{"id":1028,"date":"2025-06-20T17:29:41","date_gmt":"2025-06-20T17:29:41","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1028"},"modified":"2025-09-10T18:26:02","modified_gmt":"2025-09-10T18:26:02","slug":"area-and-arc-length-in-polar-coordinates-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/area-and-arc-length-in-polar-coordinates-learn-it-1\/","title":{"raw":"Area and Arc Length in Polar Coordinates: Learn It 1","rendered":"Area and Arc Length in Polar Coordinates: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Calculate the area of regions when they're described in polar coordinates<\/li>\r\n \t<li>Find the length of a curve that's given in polar form<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2 data-type=\"title\">Areas of Regions Bounded by Polar Curves<\/h2>\r\n<p class=\"whitespace-normal break-words\">In rectangular coordinates, you calculate the area under a curve [latex]y = f(x)[\/latex] from [latex]x = a[\/latex] to [latex]x = b[\/latex] using the definite integral [latex]A = \\int_a^b f(x) dx[\/latex]. You also found arc length using [latex]L = \\int_a^b \\sqrt{1 + (f'(x))^2} dx[\/latex].<\/p>\r\n\r\n<section class=\"textbox recall\" aria-label=\"Recall\"><strong>\u00a0Fundamental Theorem of Calculus<\/strong>: The definite integral [latex]\\int_a^b f(x) dx[\/latex] gives the exact area under the curve [latex]y = f(x)[\/latex] when [latex]f(x) &gt; 0[\/latex] on the interval [latex][a,b][\/latex].<\/section>\r\n<p class=\"whitespace-normal break-words\">The key insight is adapting the Riemann sum approach. For rectangular coordinates, we used rectangles to approximate area. For polar curves, we use <strong>circular sectors<\/strong> instead.<\/p>\r\n<p class=\"whitespace-normal break-words\">Consider a polar curve [latex]r = f(\\theta)[\/latex] where [latex]\\alpha \\leq \\theta \\leq \\beta[\/latex]. Here's how we build the area formula:<\/p>\r\n<p class=\"whitespace-normal break-words\">Divide [latex][\\alpha, \\beta][\/latex] into [latex]n[\/latex] equal subintervals, each with width: [latex]\\Delta\\theta = \\frac{\\beta - \\alpha}{n}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">The partition points are: [latex]\\theta_i = \\alpha + i\\Delta\\theta[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Each [latex]\\theta_i[\/latex] creates a line through the origin, dividing our region into sectors. The radius of each sector is [latex]r_i = f(\\theta_i)[\/latex].<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_11_04_001\"><figcaption><\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"668\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234933\/CNX_Calc_Figure_11_04_001.jpg\" alt=\"On the polar coordinate plane, a curve is drawn in the first quadrant, and there are rays from the origin that intersect this curve at a regular interval. Every time one of these rays intersects the curve, a perpendicular line is made from the ray to the next ray. The first instance of a ray-curve intersection is labeled \u03b8 = \u03b1; the last instance is labeled \u03b8 = \u03b2. The intervening ones are marked \u03b81, \u03b82, \u2026, \u03b8n\u22121.\" width=\"668\" height=\"669\" data-media-type=\"image\/jpeg\" \/> Figure 1. A partition of a typical curve in polar coordinates.[\/caption]<\/figure>\r\n<p class=\"whitespace-normal break-words\">The line segments are connected by arcs of constant radius, creating sectors whose areas we can calculate using a geometric formula. We'll use these sector areas to approximate the area between successive line segments.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_11_04_002\"><figcaption><\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234935\/CNX_Calc_Figure_11_04_002.jpg\" alt=\"A circle is drawn with radius r and a sector of angle \u03b8. It is noted that A = (1\/2) \u03b8 r2.\" width=\"325\" height=\"330\" data-media-type=\"image\/jpeg\" \/> Figure 2. The area of a sector of a circle is given by [latex]A=\\frac{1}{2}\\theta {r}^{2}[\/latex].[\/caption]<\/figure>\r\n<p class=\"whitespace-normal break-words\">A full circle has area [latex]\\pi r^2[\/latex] and central angle [latex]2\\pi[\/latex]. A sector represents the fraction [latex]\\frac{\\theta}{2\\pi}[\/latex] of the full circle, so:<\/p>\r\n\r\n<div id=\"fs-id1167793784066\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]A=\\left(\\frac{\\theta }{2\\pi }\\right)\\pi {r}^{2}=\\frac{1}{2}\\theta {r}^{2}[\/latex].<\/div>\r\nSince the radius of each sector is [latex]r_i = f(\\theta_i)[\/latex], the area of the [latex]i[\/latex]th sector becomes:\r\n<div id=\"fs-id1167794097560\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{A}_{i}=\\frac{1}{2}\\left(\\Delta \\theta \\right){\\left(f\\left({\\theta }_{i}\\right)\\right)}^{2}[\/latex].<\/div>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<p class=\"whitespace-normal break-words\">The Riemann sum that approximates the total area is:<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167794146111\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{A}_{n}=\\displaystyle\\sum _{i=1}^{n}{A}_{i}\\approx \\displaystyle\\sum _{i=1}^{n}\\frac{1}{2}\\left(\\Delta \\theta \\right){\\left(f\\left({\\theta }_{i}\\right)\\right)}^{2}[\/latex].<\/div>\r\nTaking the limit as [latex]n \\to \\infty[\/latex], we get the exact area:\r\n<div id=\"fs-id1167793983076\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]A=\\underset{n\\to \\infty }{\\text{lim}}{A}_{n}=\\frac{1}{2}{\\displaystyle\\int }_{\\alpha }^{\\beta }{\\left(f\\left(\\theta \\right)\\right)}^{2}d\\theta [\/latex].<\/div>\r\n<p class=\"whitespace-normal break-words\">This derivation leads us to our main theorem.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>area of a region bounded by a polar curve<\/h3>\r\n<p id=\"fs-id1167794045635\">Suppose [latex]f[\/latex] is continuous and nonnegative on the interval [latex]\\alpha \\le \\theta \\le \\beta [\/latex] with [latex]0&lt;\\beta -\\alpha \\le 2\\pi [\/latex]. The area of the region bounded by the graph of [latex]r=f\\left(\\theta \\right)[\/latex] between the radial lines [latex]\\theta =\\alpha [\/latex] and [latex]\\theta =\\beta [\/latex] is:<\/p>\r\n\r\n<div id=\"fs-id1167793300455\" style=\"text-align: center;\" data-type=\"equation\">[latex]A=\\frac{1}{2}{\\displaystyle\\int }_{\\alpha }^{\\beta }{\\left[f\\left(\\theta \\right)\\right]}^{2}d\\theta =\\frac{1}{2}{\\displaystyle\\int }_{\\alpha }^{\\beta }{r}^{2}d\\theta [\/latex].<\/div>\r\n<\/section>\r\n<div data-type=\"equation\"><section class=\"textbox proTip\" aria-label=\"Pro Tip\">Remember the factor of [latex]\\frac{1}{2}[\/latex]! This comes from the sector area formula and is easy to forget when setting up polar area problems.<\/section><\/div>\r\n<section class=\"textbox recall\" aria-label=\"Recall\"><strong>Recall: Solving Trigonometric Equations\r\n[latex]\\\\[\/latex]\r\n<\/strong>When finding intersection points for polar curves, you'll often need to solve trigonometric equations. Follow these steps:\r\n<ol>\r\n \t<li>Use trigonometric identities to rewrite the expression in terms of a single trigonometric function, if necessary.<\/li>\r\n \t<li>Use algebra to isolate the trigonometric expression.<\/li>\r\n \t<li>Identify all angles on the unit circle that satisfy the equation.<\/li>\r\n \t<li>Note the period of the function to state all possible solutions for the angle.<\/li>\r\n \t<li>Solve for the variable in the angle expression, and identify the angles that lie within the desired interval.<\/li>\r\n<\/ol>\r\nFor example, to solve the equation [latex] 1 + 3\\cos 2\\theta = 5 \\cos 2\\theta [\/latex] on the interval [latex] \\left[0, 2\\pi \\right) [\/latex]\r\n<ul>\r\n \t<li>Subtract the\u00a0[latex] 3\\cos 2\\theta [\/latex] term and then divide both sides of the equation by [latex] 2 [\/latex]:<\/li>\r\n<\/ul>\r\n<p style=\"padding-left: 90px;\">[latex] \\cos 2\\theta = \\frac{1}{2} [\/latex]<\/p>\r\n\r\n<ul>\r\n \t<li>There are two angles on the unit circle where cosine is [latex] \\frac{1}{2} [\/latex]:<\/li>\r\n<\/ul>\r\n<p style=\"padding-left: 90px;\">[latex] 2\\theta = \\frac{\\pi}{3} [\/latex] and [latex] 2\\theta = \\frac{5\\pi}{3} [\/latex]<\/p>\r\n\r\n<ul>\r\n \t<li>Since [latex] \\cos x [\/latex] has a period of [latex] 2\\pi [\/latex], all possible solutions are given by:<\/li>\r\n<\/ul>\r\n<p style=\"padding-left: 90px;\">[latex] 2\\theta = \\frac{\\pi}{3} + 2\\pi k, 2\\theta = \\frac{5\\pi}{3} + 2\\pi k [\/latex] for any integer [latex] k [\/latex].<\/p>\r\n\r\n<ul>\r\n \t<li>Divide by [latex] 2 [\/latex] to isolate [latex] \\theta [\/latex]:<\/li>\r\n<\/ul>\r\n<p style=\"padding-left: 90px;\">[latex] \\theta = \\frac{\\pi}{6} + \\pi k, \\theta = \\frac{5\\pi}{6} + \\pi k [\/latex] for any integer [latex] k [\/latex]<\/p>\r\n\r\n<ul>\r\n \t<li>The possible values for [latex] k [\/latex] that result in angles within the interval [latex] \\left[0, 2\\pi \\right) [\/latex] are [latex] k = 0 \\: \\text{and} \\: 1 [\/latex], yielding the solution:<\/li>\r\n<\/ul>\r\n<p style=\"padding-left: 90px;\">[latex] \\theta = \\frac{\\pi}{6}, \\frac{5\\pi}{6}, \\frac{7\\pi}{6}, \\frac{11\\pi}{6} [\/latex]<\/p>\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1167794077086\" data-type=\"problem\">\r\n<p id=\"fs-id1167794054546\">Find the area of one petal of the rose defined by the equation [latex]r=3\\sin\\left(2\\theta \\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558899\"]\r\n<div id=\"fs-id1167793966078\" data-type=\"solution\">\r\n<p id=\"fs-id1167793970649\">The graph of [latex]r=3\\sin\\left(2\\theta \\right)[\/latex] follows.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_11_04_003\"><figcaption><\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"354\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234937\/CNX_Calc_Figure_11_04_003.jpg\" alt=\"A four-petaled rose with furthest extent 3 from the origin at \u03c0\/4, 3\u03c0\/4, 5\u03c0\/4, and 7\u03c0\/4.\" width=\"354\" height=\"354\" data-media-type=\"image\/jpeg\" \/> Figure 3. The graph of [latex]r=3\\sin\\left(2\\theta \\right)[\/latex].[\/caption]<\/figure>\r\n<p id=\"fs-id1167793369342\">When [latex]\\theta =0[\/latex] we have [latex]r=3\\sin\\left(2\\left(0\\right)\\right)=0[\/latex]. The next value for which [latex]r=0[\/latex] is [latex]\\theta =\\frac{\\pi}{2}[\/latex]. This can be seen by solving the equation [latex]3\\sin\\left(2\\theta \\right)=0[\/latex] for [latex]\\theta [\/latex]. Therefore the values [latex]\\theta =0[\/latex] to [latex]\\theta =\\frac{\\pi}{2}[\/latex] trace out the first petal of the rose. To find the area inside this petal, use the theorem with [latex]f\\left(\\theta \\right)=3\\sin\\left(2\\theta \\right)[\/latex], [latex]\\alpha =0[\/latex], and [latex]\\beta =\\frac{\\pi}{2}\\text{:}[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1167793990528\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill A&amp; =\\frac{1}{2}{\\displaystyle\\int }_{\\alpha }^{\\beta }{\\left[f\\left(\\theta \\right)\\right]}^{2}d\\theta \\hfill \\\\ &amp; =\\frac{1}{2}{\\displaystyle\\int }_{0}^{\\frac{\\pi}{2}}{\\left[3\\sin\\left(2\\theta \\right)\\right]}^{2}d\\theta \\hfill \\\\ &amp; =\\frac{1}{2}{\\displaystyle\\int }_{0}^{\\frac{\\pi}{2}}9{\\sin}^{2}\\left(2\\theta \\right)d\\theta .\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794034637\">To evaluate this integral, use the formula [latex]{\\sin}^{2}\\alpha =\\frac{\\left(1-\\cos\\left(2\\alpha \\right)\\right)}{2}[\/latex] with [latex]\\alpha =2\\theta \\text{:}[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1167793876806\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill A&amp; =\\frac{1}{2}{\\displaystyle\\int }_{0}^{\\frac{\\pi}{2}}9{\\sin}^{2}\\left(2\\theta \\right)d\\theta \\hfill \\\\ &amp; =\\frac{9}{2}{\\displaystyle\\int }_{0}^{\\frac{\\pi}{2}}\\frac{\\left(1-\\cos\\left(4\\theta \\right)\\right)}{2}d\\theta \\hfill \\\\ &amp; =\\frac{9}{4}\\left({\\displaystyle\\int }_{0}^{\\frac{\\pi}{2}}1-\\cos\\left(4\\theta \\right)d\\theta \\right)\\hfill \\\\ &amp; =\\frac{9}{4}{\\left(\\theta -\\frac{\\sin\\left(4\\theta \\right)}{4}\\right)}_{0}^{\\frac{\\pi}{2}}\\hfill \\\\ &amp; =\\frac{9}{4}\\left(\\frac{\\pi }{2}-\\frac{\\sin2\\pi }{4}\\right)-\\frac{9}{4}\\left(0-\\frac{\\sin4\\left(0\\right)}{4}\\right)\\hfill \\\\ &amp; =\\frac{9\\pi }{8}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the example above.<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/F10azktoCBI?controls=0&amp;start=38&amp;end=258&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/7.4AreaAndArcLength38to258_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"7.4 Area and Arc Length in Polar Coordinates\" here (opens in new window)<\/a>.\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]311325[\/ohm_question]<\/section>\r\n<p class=\"whitespace-normal break-words\">The previous example showed how to find area inside one curve. You can also find the area <strong>between <\/strong>two polar curves using the same formula, but this requires some additional steps.<\/p>\r\n<p class=\"whitespace-normal break-words\">When finding the area between two polar curves, you need to:<\/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-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Find the intersection points of the curves<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Determine which curve is the outer curve and which is the inner curve in each region<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Set up the integral as: [latex]A = \\frac{1}{2}\\int_{\\alpha}^{\\beta} [r_{\\text{outer}}^2 - r_{\\text{inner}}^2] d\\theta[\/latex]<\/li>\r\n<\/ul>\r\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\"><strong>Identifying outer vs. inner curves:<\/strong> At any given angle [latex]\\theta[\/latex], the curve with the larger [latex]r[\/latex]-value is the outer curve. This can change between intersection points, so check each region separately.<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1167793261062\" data-type=\"problem\">\r\n<p id=\"fs-id1167794334368\">Find the area outside the cardioid [latex]r=2+2\\sin\\theta [\/latex] and inside the circle [latex]r=6\\sin\\theta [\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558896\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558896\"]\r\n<div id=\"fs-id1167794038690\" data-type=\"solution\">\r\n<p id=\"fs-id1167793969730\">First draw a graph containing both curves as shown.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_11_04_004\"><figcaption><\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"354\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234939\/CNX_Calc_Figure_11_04_004.jpg\" alt=\"A cardioid with equation r = 2 + 2 sin\u03b8 is shown, so it has its upper heart part at the origin and the rest of the cardioid is pointed up. There is a circle with radius 6 centered at (3, \u03c0\/2). The area above the cardioid but below the circle is shaded orange.\" width=\"354\" height=\"356\" data-media-type=\"image\/jpeg\" \/> Figure 4. The region between the curves [latex]r=2+2\\sin\\theta [\/latex] and [latex]r=6\\sin\\theta [\/latex].[\/caption]<\/figure>\r\n<p id=\"fs-id1167793269240\">To determine the limits of integration, first find the points of intersection by setting the two functions equal to each other and solving for [latex]\\theta \\text{:}[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1167794188708\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill 6\\sin\\theta &amp; =\\hfill &amp; 2+2\\sin\\theta \\hfill \\\\ \\hfill 4\\sin\\theta &amp; =\\hfill &amp; 2\\hfill \\\\ \\hfill \\sin\\theta &amp; =\\hfill &amp; \\frac{1}{2}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793262008\">This gives the solutions [latex]\\theta =\\frac{\\pi }{6}[\/latex] and [latex]\\theta =\\frac{5\\pi }{6}[\/latex], which are the limits of integration. The circle [latex]r=3\\sin\\theta [\/latex] is the red graph, which is the outer function, and the cardioid [latex]r=2+2\\sin\\theta [\/latex] is the blue graph, which is the inner function. To calculate the area between the curves, start with the area inside the circle between [latex]\\theta =\\frac{\\pi }{6}[\/latex] and [latex]\\theta =\\frac{5\\pi }{6}[\/latex], then subtract the area inside the cardioid between [latex]\\theta =\\frac{\\pi }{6}[\/latex] and [latex]\\theta =\\frac{5\\pi }{6}\\text{:}[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1167794331207\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill A&amp; =\\text{circle}-\\text{cardioid}\\hfill \\\\ &amp; =\\frac{1}{2}{\\displaystyle\\int }_{\\frac{\\pi}{6}}^{\\frac{5\\pi}{6}}{\\left[6\\sin\\theta \\right]}^{2}d\\theta -\\frac{1}{2}{\\displaystyle\\int }_{\\frac{\\pi}{6}}^{\\frac{5\\pi}{6}}{\\left[2+2\\sin\\theta \\right]}^{2}d\\theta \\hfill \\\\ &amp; =\\frac{1}{2}{\\displaystyle\\int }_{\\frac{\\pi}{6}}^{\\frac{5\\pi}{6}}36{\\sin}^{2}\\theta d\\theta -\\frac{1}{2}{\\displaystyle\\int }_{\\frac{\\pi}{6}}^{\\frac{5\\pi}{6}}4+8\\sin\\theta +4{\\sin}^{2}\\theta d\\theta \\hfill \\\\ &amp; =18{\\displaystyle\\int }_{\\frac{\\pi}{6}}^{\\frac{5\\pi}{6}}\\frac{1-\\cos\\left(2\\theta \\right)}{2}d\\theta -2{\\displaystyle\\int }_{\\frac{\\pi}{6}}^{\\frac{5\\pi}{6}}1+2\\sin\\theta +\\frac{1-\\cos\\left(2\\theta \\right)}{2}d\\theta \\hfill \\\\ &amp; =9{\\left[\\theta -\\frac{\\sin\\left(2\\theta \\right)}{2}\\right]}_{\\frac{\\pi}{6}}^{\\frac{5\\pi}{6}}-2{\\left[\\frac{3\\theta }{2}-2\\cos\\theta -\\frac{\\sin\\left(2\\theta \\right)}{4}\\right]}_{\\frac{\\pi}{6}}^{\\frac{5\\pi}{6}}\\hfill \\\\ &amp; =9\\left(\\frac{5\\pi }{6}-\\frac{\\sin2\\left(\\frac{5\\pi}{6}\\right)}{2}\\right)-9\\left(\\frac{\\pi }{6}-\\frac{\\sin2\\left(\\frac{\\pi}{6}\\right)}{2}\\right)\\hfill \\\\ &amp; \\text{-}\\left(3\\left(\\frac{5\\pi }{6}\\right)-4\\cos\\frac{5\\pi }{6}-\\frac{\\sin2\\left(\\frac{5\\pi}{6}\\right)}{2}\\right)+\\left(3\\left(\\frac{\\pi }{6}\\right)-4\\cos\\frac{\\pi }{6}-\\frac{\\sin2\\left(\\frac{\\pi}{6}\\right)}{2}\\right)\\hfill \\\\ &amp; =4\\pi .\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the example above.<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/F10azktoCBI?controls=0&amp;start=260&amp;end=650&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/7.4AreaAndArcLength260to650_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"7.4 Area and Arc Length in Polar Coordinates\" here (opens in new window)<\/a>.\r\n\r\n<\/section>\r\n<p class=\"whitespace-normal break-words\">In the previous example, we found the area inside the circle and outside the cardioid by first finding their intersection points. When we solved [latex]2 + 2\\sin\\theta = 6\\sin\\theta[\/latex] directly, we got two solutions: [latex]\\theta = \\frac{\\pi}{6}[\/latex] and [latex]\\theta = \\frac{5\\pi}{6}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">However, looking at the graph reveals three intersection points. The third intersection point is the origin.<\/p>\r\n<p class=\"whitespace-normal break-words\">The reason why this point did not show up as a solution is because the origin is on both graphs but for different values of [latex]\\theta [\/latex].<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p class=\"whitespace-pre-wrap break-words\">For the cardioid [latex]r = 2 + 2\\sin\\theta[\/latex]: [latex]2 + 2\\sin\\theta = 0[\/latex] [latex]\\sin\\theta = -1[\/latex] [latex]\\theta = \\frac{3\\pi}{2} + 2n\\pi[\/latex] (where [latex]n[\/latex] is any integer)<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">For the circle [latex]r = 6\\sin\\theta[\/latex]: [latex]6\\sin\\theta = 0[\/latex] [latex]\\theta = n\\pi[\/latex] (where [latex]n[\/latex] is any integer)<\/p>\r\n<p class=\"whitespace-normal break-words\">These two solution sets have <strong>no values in common<\/strong>, yet the curves still intersect at the origin.<\/p>\r\n\r\n<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\">Always check for intersection at the origin! When polar curves pass through the origin, they may do so at different [latex]\\theta[\/latex] values, so the origin won't appear when you solve [latex]r_1(\\theta) = r_2(\\theta)[\/latex] algebraically.<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Calculate the area of regions when they&#8217;re described in polar coordinates<\/li>\n<li>Find the length of a curve that&#8217;s given in polar form<\/li>\n<\/ul>\n<\/section>\n<h2 data-type=\"title\">Areas of Regions Bounded by Polar Curves<\/h2>\n<p class=\"whitespace-normal break-words\">In rectangular coordinates, you calculate the area under a curve [latex]y = f(x)[\/latex] from [latex]x = a[\/latex] to [latex]x = b[\/latex] using the definite integral [latex]A = \\int_a^b f(x) dx[\/latex]. You also found arc length using [latex]L = \\int_a^b \\sqrt{1 + (f'(x))^2} dx[\/latex].<\/p>\n<section class=\"textbox recall\" aria-label=\"Recall\"><strong>\u00a0Fundamental Theorem of Calculus<\/strong>: The definite integral [latex]\\int_a^b f(x) dx[\/latex] gives the exact area under the curve [latex]y = f(x)[\/latex] when [latex]f(x) > 0[\/latex] on the interval [latex][a,b][\/latex].<\/section>\n<p class=\"whitespace-normal break-words\">The key insight is adapting the Riemann sum approach. For rectangular coordinates, we used rectangles to approximate area. For polar curves, we use <strong>circular sectors<\/strong> instead.<\/p>\n<p class=\"whitespace-normal break-words\">Consider a polar curve [latex]r = f(\\theta)[\/latex] where [latex]\\alpha \\leq \\theta \\leq \\beta[\/latex]. Here&#8217;s how we build the area formula:<\/p>\n<p class=\"whitespace-normal break-words\">Divide [latex][\\alpha, \\beta][\/latex] into [latex]n[\/latex] equal subintervals, each with width: [latex]\\Delta\\theta = \\frac{\\beta - \\alpha}{n}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">The partition points are: [latex]\\theta_i = \\alpha + i\\Delta\\theta[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Each [latex]\\theta_i[\/latex] creates a line through the origin, dividing our region into sectors. The radius of each sector is [latex]r_i = f(\\theta_i)[\/latex].<\/p>\n<figure id=\"CNX_Calc_Figure_11_04_001\"><figcaption><\/figcaption><figure style=\"width: 668px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234933\/CNX_Calc_Figure_11_04_001.jpg\" alt=\"On the polar coordinate plane, a curve is drawn in the first quadrant, and there are rays from the origin that intersect this curve at a regular interval. Every time one of these rays intersects the curve, a perpendicular line is made from the ray to the next ray. The first instance of a ray-curve intersection is labeled \u03b8 = \u03b1; the last instance is labeled \u03b8 = \u03b2. The intervening ones are marked \u03b81, \u03b82, \u2026, \u03b8n\u22121.\" width=\"668\" height=\"669\" data-media-type=\"image\/jpeg\" \/><figcaption class=\"wp-caption-text\">Figure 1. A partition of a typical curve in polar coordinates.<\/figcaption><\/figure>\n<\/figure>\n<p class=\"whitespace-normal break-words\">The line segments are connected by arcs of constant radius, creating sectors whose areas we can calculate using a geometric formula. We&#8217;ll use these sector areas to approximate the area between successive line segments.<\/p>\n<figure id=\"CNX_Calc_Figure_11_04_002\"><figcaption><\/figcaption><figure style=\"width: 325px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234935\/CNX_Calc_Figure_11_04_002.jpg\" alt=\"A circle is drawn with radius r and a sector of angle \u03b8. It is noted that A = (1\/2) \u03b8 r2.\" width=\"325\" height=\"330\" data-media-type=\"image\/jpeg\" \/><figcaption class=\"wp-caption-text\">Figure 2. The area of a sector of a circle is given by [latex]A=\\frac{1}{2}\\theta {r}^{2}[\/latex].<\/figcaption><\/figure>\n<\/figure>\n<p class=\"whitespace-normal break-words\">A full circle has area [latex]\\pi r^2[\/latex] and central angle [latex]2\\pi[\/latex]. A sector represents the fraction [latex]\\frac{\\theta}{2\\pi}[\/latex] of the full circle, so:<\/p>\n<div id=\"fs-id1167793784066\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]A=\\left(\\frac{\\theta }{2\\pi }\\right)\\pi {r}^{2}=\\frac{1}{2}\\theta {r}^{2}[\/latex].<\/div>\n<p>Since the radius of each sector is [latex]r_i = f(\\theta_i)[\/latex], the area of the [latex]i[\/latex]th sector becomes:<\/p>\n<div id=\"fs-id1167794097560\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{A}_{i}=\\frac{1}{2}\\left(\\Delta \\theta \\right){\\left(f\\left({\\theta }_{i}\\right)\\right)}^{2}[\/latex].<\/div>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\n<p class=\"whitespace-normal break-words\">The Riemann sum that approximates the total area is:<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1167794146111\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{A}_{n}=\\displaystyle\\sum _{i=1}^{n}{A}_{i}\\approx \\displaystyle\\sum _{i=1}^{n}\\frac{1}{2}\\left(\\Delta \\theta \\right){\\left(f\\left({\\theta }_{i}\\right)\\right)}^{2}[\/latex].<\/div>\n<p>Taking the limit as [latex]n \\to \\infty[\/latex], we get the exact area:<\/p>\n<div id=\"fs-id1167793983076\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]A=\\underset{n\\to \\infty }{\\text{lim}}{A}_{n}=\\frac{1}{2}{\\displaystyle\\int }_{\\alpha }^{\\beta }{\\left(f\\left(\\theta \\right)\\right)}^{2}d\\theta[\/latex].<\/div>\n<p class=\"whitespace-normal break-words\">This derivation leads us to our main theorem.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>area of a region bounded by a polar curve<\/h3>\n<p id=\"fs-id1167794045635\">Suppose [latex]f[\/latex] is continuous and nonnegative on the interval [latex]\\alpha \\le \\theta \\le \\beta[\/latex] with [latex]0<\\beta -\\alpha \\le 2\\pi[\/latex]. The area of the region bounded by the graph of [latex]r=f\\left(\\theta \\right)[\/latex] between the radial lines [latex]\\theta =\\alpha[\/latex] and [latex]\\theta =\\beta[\/latex] is:<\/p>\n<div id=\"fs-id1167793300455\" style=\"text-align: center;\" data-type=\"equation\">[latex]A=\\frac{1}{2}{\\displaystyle\\int }_{\\alpha }^{\\beta }{\\left[f\\left(\\theta \\right)\\right]}^{2}d\\theta =\\frac{1}{2}{\\displaystyle\\int }_{\\alpha }^{\\beta }{r}^{2}d\\theta[\/latex].<\/div>\n<\/section>\n<div data-type=\"equation\">\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">Remember the factor of [latex]\\frac{1}{2}[\/latex]! This comes from the sector area formula and is easy to forget when setting up polar area problems.<\/section>\n<\/div>\n<section class=\"textbox recall\" aria-label=\"Recall\"><strong>Recall: Solving Trigonometric Equations<br \/>\n[latex]\\\\[\/latex]<br \/>\n<\/strong>When finding intersection points for polar curves, you&#8217;ll often need to solve trigonometric equations. Follow these steps:<\/p>\n<ol>\n<li>Use trigonometric identities to rewrite the expression in terms of a single trigonometric function, if necessary.<\/li>\n<li>Use algebra to isolate the trigonometric expression.<\/li>\n<li>Identify all angles on the unit circle that satisfy the equation.<\/li>\n<li>Note the period of the function to state all possible solutions for the angle.<\/li>\n<li>Solve for the variable in the angle expression, and identify the angles that lie within the desired interval.<\/li>\n<\/ol>\n<p>For example, to solve the equation [latex]1 + 3\\cos 2\\theta = 5 \\cos 2\\theta[\/latex] on the interval [latex]\\left[0, 2\\pi \\right)[\/latex]<\/p>\n<ul>\n<li>Subtract the\u00a0[latex]3\\cos 2\\theta[\/latex] term and then divide both sides of the equation by [latex]2[\/latex]:<\/li>\n<\/ul>\n<p style=\"padding-left: 90px;\">[latex]\\cos 2\\theta = \\frac{1}{2}[\/latex]<\/p>\n<ul>\n<li>There are two angles on the unit circle where cosine is [latex]\\frac{1}{2}[\/latex]:<\/li>\n<\/ul>\n<p style=\"padding-left: 90px;\">[latex]2\\theta = \\frac{\\pi}{3}[\/latex] and [latex]2\\theta = \\frac{5\\pi}{3}[\/latex]<\/p>\n<ul>\n<li>Since [latex]\\cos x[\/latex] has a period of [latex]2\\pi[\/latex], all possible solutions are given by:<\/li>\n<\/ul>\n<p style=\"padding-left: 90px;\">[latex]2\\theta = \\frac{\\pi}{3} + 2\\pi k, 2\\theta = \\frac{5\\pi}{3} + 2\\pi k[\/latex] for any integer [latex]k[\/latex].<\/p>\n<ul>\n<li>Divide by [latex]2[\/latex] to isolate [latex]\\theta[\/latex]:<\/li>\n<\/ul>\n<p style=\"padding-left: 90px;\">[latex]\\theta = \\frac{\\pi}{6} + \\pi k, \\theta = \\frac{5\\pi}{6} + \\pi k[\/latex] for any integer [latex]k[\/latex]<\/p>\n<ul>\n<li>The possible values for [latex]k[\/latex] that result in angles within the interval [latex]\\left[0, 2\\pi \\right)[\/latex] are [latex]k = 0 \\: \\text{and} \\: 1[\/latex], yielding the solution:<\/li>\n<\/ul>\n<p style=\"padding-left: 90px;\">[latex]\\theta = \\frac{\\pi}{6}, \\frac{5\\pi}{6}, \\frac{7\\pi}{6}, \\frac{11\\pi}{6}[\/latex]<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1167794077086\" data-type=\"problem\">\n<p id=\"fs-id1167794054546\">Find the area of one petal of the rose defined by the equation [latex]r=3\\sin\\left(2\\theta \\right)[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558899\">Show Solution<\/button><\/p>\n<div id=\"q44558899\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167793966078\" data-type=\"solution\">\n<p id=\"fs-id1167793970649\">The graph of [latex]r=3\\sin\\left(2\\theta \\right)[\/latex] follows.<\/p>\n<figure id=\"CNX_Calc_Figure_11_04_003\"><figcaption><\/figcaption><figure style=\"width: 354px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234937\/CNX_Calc_Figure_11_04_003.jpg\" alt=\"A four-petaled rose with furthest extent 3 from the origin at \u03c0\/4, 3\u03c0\/4, 5\u03c0\/4, and 7\u03c0\/4.\" width=\"354\" height=\"354\" data-media-type=\"image\/jpeg\" \/><figcaption class=\"wp-caption-text\">Figure 3. The graph of [latex]r=3\\sin\\left(2\\theta \\right)[\/latex].<\/figcaption><\/figure>\n<\/figure>\n<p id=\"fs-id1167793369342\">When [latex]\\theta =0[\/latex] we have [latex]r=3\\sin\\left(2\\left(0\\right)\\right)=0[\/latex]. The next value for which [latex]r=0[\/latex] is [latex]\\theta =\\frac{\\pi}{2}[\/latex]. This can be seen by solving the equation [latex]3\\sin\\left(2\\theta \\right)=0[\/latex] for [latex]\\theta[\/latex]. Therefore the values [latex]\\theta =0[\/latex] to [latex]\\theta =\\frac{\\pi}{2}[\/latex] trace out the first petal of the rose. To find the area inside this petal, use the theorem with [latex]f\\left(\\theta \\right)=3\\sin\\left(2\\theta \\right)[\/latex], [latex]\\alpha =0[\/latex], and [latex]\\beta =\\frac{\\pi}{2}\\text{:}[\/latex]<\/p>\n<div id=\"fs-id1167793990528\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill A& =\\frac{1}{2}{\\displaystyle\\int }_{\\alpha }^{\\beta }{\\left[f\\left(\\theta \\right)\\right]}^{2}d\\theta \\hfill \\\\ & =\\frac{1}{2}{\\displaystyle\\int }_{0}^{\\frac{\\pi}{2}}{\\left[3\\sin\\left(2\\theta \\right)\\right]}^{2}d\\theta \\hfill \\\\ & =\\frac{1}{2}{\\displaystyle\\int }_{0}^{\\frac{\\pi}{2}}9{\\sin}^{2}\\left(2\\theta \\right)d\\theta .\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794034637\">To evaluate this integral, use the formula [latex]{\\sin}^{2}\\alpha =\\frac{\\left(1-\\cos\\left(2\\alpha \\right)\\right)}{2}[\/latex] with [latex]\\alpha =2\\theta \\text{:}[\/latex]<\/p>\n<div id=\"fs-id1167793876806\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill A& =\\frac{1}{2}{\\displaystyle\\int }_{0}^{\\frac{\\pi}{2}}9{\\sin}^{2}\\left(2\\theta \\right)d\\theta \\hfill \\\\ & =\\frac{9}{2}{\\displaystyle\\int }_{0}^{\\frac{\\pi}{2}}\\frac{\\left(1-\\cos\\left(4\\theta \\right)\\right)}{2}d\\theta \\hfill \\\\ & =\\frac{9}{4}\\left({\\displaystyle\\int }_{0}^{\\frac{\\pi}{2}}1-\\cos\\left(4\\theta \\right)d\\theta \\right)\\hfill \\\\ & =\\frac{9}{4}{\\left(\\theta -\\frac{\\sin\\left(4\\theta \\right)}{4}\\right)}_{0}^{\\frac{\\pi}{2}}\\hfill \\\\ & =\\frac{9}{4}\\left(\\frac{\\pi }{2}-\\frac{\\sin2\\pi }{4}\\right)-\\frac{9}{4}\\left(0-\\frac{\\sin4\\left(0\\right)}{4}\\right)\\hfill \\\\ & =\\frac{9\\pi }{8}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/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 example above.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/F10azktoCBI?controls=0&amp;start=38&amp;end=258&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/7.4AreaAndArcLength38to258_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;7.4 Area and Arc Length in Polar Coordinates&#8221; here (opens in new window)<\/a>.<\/p>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm311325\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=311325&theme=lumen&iframe_resize_id=ohm311325&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<p class=\"whitespace-normal break-words\">The previous example showed how to find area inside one curve. You can also find the area <strong>between <\/strong>two polar curves using the same formula, but this requires some additional steps.<\/p>\n<p class=\"whitespace-normal break-words\">When finding the area between two polar curves, you need to:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Find the intersection points of the curves<\/li>\n<li class=\"whitespace-normal break-words\">Determine which curve is the outer curve and which is the inner curve in each region<\/li>\n<li class=\"whitespace-normal break-words\">Set up the integral as: [latex]A = \\frac{1}{2}\\int_{\\alpha}^{\\beta} [r_{\\text{outer}}^2 - r_{\\text{inner}}^2] d\\theta[\/latex]<\/li>\n<\/ul>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\"><strong>Identifying outer vs. inner curves:<\/strong> At any given angle [latex]\\theta[\/latex], the curve with the larger [latex]r[\/latex]-value is the outer curve. This can change between intersection points, so check each region separately.<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1167793261062\" data-type=\"problem\">\n<p id=\"fs-id1167794334368\">Find the area outside the cardioid [latex]r=2+2\\sin\\theta[\/latex] and inside the circle [latex]r=6\\sin\\theta[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558896\">Show Solution<\/button><\/p>\n<div id=\"q44558896\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167794038690\" data-type=\"solution\">\n<p id=\"fs-id1167793969730\">First draw a graph containing both curves as shown.<\/p>\n<figure id=\"CNX_Calc_Figure_11_04_004\"><figcaption><\/figcaption><figure style=\"width: 354px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234939\/CNX_Calc_Figure_11_04_004.jpg\" alt=\"A cardioid with equation r = 2 + 2 sin\u03b8 is shown, so it has its upper heart part at the origin and the rest of the cardioid is pointed up. There is a circle with radius 6 centered at (3, \u03c0\/2). The area above the cardioid but below the circle is shaded orange.\" width=\"354\" height=\"356\" data-media-type=\"image\/jpeg\" \/><figcaption class=\"wp-caption-text\">Figure 4. The region between the curves [latex]r=2+2\\sin\\theta [\/latex] and [latex]r=6\\sin\\theta [\/latex].<\/figcaption><\/figure>\n<\/figure>\n<p id=\"fs-id1167793269240\">To determine the limits of integration, first find the points of intersection by setting the two functions equal to each other and solving for [latex]\\theta \\text{:}[\/latex]<\/p>\n<div id=\"fs-id1167794188708\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill 6\\sin\\theta & =\\hfill & 2+2\\sin\\theta \\hfill \\\\ \\hfill 4\\sin\\theta & =\\hfill & 2\\hfill \\\\ \\hfill \\sin\\theta & =\\hfill & \\frac{1}{2}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793262008\">This gives the solutions [latex]\\theta =\\frac{\\pi }{6}[\/latex] and [latex]\\theta =\\frac{5\\pi }{6}[\/latex], which are the limits of integration. The circle [latex]r=3\\sin\\theta[\/latex] is the red graph, which is the outer function, and the cardioid [latex]r=2+2\\sin\\theta[\/latex] is the blue graph, which is the inner function. To calculate the area between the curves, start with the area inside the circle between [latex]\\theta =\\frac{\\pi }{6}[\/latex] and [latex]\\theta =\\frac{5\\pi }{6}[\/latex], then subtract the area inside the cardioid between [latex]\\theta =\\frac{\\pi }{6}[\/latex] and [latex]\\theta =\\frac{5\\pi }{6}\\text{:}[\/latex]<\/p>\n<div id=\"fs-id1167794331207\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill A& =\\text{circle}-\\text{cardioid}\\hfill \\\\ & =\\frac{1}{2}{\\displaystyle\\int }_{\\frac{\\pi}{6}}^{\\frac{5\\pi}{6}}{\\left[6\\sin\\theta \\right]}^{2}d\\theta -\\frac{1}{2}{\\displaystyle\\int }_{\\frac{\\pi}{6}}^{\\frac{5\\pi}{6}}{\\left[2+2\\sin\\theta \\right]}^{2}d\\theta \\hfill \\\\ & =\\frac{1}{2}{\\displaystyle\\int }_{\\frac{\\pi}{6}}^{\\frac{5\\pi}{6}}36{\\sin}^{2}\\theta d\\theta -\\frac{1}{2}{\\displaystyle\\int }_{\\frac{\\pi}{6}}^{\\frac{5\\pi}{6}}4+8\\sin\\theta +4{\\sin}^{2}\\theta d\\theta \\hfill \\\\ & =18{\\displaystyle\\int }_{\\frac{\\pi}{6}}^{\\frac{5\\pi}{6}}\\frac{1-\\cos\\left(2\\theta \\right)}{2}d\\theta -2{\\displaystyle\\int }_{\\frac{\\pi}{6}}^{\\frac{5\\pi}{6}}1+2\\sin\\theta +\\frac{1-\\cos\\left(2\\theta \\right)}{2}d\\theta \\hfill \\\\ & =9{\\left[\\theta -\\frac{\\sin\\left(2\\theta \\right)}{2}\\right]}_{\\frac{\\pi}{6}}^{\\frac{5\\pi}{6}}-2{\\left[\\frac{3\\theta }{2}-2\\cos\\theta -\\frac{\\sin\\left(2\\theta \\right)}{4}\\right]}_{\\frac{\\pi}{6}}^{\\frac{5\\pi}{6}}\\hfill \\\\ & =9\\left(\\frac{5\\pi }{6}-\\frac{\\sin2\\left(\\frac{5\\pi}{6}\\right)}{2}\\right)-9\\left(\\frac{\\pi }{6}-\\frac{\\sin2\\left(\\frac{\\pi}{6}\\right)}{2}\\right)\\hfill \\\\ & \\text{-}\\left(3\\left(\\frac{5\\pi }{6}\\right)-4\\cos\\frac{5\\pi }{6}-\\frac{\\sin2\\left(\\frac{5\\pi}{6}\\right)}{2}\\right)+\\left(3\\left(\\frac{\\pi }{6}\\right)-4\\cos\\frac{\\pi }{6}-\\frac{\\sin2\\left(\\frac{\\pi}{6}\\right)}{2}\\right)\\hfill \\\\ & =4\\pi .\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/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 example above.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/F10azktoCBI?controls=0&amp;start=260&amp;end=650&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/7.4AreaAndArcLength260to650_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;7.4 Area and Arc Length in Polar Coordinates&#8221; here (opens in new window)<\/a>.<\/p>\n<\/section>\n<p class=\"whitespace-normal break-words\">In the previous example, we found the area inside the circle and outside the cardioid by first finding their intersection points. When we solved [latex]2 + 2\\sin\\theta = 6\\sin\\theta[\/latex] directly, we got two solutions: [latex]\\theta = \\frac{\\pi}{6}[\/latex] and [latex]\\theta = \\frac{5\\pi}{6}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">However, looking at the graph reveals three intersection points. The third intersection point is the origin.<\/p>\n<p class=\"whitespace-normal break-words\">The reason why this point did not show up as a solution is because the origin is on both graphs but for different values of [latex]\\theta[\/latex].<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-pre-wrap break-words\">For the cardioid [latex]r = 2 + 2\\sin\\theta[\/latex]: [latex]2 + 2\\sin\\theta = 0[\/latex] [latex]\\sin\\theta = -1[\/latex] [latex]\\theta = \\frac{3\\pi}{2} + 2n\\pi[\/latex] (where [latex]n[\/latex] is any integer)<\/p>\n<p class=\"whitespace-pre-wrap break-words\">For the circle [latex]r = 6\\sin\\theta[\/latex]: [latex]6\\sin\\theta = 0[\/latex] [latex]\\theta = n\\pi[\/latex] (where [latex]n[\/latex] is any integer)<\/p>\n<p class=\"whitespace-normal break-words\">These two solution sets have <strong>no values in common<\/strong>, yet the curves still intersect at the origin.<\/p>\n<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">Always check for intersection at the origin! When polar curves pass through the origin, they may do so at different [latex]\\theta[\/latex] values, so the origin won&#8217;t appear when you solve [latex]r_1(\\theta) = r_2(\\theta)[\/latex] algebraically.<\/section>\n","protected":false},"author":15,"menu_order":11,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":677,"module-header":"- Select Header -","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\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1028"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":8,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1028\/revisions"}],"predecessor-version":[{"id":2308,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1028\/revisions\/2308"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/677"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1028\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1028"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1028"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1028"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1028"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}