mce-fragment=\"1\"><\/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><\/section>\r\n<h2 data-type=\"title\">General Polar Regions of Integration<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Just as rectangular regions generalize beyond rectangles to curves, polar regions extend beyond simple sectors. For regions with curved polar boundaries, you integrate with variable limits that depend on [latex]\\theta[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>General Polar Regions<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">A general polar region has the form:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]R = {(r,\\theta) \\mid \\alpha \\leq \\theta \\leq \\beta, h_1(\\theta) \\leq r \\leq h_2(\\theta)}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">The angle [latex]\\theta[\/latex] ranges between constants, while the radius [latex]r[\/latex] varies between two functions of [latex]\\theta[\/latex]. Think of it as sweeping an angle from [latex]\\alpha[\/latex] to [latex]\\beta[\/latex], where at each angle the radius extends from inner curve [latex]h_1(\\theta)[\/latex] to outer curve [latex]h_2(\\theta)[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Iterated Integral<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{R}{\\displaystyle\\iint}f(r,\\theta)dA = \\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<p class=\"whitespace-normal break-words\">The outer integral spans the angular range (constants). The inner integral spans the radial range (functions of [latex]\\theta[\/latex]).<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Sketch the region<\/strong>: Draw the polar curves to visualize the boundaries<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Find the angular range<\/strong>: Determine [latex]\\alpha[\/latex] and [latex]\\beta[\/latex] where the region exists<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Identify radial bounds<\/strong>: At each angle [latex]\\theta[\/latex], find where [latex]r[\/latex] starts ([latex]h_1(\\theta)[\/latex]) and ends ([latex]h_2(\\theta)[\/latex])<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Set up the integral<\/strong>: Place angle limits outside, radial limits (as functions) inside<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Include the [latex]r[\/latex]<\/strong>: Always multiply by [latex]r[\/latex] in [latex]r\\,dr\\,d\\theta[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Integrate<\/strong>: Work inside out\u2014first with respect to [latex]r[\/latex], treating [latex]\\theta[\/latex] as constant<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>Key Reminder<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">When radial bounds depend on [latex]\\theta[\/latex], you're describing how far out the region extends at each particular angle. This is the polar equivalent of Type I regions in rectangular coordinates.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">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].[reveal-answer q=\"098544213\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"098544213\"][latex]\\pi\/4[\/latex][\/hidden-answer]<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<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\" data-mce-fragment=\"1\"><\/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><\/section>\r\n<h2>Polar Areas and Volumes<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Polar coordinates excel at computing volumes and areas for circular or radial regions. The same double integral framework applies, but the geometry often makes calculations dramatically simpler than rectangular coordinates.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Volume Under a Surface<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">For a solid under [latex]z = f(r,\\theta)[\/latex] over a polar region [latex]D[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]V = \\underset{D}{\\displaystyle\\iint}f(r,\\theta)r\\,dr\\,d\\theta[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Convert [latex]x^2 + y^2[\/latex] to [latex]r^2[\/latex] in the function, and don't forget the [latex]r[\/latex] in [latex]r,dr,d\\theta[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Area of a Polar Region<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">To find the area of region [latex]D[\/latex], integrate 1:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]A = \\displaystyle\\int_\\alpha^\\beta\\displaystyle\\int_{h_1(\\theta)}^{h_2(\\theta)}r\\,dr\\,d\\theta[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">The [latex]r[\/latex] factor handles the \"spreading out\" of area elements as radius increases.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Strategic Decisions<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Not every problem is easier in polar coordinates. If the region is triangular or the function doesn't simplify with [latex]r^2 = x^2 + y^2[\/latex], stick with rectangular coordinates. Compare the complexity of both setups before committing.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Sketch the region<\/strong>: Identify whether it's naturally circular, radial, or involves [latex]x^2 + y^2[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Find polar bounds<\/strong>: Determine [latex]\\alpha \\leq \\theta \\leq \\beta[\/latex] and [latex]h_1(\\theta) \\leq r \\leq h_2(\\theta)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Convert the function<\/strong>: Replace [latex]x[\/latex] and [latex]y[\/latex] with [latex]r\\cos\\theta[\/latex] and [latex]r\\sin\\theta[\/latex]; simplify [latex]x^2 + y^2[\/latex] to [latex]r^2[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Set up carefully<\/strong>: Include the [latex]r[\/latex] in [latex]r,dr,d\\theta[\/latex]\u2014this is non-negotiable<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Integrate<\/strong>: Usually [latex]r[\/latex] first, then [latex]\\theta[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>Common Applications<\/strong><\/p>\r\n\r\n<ul>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Cones<\/strong>: For [latex]z = h - \\frac{h}{a}\\sqrt{x^2 + y^2}[\/latex] over a circle, convert to [latex]z = h - \\frac{h}{a}r[\/latex]\u2014much simpler.<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Polar roses<\/strong>: For [latex]r = \\cos(n\\theta)[\/latex], use symmetry. Find the area of one petal and multiply by the number of petals.<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Intersecting regions<\/strong>: When a circle and cardioid overlap, split the integral at intersection points, integrating different radial bounds over different angular ranges.<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Improper Integrals<\/strong>: The entire plane becomes [latex]0 \\leq \\theta \\leq 2\\pi[\/latex], [latex]0 \\leq r &lt; \\infty[\/latex], requiring limits as [latex]r \\to \\infty[\/latex].<\/li>\r\n<\/ul>\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Use polar coordinates to find an iterated integral for finding the volume of the solid enclosed by the paraboloids [latex]{z} = {x^2} + {y^2}[\/latex] and\u00a0[latex]{z} = {16} - {x^2} - {y^2}[\/latex].[reveal-answer q=\"512348890\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"512348890\"][latex]V=\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^{2\\sqrt2}(16-2r^2)r \\ dr \\ d\\theta=64\\pi[\/latex] cubic units.[\/hidden-answer]<\/section><section class=\"textbox example\" aria-label=\"Example\">Find the area enclosed inside the cardioid [latex]{r} = {3} - {{3}{\\sin}{\\theta}}[\/latex] and outside the cardioid [latex]{r} = {1} + {{\\sin}{\\theta}}[\/latex].[reveal-answer q=\"579921064\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"579921064\"][latex]A=2\\displaystyle\\int_{-\\pi\/2}^{\\pi\/6}\\displaystyle\\int_{1+\\sin\\theta}^{3-3\\sin\\theta}r \\ dr \\ d\\theta = 8\\pi+9\\sqrt3[\/latex][\/hidden-answer]<\/section><section class=\"textbox example\" aria-label=\"Example\">Evaluate the integral\u00a0[latex]\\underset{R^2}{\\displaystyle\\iint}{e^{{-4}{(x^2+y^2)}}}{dx^{}}{dy}[\/latex].[reveal-answer q=\"967834547\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"967834547\"][latex]\\frac{\\pi}4[\/latex][\/hidden-answer]<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8197104&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=TdDx0cLReoI&amp;video_target=tpm-plugin-e10unxm4-TdDx0cLReoI\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP5.22_transcript.html\">transcript for \u201cCP 5.22\u201d here (opens in new window).<\/a><\/center><\/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<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Some regions and functions are naturally suited to polar coordinates rather than rectangular coordinates. Converting to polar coordinates can transform a difficult integral into a manageable one, especially for circular or radial regions.<\/p>\n<p class=\"whitespace-normal break-words\">A polar rectangular region is bounded by constant [latex]r[\/latex]-values and constant [latex]\\theta[\/latex]-values:<\/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\">This describes a sector of an annulus (ring) with inner radius [latex]a[\/latex], outer radius [latex]b[\/latex], and angular span from [latex]\\alpha[\/latex] to [latex]\\beta[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Critical Formula<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">When converting to polar coordinates, the area element changes:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]dA = r\\,dr\\,d\\theta[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">The extra [latex]r[\/latex] factor is essential\u2014it accounts for how polar &#8220;rectangles&#8221; get wider as you move away from the origin. Never forget this [latex]r[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Double Integral in Polar Form<\/strong><\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{R}{\\displaystyle\\iint}f(r,\\theta)dA = \\displaystyle\\int_{\\theta=\\alpha}^{\\theta=\\beta}\\displaystyle\\int_{r=a}^{r=b}f(r,\\theta)r\\,dr\\,d\\theta[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Converting from Rectangular<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">When [latex]f[\/latex] is given in [latex]x[\/latex] and [latex]y[\/latex], use the conversions:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]x = r\\cos\\theta[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]y = r\\sin\\theta[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]dA = r,dr,d\\theta[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">For example, [latex]x^2 + y^2[\/latex] becomes simply [latex]r^2[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Identify if polar is appropriate<\/strong>: Look for circles, sectors, or expressions involving [latex]x^2 + y^2[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Determine bounds<\/strong>: Find the [latex]r[\/latex]-range (radii) and [latex]\\theta[\/latex]-range (angles) that describe your region<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Convert the integrand<\/strong>: Replace [latex]x[\/latex] and [latex]y[\/latex] with [latex]r\\cos\\theta[\/latex] and [latex]r\\sin\\theta[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Don&#8217;t forget the [latex]r[\/latex]<\/strong>: Include it in [latex]r\\,dr\\,d\\theta[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Integrate<\/strong>: Usually integrate with respect to [latex]r[\/latex] first, then [latex]\\theta[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\">Use polar coordinates when:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">The region is circular, semicircular, or a sector<\/li>\n<li class=\"whitespace-normal break-words\">The integrand contains [latex]x^2 + y^2[\/latex] (which becomes [latex]r^2[\/latex])<\/li>\n<li class=\"whitespace-normal break-words\">Boundaries are easier to express as [latex]r = \\text{constant}[\/latex] or [latex]\\theta = \\text{constant}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q733260947\">Show Solution<\/button><\/p>\n<div id=\"q733260947\" class=\"hidden-answer\" style=\"display: none\">[latex]\\frac{14}3[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q784312663\">Show Solution<\/button><\/p>\n<div id=\"q784312663\" class=\"hidden-answer\" style=\"display: none\">[latex]8\\pi[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">\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\" data-mce-fragment=\"1\"><\/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<\/section>\n<h2 data-type=\"title\">General Polar Regions of Integration<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Just as rectangular regions generalize beyond rectangles to curves, polar regions extend beyond simple sectors. For regions with curved polar boundaries, you integrate with variable limits that depend on [latex]\\theta[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>General Polar Regions<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">A general polar region has the form:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]R = {(r,\\theta) \\mid \\alpha \\leq \\theta \\leq \\beta, h_1(\\theta) \\leq r \\leq h_2(\\theta)}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">The angle [latex]\\theta[\/latex] ranges between constants, while the radius [latex]r[\/latex] varies between two functions of [latex]\\theta[\/latex]. Think of it as sweeping an angle from [latex]\\alpha[\/latex] to [latex]\\beta[\/latex], where at each angle the radius extends from inner curve [latex]h_1(\\theta)[\/latex] to outer curve [latex]h_2(\\theta)[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Iterated Integral<\/strong><\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{R}{\\displaystyle\\iint}f(r,\\theta)dA = \\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<p class=\"whitespace-normal break-words\">The outer integral spans the angular range (constants). The inner integral spans the radial range (functions of [latex]\\theta[\/latex]).<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Sketch the region<\/strong>: Draw the polar curves to visualize the boundaries<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Find the angular range<\/strong>: Determine [latex]\\alpha[\/latex] and [latex]\\beta[\/latex] where the region exists<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Identify radial bounds<\/strong>: At each angle [latex]\\theta[\/latex], find where [latex]r[\/latex] starts ([latex]h_1(\\theta)[\/latex]) and ends ([latex]h_2(\\theta)[\/latex])<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Set up the integral<\/strong>: Place angle limits outside, radial limits (as functions) inside<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Include the [latex]r[\/latex]<\/strong>: Always multiply by [latex]r[\/latex] in [latex]r\\,dr\\,d\\theta[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Integrate<\/strong>: Work inside out\u2014first with respect to [latex]r[\/latex], treating [latex]\\theta[\/latex] as constant<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>Key Reminder<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">When radial bounds depend on [latex]\\theta[\/latex], you&#8217;re describing how far out the region extends at each particular angle. This is the polar equivalent of Type I regions in rectangular coordinates.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q098544213\">Show Solution<\/button><\/p>\n<div id=\"q098544213\" class=\"hidden-answer\" style=\"display: none\">[latex]\\pi\/4[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/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\" data-mce-fragment=\"1\"><\/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<\/section>\n<h2>Polar Areas and Volumes<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Polar coordinates excel at computing volumes and areas for circular or radial regions. The same double integral framework applies, but the geometry often makes calculations dramatically simpler than rectangular coordinates.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Volume Under a Surface<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">For a solid under [latex]z = f(r,\\theta)[\/latex] over a polar region [latex]D[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]V = \\underset{D}{\\displaystyle\\iint}f(r,\\theta)r\\,dr\\,d\\theta[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Convert [latex]x^2 + y^2[\/latex] to [latex]r^2[\/latex] in the function, and don&#8217;t forget the [latex]r[\/latex] in [latex]r,dr,d\\theta[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Area of a Polar Region<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">To find the area of region [latex]D[\/latex], integrate 1:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]A = \\displaystyle\\int_\\alpha^\\beta\\displaystyle\\int_{h_1(\\theta)}^{h_2(\\theta)}r\\,dr\\,d\\theta[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">The [latex]r[\/latex] factor handles the &#8220;spreading out&#8221; of area elements as radius increases.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Strategic Decisions<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Not every problem is easier in polar coordinates. If the region is triangular or the function doesn&#8217;t simplify with [latex]r^2 = x^2 + y^2[\/latex], stick with rectangular coordinates. Compare the complexity of both setups before committing.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Sketch the region<\/strong>: Identify whether it&#8217;s naturally circular, radial, or involves [latex]x^2 + y^2[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Find polar bounds<\/strong>: Determine [latex]\\alpha \\leq \\theta \\leq \\beta[\/latex] and [latex]h_1(\\theta) \\leq r \\leq h_2(\\theta)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Convert the function<\/strong>: Replace [latex]x[\/latex] and [latex]y[\/latex] with [latex]r\\cos\\theta[\/latex] and [latex]r\\sin\\theta[\/latex]; simplify [latex]x^2 + y^2[\/latex] to [latex]r^2[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Set up carefully<\/strong>: Include the [latex]r[\/latex] in [latex]r,dr,d\\theta[\/latex]\u2014this is non-negotiable<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Integrate<\/strong>: Usually [latex]r[\/latex] first, then [latex]\\theta[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>Common Applications<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\"><strong>Cones<\/strong>: For [latex]z = h - \\frac{h}{a}\\sqrt{x^2 + y^2}[\/latex] over a circle, convert to [latex]z = h - \\frac{h}{a}r[\/latex]\u2014much simpler.<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Polar roses<\/strong>: For [latex]r = \\cos(n\\theta)[\/latex], use symmetry. Find the area of one petal and multiply by the number of petals.<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Intersecting regions<\/strong>: When a circle and cardioid overlap, split the integral at intersection points, integrating different radial bounds over different angular ranges.<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Improper Integrals<\/strong>: The entire plane becomes [latex]0 \\leq \\theta \\leq 2\\pi[\/latex], [latex]0 \\leq r < \\infty[\/latex], requiring limits as [latex]r \\to \\infty[\/latex].<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Use polar coordinates to find an iterated integral for finding the volume of the solid enclosed by the paraboloids [latex]{z} = {x^2} + {y^2}[\/latex] and\u00a0[latex]{z} = {16} - {x^2} - {y^2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q512348890\">Show Solution<\/button><\/p>\n<div id=\"q512348890\" class=\"hidden-answer\" style=\"display: none\">[latex]V=\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^{2\\sqrt2}(16-2r^2)r \\ dr \\ d\\theta=64\\pi[\/latex] cubic units.<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Find the area enclosed inside the cardioid [latex]{r} = {3} - {{3}{\\sin}{\\theta}}[\/latex] and outside the cardioid [latex]{r} = {1} + {{\\sin}{\\theta}}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q579921064\">Show Solution<\/button><\/p>\n<div id=\"q579921064\" class=\"hidden-answer\" style=\"display: none\">[latex]A=2\\displaystyle\\int_{-\\pi\/2}^{\\pi\/6}\\displaystyle\\int_{1+\\sin\\theta}^{3-3\\sin\\theta}r \\ dr \\ d\\theta = 8\\pi+9\\sqrt3[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Evaluate the integral\u00a0[latex]\\underset{R^2}{\\displaystyle\\iint}{e^{{-4}{(x^2+y^2)}}}{dx^{}}{dy}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q967834547\">Show Solution<\/button><\/p>\n<div id=\"q967834547\" class=\"hidden-answer\" style=\"display: none\">[latex]\\frac{\\pi}4[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8197104&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=TdDx0cLReoI&amp;video_target=tpm-plugin-e10unxm4-TdDx0cLReoI\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/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\/CP5.22_transcript.html\">transcript for \u201cCP 5.22\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":23,"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-691","chapter","type-chapter","status-publish","hentry"],"part":432,"module-header":"fresh_take","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\/691","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":11,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/691\/revisions"}],"predecessor-version":[{"id":2185,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/691\/revisions\/2185"}],"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\/691\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=691"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=691"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=691"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=691"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}