{"id":1066,"date":"2021-11-01T19:15:33","date_gmt":"2021-11-01T19:15:33","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=1066"},"modified":"2022-11-01T04:23:36","modified_gmt":"2022-11-01T04:23:36","slug":"polar-areas-and-volumes","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/polar-areas-and-volumes\/","title":{"raw":"Polar Areas and Volumes","rendered":"Polar Areas and Volumes"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul class=\"os-abstract\">\r\n \t<li><span class=\"os-abstract-content\">Use double integrals in polar coordinates to calculate areas and volumes.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1167793277568\">As in rectangular coordinates, if a solid [latex]S[\/latex] is bounded by the surface [latex]{z} = {f}{(r,{\\theta})}[\/latex], as well as by the surfaces [latex]{r} = {a}, {r} = {b}, {\\theta} = {\\alpha}[\/latex], and [latex]{\\theta} = {\\beta}[\/latex] we can find the volume [latex]V[\/latex] of [latex]S[\/latex] by double integration, as<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{{V} = \\underset{R}{\\displaystyle\\iint}{f}{(r,{\\theta})} \\ {r} \\ {dr} \\ {d{\\theta}} = \\displaystyle\\int_{\\theta=\\alpha}^{\\theta=\\beta}\\displaystyle\\int_{r=a}^{r=b}f(r,\\theta)r \\ dr \\ d\\theta}[\/latex].<\/p>\r\nIf the base of the solid can be described as [latex]{D} = {\\left \\{{(r,{\\theta})}{\\mid}{\\alpha} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {{\\beta},{h_1}}{(\\theta)} \\ {\\leq} \\ {r} \\ {\\leq} \\ {h_2{(\\theta)}} \\right \\}}[\/latex], then the double integral for the volume becomes\r\n<p style=\"text-align: center;\">[latex]\\large{{V} = \\underset{D}{\\displaystyle\\iint}{f}{(r,{\\theta})} \\ {r} \\ {dr} \\ {d{\\theta}} = \\displaystyle\\int_{\\theta=\\alpha}^{\\theta=\\beta}\\displaystyle\\int_{r=h_1(\\theta)}^{r=h_2(\\theta)}f(r,\\theta)r \\ dr \\ d\\theta}[\/latex]<\/p>\r\n<span style=\"font-size: 1rem; text-align: initial;\">We illustrate this idea with some examples.<\/span>\r\n<div id=\"fs-id1167793949568\" class=\"ui-has-child-title\" data-type=\"example\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding a volume using a double integral<\/h3>\r\nFind the volume of the solid that lies under the paraboloid [latex]z=1-x^{2}-y^{2}[\/latex] and above the unit circle on the [latex]xy[\/latex]-plane (see the following figure).\r\n\r\n[caption id=\"attachment_1364\" align=\"aligncenter\" width=\"456\"]<img class=\"size-full wp-image-1364\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25143225\/5-3-7.jpeg\" alt=\"The paraboloid z = 1 minus x squared minus y squared is shown, which in this graph looks like a sheet with the middle gently puffed up and the corners anchored.\" width=\"456\" height=\"371\" \/> Figure 1. The paraboloid\u00a0[latex]z=1-x^{2}-y^{2}[\/latex].[\/caption][reveal-answer q=\"613772890\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"613772890\"]\r\n<p id=\"fs-id1167793416705\">By the method of double integration, we can see that the volume is the iterated integral of the form [latex]\\underset{R}{\\displaystyle\\iint}{(1-{x^2}-{y^2})}{dA}[\/latex] where [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-id1167793936934\">This integration was shown before in\u00a0Example \"Evaluating a Double Integral by Converting from Rectangular Coordinates\", so the volume is [latex]{\\frac{\\pi}{2}}[\/latex] cubic units.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding a volume using double integration<\/h3>\r\nFind the volume of the solid that lies under the paraboloid [latex]{z} = {4} - {x^2} - {y^2}[\/latex] and above the disk [latex]{(x-1)^2} + {y^2} = {1}[\/latex] on the [latex]xy[\/latex]-plane. See the paraboloid in\u00a0Figure 2\u00a0intersecting the cylinder [latex]{(x-1)^2} + {y^2} = {1}[\/latex] above the [latex]xy[\/latex]-plane.\r\n\r\n[caption id=\"attachment_1365\" align=\"aligncenter\" width=\"419\"]<img class=\"size-full wp-image-1365\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25143256\/5-3-8.jpeg\" alt=\"A paraboloid with equation z = 4 minus x squared minus y squared is intersected by a cylinder with equation (x minus 1) squared + y squared = 1.\" width=\"419\" height=\"500\" \/> Figure 2.\u00a0Finding the volume of a solid with a paraboloid cap and a circular base.[\/caption]\r\n\r\n[reveal-answer q=\"143368392\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"143368392\"]\r\n<p id=\"fs-id1167793938250\">First change the disk [latex]{(x-1)^2} + {y^2} = {1}[\/latex] to polar coordinates. Expanding the square term, we have [latex]{x^2} - {2x} + {1} + {y^2} = {1}[\/latex]. Then simplify to get [latex]{x^2} + {y^2} = {2x}[\/latex], which in polar coordinates becomes [latex]{r^2} = {2r}{\\cos}{\\theta}[\/latex] and then either [latex]{r} = {0}[\/latex] or [latex]{r} = {2}{\\cos}{\\theta}[\/latex]. Similarly, the equation of the paraboloid changes to [latex]{z} = {4} - {r^2}[\/latex]. Therefore we can describe the disk [latex]{(x-1)^2} + {y^2} = {1}[\/latex] on the [latex]xy[\/latex]-plane as the region<\/p>\r\n<p style=\"text-align: center;\">[latex]{D} = {\\left \\{{(r,{\\theta})}{\\mid}{0} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {{\\pi},{0}} \\ {\\leq} \\ {r} \\ {\\leq} \\ {2{\\cos}{\\theta}} \\right \\}}[\/latex].<\/p>\r\n<p id=\"fs-id1167793547432\">Hence the volume of the solid bounded above by the paraboloid [latex]{z} = {4} - {x^2} - {y^2}[\/latex] and below by [latex]{r} = {2}{\\cos}{\\theta}[\/latex] is<\/p>\r\n<p style=\"text-align: left;\">[latex]\\hspace{3cm}\\begin{align}\r\nV&amp;=\\underset{D}{\\displaystyle\\iint}f(r,\\theta)r \\ dr \\ d\\theta =\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi}\\displaystyle\\int_{r=0}^{r=2\\cos\\theta}(4-r^2)r \\ dr \\ d\\theta \\\\\r\n&amp;=\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi}\\left[4\\frac{r^2}2-\\frac{r^4}4\\Bigg|_{r=0}^{r=2\\cos\\theta}\\right]d\\theta \\\\\r\n&amp;=\\displaystyle\\int_{0}^{\\pi}[8\\cos^2\\theta-4\\cos^2\\theta]d\\theta=\\left[\\frac52\\theta+\\frac52\\sin\\theta\\cos\\theta-\\sin\\theta\\cos^3\\theta\\right]_0^{\\pi}=\\frac52\\pi\r\n\\end{align}[\/latex].\r\n[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\nNotice in the next example that integration is not always easy with polar coordinates. Complexity of integration depends on the function and also on the region over which we need to perform the integration. If the region has a more natural expression in polar coordinates or if [latex]f[\/latex] has a simpler antiderivative in polar coordinates, then the change in polar coordinates is appropriate; otherwise, use rectangular coordinates.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding a volume using a double integral<\/h3>\r\nFind the volume of the region that lies under the paraboloid [latex]z=x^{2}+y^{2}[\/latex] and above the triangle enclosed by the lines [latex]y=x[\/latex], [latex]x=0[\/latex] and [latex]x+y=2[\/latex] in the [latex]xy[\/latex]-plane (Figure 3).\r\n\r\n[reveal-answer q=\"300824617\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"300824617\"]\r\n<p id=\"fs-id1167793948459\">First examine the region over which we need to set up the double integral and the accompanying paraboloid.<\/p>\r\n\r\n\r\n[caption id=\"attachment_1366\" align=\"aligncenter\" width=\"848\"]<img class=\"size-full wp-image-1366\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25143350\/5-3-9.jpeg\" alt=\"This figure consists of three figures. The first is simply a paraboloid that opens up. The second shows the region D bounded by x = 0, y = x, and x + y = 2 with a vertical double-sided arrow within the region. The second shows the same region but in polar coordinates, so the lines bounding D are theta = pi\/2, r = 2\/(cos theta + sin theta), and theta = pi\/4, with a double-sided arrow that has one side pointed at the origin.\" width=\"848\" height=\"305\" \/> Figure 3.\u00a0Finding the volume of a solid under a paraboloid and above a given triangle.[\/caption]\r\n\r\nThe region [latex]D[\/latex] is [latex]{\\left \\{{(x,y)}{\\mid}{0} \\ {\\leq} \\ {x} \\ {\\leq} \\ {1,x} \\ {\\leq} \\ {y} \\ {\\leq} \\ {2-x} \\right \\}}[\/latex]. Converting the lines [latex]y=x[\/latex], [latex]x=0[\/latex], and [latex]x+y=2[\/latex] in the [latex]xy[\/latex]-plane to functions of [latex]r[\/latex] and [latex]\\theta[\/latex], we have [latex]{{\\theta} = {\\pi}\/{4}}, \\ {{\\theta} = {\\pi}\/{2}},[\/latex] and [latex]{r} = {2}\/{({cos}{\\theta}+{sin}{\\theta})}[\/latex], respectively. Graphing the region on the [latex]xy[\/latex]-plane, we see that it looks like [latex]{D}= {\\left \\{{(r,{\\theta})}{\\mid}{{\\pi}\/{4}} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {{\\pi}\/{2},0} \\ {\\leq} \\ {r} \\ {\\leq} \\ {2\/{({\\cos}{\\theta}+{\\sin}{\\theta})}} \\right \\}}[\/latex]. Now converting the equation of the surface gives [latex]{z} = {x^2} + {y^2} = {r^2}[\/latex]. Therefore, the volume of the solid is given by the double integral\r\n[latex]\\hspace{2cm}\\begin{align}\r\n\r\nV&amp;=\\underset{D}{\\displaystyle\\iint}f(r,\\theta)r \\ dr \\ d\\theta=\\displaystyle\\int_{\\theta=\\pi\/4}^{\\theta=\\pi\/2}\\displaystyle\\int_{r=0}^{r=2\/(\\cos\\theta+\\sin\\theta}r^2r \\ dr \\ d\\theta=\\displaystyle\\int_{\\pi\/4}^{\\pi\/2}\\left[\\frac{r^4}4\\right]_{0}^{2\/(\\cos\\theta+\\sin\\theta)} \\ d\\theta \\\\\r\n\r\n&amp;=\\frac14\\displaystyle\\int_{\\pi\/4}^{\\pi\/2}\\left(\\frac2{(\\cos\\theta+\\sin\\theta)}\\right)^4 \\ d\\theta = \\frac{16}4\\displaystyle\\int_{\\pi\/4}^{\\pi\/2}\\left(\\frac1{(\\cos\\theta+\\sin\\theta)}\\right)^4 \\ d\\theta =4\\displaystyle\\int_{\\pi\/4}^{\\pi\/2}\\left(\\frac1{(\\cos\\theta+\\sin\\theta)}\\right)^4 \\ d\\theta.\r\n\r\n\\end{align}[\/latex]\r\nAs you can see, this integral is very complicated. So, we can instead evaluate this double integral in rectangular coordinates as\r\n<p style=\"text-align: center;\">[latex]\\large{V=\\displaystyle\\int_0^1\\displaystyle\\int_x^{2-x}(x^2+y^2)dydx}[\/latex].<\/p>\r\nEvaluating gives\r\n[latex]\\hspace{6cm}\\begin{align}\r\n\r\nV&amp;=\\displaystyle\\int_0^1\\displaystyle\\int_x^{2-x}(x^2+y^2)dydx=\\displaystyle\\int_0^1\\left[x^2y+\\frac{y^3}3\\right]\\Bigg|_x^{2-x}dx \\\\\r\n\r\n&amp;=\\displaystyle\\int_0^1\\frac83-4x+4x^2-\\frac{8x^3}3dx \\\\\r\n\r\n&amp;=\\left[\\frac{8x}3-2x^2+\\frac{4x^3}3-\\frac{2x^4}3\\right]\\bigg|_0^1=\\frac43.\r\n\r\n\\end{align}[\/latex]\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nTo answer the question of how the formulas for the volumes of different standard solids such as a sphere, a cone, or a cylinder are found, we want to demonstrate an example and find the volume of an arbitrary cone.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding a volume using a double integral<\/h3>\r\nUse polar coordinates to find the volume inside the cone [latex]{z} = {2} - {\\sqrt{{x^2}+{y^2}}}[\/latex] and above the [latex]xy[\/latex]-<span class=\"os-math-in-para\"><span id=\"MathJax-Element-550-Frame\" class=\"MathJax\" style=\"box-sizing: border-box; overflow: initial; display: inline-table; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;\" tabindex=\"0\" role=\"presentation\" data-mathml=\"&lt;math xmlns=&quot;http:\/\/www.w3.org\/1998\/Math\/MathML&quot; display=&quot;inline&quot;&gt;&lt;semantics&gt;&lt;mrow&gt;&lt;mrow&gt;&lt;mi&gt;x&lt;\/mi&gt;&lt;mi&gt;y&lt;\/mi&gt;&lt;mtext&gt;-plane&lt;\/mtext&gt;&lt;mtext&gt;.&lt;\/mtext&gt;&lt;\/mrow&gt;&lt;\/mrow&gt;&lt;annotation-xml encoding=&quot;MathML-Content&quot;&gt;&lt;mrow&gt;&lt;mi&gt;x&lt;\/mi&gt;&lt;mi&gt;y&lt;\/mi&gt;&lt;mtext&gt;-plane&lt;\/mtext&gt;&lt;mtext&gt;.&lt;\/mtext&gt;&lt;\/mrow&gt;&lt;\/annotation-xml&gt;&lt;\/semantics&gt;&lt;\/math&gt;\"><span id=\"MathJax-Span-13451\" class=\"math\"><span id=\"MathJax-Span-13452\" class=\"mrow\"><span id=\"MathJax-Span-13453\" class=\"semantics\"><span id=\"MathJax-Span-13454\" class=\"mrow\"><span id=\"MathJax-Span-13455\" class=\"mrow\"><span id=\"MathJax-Span-13458\" class=\"mtext\">plane<\/span><span id=\"MathJax-Span-13459\" class=\"mtext\">.<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>\r\n\r\n[reveal-answer q=\"135668234\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"135668234\"]\r\n<h2 data-type=\"solution-title\"><span class=\"os-title-label\">Solution<\/span><\/h2>\r\n<div class=\"os-solution-container\">\r\n<p id=\"fs-id1167793887539\">The region [latex]D[\/latex] for the integration is the base of the cone, which appears to be a circle on the [latex]xy[\/latex]-plane (see the following figure).<\/p>\r\n\r\n[caption id=\"attachment_1367\" align=\"aligncenter\" width=\"556\"]<img class=\"size-full wp-image-1367\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25143428\/5-3-10.jpeg\" alt=\"A cone given by z = 2 minus the square root of (x squared plus y squared) and a circle given by x squared plus y squared = 4. The cone is above the circle in xyz space.\" width=\"556\" height=\"533\" \/> Figure 4.\u00a0Finding the volume of a solid inside the cone and above the [latex]xy[\/latex]-plane.[\/caption]\r\n<div id=\"fs-id1167793958280\" class=\" ui-solution-visible\" data-type=\"solution\" aria-label=\"hide solution\" aria-expanded=\"true\"><section class=\"ui-body\" role=\"alert\">\r\n<div class=\"os-solution-container\">\r\n<p id=\"fs-id1167794049092\">We find the equation of the circle by setting [latex]z=0[\/latex]:<\/p>\r\n[latex]\\hspace{10cm}\\large{\\begin{align}\r\n{0} &amp;= {2} - {\\sqrt{{x^2}+{y^2}}} \\\\\r\n{2} &amp;= {\\sqrt{{x^2}+{y^2}}} \\\\\r\n{x^2} + {y^2} &amp;= {4}\r\n\\end{align}}[\/latex]\r\n<p id=\"fs-id1167794123037\">This means the radius of the circle is 2, so for the integration we have [latex]{0} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {2{\\pi}}[\/latex] and [latex]{0} \\ {\\leq} \\ {r} \\ {\\leq} \\ {2}[\/latex]. Substituting [latex]{x} = {r}{cos}{\\theta}[\/latex] and [latex]{y} = {r}{sin}{\\theta}[\/latex] in the equation [latex]{z} = {2} - {\\sqrt{{x^2}+{y^2}}}[\/latex] we have [latex]z=2-r[\/latex]. Therefore, the volume of the cone is<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\int_{\\theta=0}^{\\theta=2\\pi}\\displaystyle\\int_{r=0}^{r=2}(2-r)r \\ dr \\ d\\theta = 2\\pi\\frac43=\\frac{8\\pi}3}[\/latex] cubic units.<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1167794172290\" data-type=\"commentary\">\r\n<h2 id=\"26\" data-type=\"commentary-title\"><span class=\"os-title-label\">Analysis<\/span><\/h2>\r\n<p id=\"fs-id1167794172296\">Note that if we were to find the volume of an arbitrary cone with radius [latex]a[\/latex] and height [latex]h[\/latex] units, then the equation of the cone would be [latex]{z} = {h} - {\\frac{h}{a}}{\\sqrt{{x^2}+{y^2}}}[\/latex].<\/p>\r\n<p id=\"fs-id1167793895360\">We can still use\u00a0Figure 4\u00a0and set up the integral as [latex]\\displaystyle\\int_{\\theta=0}^{\\theta=2\\pi}\\displaystyle\\int_{r=0}^{r=a}\\left(h-\\frac{h}{a}\\right)r \\ dr \\ d\\theta[\/latex].<\/p>\r\n<p id=\"fs-id1167793419012\">Evaluating the integral, we get [latex]{\\frac{1}{3}}{\\pi}{a^2}{h}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nUse 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].\r\n\r\n[reveal-answer q=\"512348890\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"512348890\"]\r\n\r\n[latex]V=\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^{2\\sqrt2}(16-2r^2)r \\ dr \\ d\\theta=64\\pi[\/latex] cubic units.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793605536\">As with rectangular coordinates, we can also use polar coordinates to find areas of certain regions using a double integral. As before, we need to understand the region whose area we want to compute. Sketching a graph and identifying the region can be helpful to realize the limits of integration. Generally, the area formula in double integration will look like<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\text{Area }A = \\displaystyle\\int_\\alpha^\\beta\\displaystyle\\int_{h_1(\\theta)}^{h_2(\\theta)}{1r} \\ {dr} \\ {d{\\theta}}}[\/latex].<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding an area using a double integral in polar coordinates<\/h3>\r\nEvaluate the area bounded by the curve\u00a0[latex]{r} = {\\cos}{4}{\\theta}[\/latex].\r\n\r\n[reveal-answer q=\"411567920\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"411567920\"]\r\n\r\nSketching the graph of the function [latex]{r} = {\\cos}{4}{\\theta}[\/latex] reveals that it is a polar rose with eight petals (see the following figure).\r\n\r\n[caption id=\"attachment_1368\" align=\"aligncenter\" width=\"360\"]<img class=\"size-full wp-image-1368\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25143517\/5-3-11.jpeg\" alt=\"A rose with eight petals given by r = cos (4 theta).\" width=\"360\" height=\"360\" \/> Figure 5.\u00a0Finding the area of a polar rose with eight petals.[\/caption]\r\n\r\nUsing\u00a0<span id=\"82852cef-a933-495b-9be8-6b3aa0676026_term218\" class=\"no-emphasis\" data-type=\"term\">symmetry<\/span>, we can see that we need to find the area of one petal and then multiply it by 8. Notice that the values of [latex]\\theta[\/latex] for which the graph passes through the origin are the zeros of the function [latex]{\\cos}{4}{\\theta}[\/latex], and these are odd multiples of [latex]{\\pi}\/{8}[\/latex]. Thus, one of the petals corresponds to the values of [latex]\\theta[\/latex] in the interval [latex][{{-\\pi}\/{8}},{{\\pi}\/{8}}][\/latex]. Therefore, the area bounded by the curve [latex]{r} = {\\cos}{4}{\\theta}[\/latex] is\r\n\r\n<center>\r\n[latex]\\begin{align}<\/center>A&amp;=8\\displaystyle\\int_{\\theta=-\\pi\/8}^{\\theta=\\pi\/8}\\displaystyle\\int_{r=0}^{r=\\cos4\\theta}1 r \\ dr \\ d\\theta \\\\\r\n\r\n&amp;=8\\displaystyle\\int_{-\\pi\/8}^{\\pi\/8}\\left[\\frac12r^2\\bigg|_{0}^{\\cos4\\theta}\\right]d\\theta=8\\displaystyle\\int_{-\\pi\/8}^{\\pi\/8}\\frac12\\cos^2{4}\\theta \\ d\\theta=8\\left[\\frac14\\theta+\\frac1{16}\\sin4\\theta\\cos4\\theta\\bigg|_{-\\pi\/8}^{\\pi\/8}\\right]=8[\\frac{\\pi}{16}]=\\frac{\\pi}2.\r\n\r\n\\end{align}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding area between two polar curves<\/h3>\r\nFind the area enclosed by the circle [latex]{r} = {3}{\\cos}{\\theta}[\/latex] and the cardioid [latex]{r} = {1} + {\\cos}{\\theta}[\/latex].\r\n\r\n[reveal-answer q=\"351268027\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"351268027\"]\r\n<p id=\"fs-id1167793544550\">First and foremost, sketch the graphs of the region (Figure 6).<\/p>\r\n\r\n\r\n[caption id=\"attachment_1369\" align=\"alignnone\" width=\"657\"]<img class=\"size-full wp-image-1369\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25143555\/5-3-12.jpeg\" alt=\"A cardioid with equation 1 + cos theta is shown overlapping a circle given by r = 3 cos theta, which is a circle of radius 3 with center (1.5, 0). The area bounded by the x axis, the cardioid, and the dashed line connecting the origin to the intersection of the cardioid and circle on the r = 2 line is shaded.\" width=\"657\" height=\"644\" \/> Figure 6.\u00a0Finding the area enclosed by both a circle and a cardioid.[\/caption]\r\n<p id=\"fs-id1167793589743\">We can from see the symmetry of the graph that we need to find the points of intersection. Setting the two equations equal to each other gives<\/p>\r\n<p style=\"text-align: center;\">[latex]{3}{cos}{\\theta} = {1} + {\\cos}{\\theta}[\/latex].<\/p>\r\n<span style=\"font-size: 1rem; text-align: initial;\">One of the points of intersection is [latex]{\\theta} = {{\\pi}\/{3}}[\/latex]. The area above the polar axis consists of two parts, with one part defined by the cardioid from [latex]{\\theta} = {0}[\/latex] to [latex]{\\theta} = {{\\pi}\/{3}}[\/latex] and the other part defined by the circle from [latex]{\\theta} = {{\\pi}\/{3}}[\/latex] to [latex]{\\theta} = {{\\pi}\/{2}}[\/latex]. By symmetry, the total area is twice the area above the polar axis. Thus, we have<\/span>\r\n<p style=\"text-align: center;\">[latex]\\large{A=2\\left[\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi\/3}\\displaystyle\\int_{r=0}^{r=1+\\cos\\theta}1r \\ dr \\ d\\theta+\\displaystyle\\int_{\\theta=\\pi\/3}^{\\theta=\\pi\/2}\\displaystyle\\int_{r=0}^{r=3\\cos\\theta}1r \\ dr \\ d\\theta\\right]}[\/latex].<\/p>\r\nEvaluating each piece separately, we find that the area is\r\n<p style=\"text-align: center;\">[latex]{A} = {2}{\\left ({{\\frac{1}{4}}{\\pi}} + {{\\frac{9}{16}}{\\sqrt{3}}} + {{\\frac{3}{8}}{\\pi}} - {{\\frac{9}{16}}{\\sqrt{3}}} \\right)} = {2}{\\left ({\\frac{5}{8}}{\\pi} \\right )} = {{\\frac{5}{4}}{\\pi}}[\/latex] square units.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nFind the area enclosed inside the cardioid [latex]{r} = {3} - {{3}{\\sin}{\\theta}}[\/latex] and outside the cardioid [latex]{r} = {1} + {{\\sin}{\\theta}}[\/latex].\r\n\r\n[reveal-answer q=\"579921064\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"579921064\"]\r\n\r\n[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]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: evaluating an improper double integral in polar coordinates<\/h3>\r\nEvaluate the integral\u00a0[latex]\\underset{R^2}{\\displaystyle\\iint}e^{-10(x^2+y^2)}dx \\ dy[\/latex].\r\n\r\n[reveal-answer q=\"271367529\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"271367529\"]\r\n<p id=\"fs-id1167793444452\">This is an improper integral because we are integrating over an unbounded region [latex]R^2[\/latex]. In polar coordinates, the entire plane [latex]R^2[\/latex] can be seen as [latex]{0} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {{{2}{\\pi}},0} \\ {\\leq} \\ {r} \\ {\\leq} \\ {\\infty}[\/latex].<\/p>\r\n<p id=\"fs-id1167793387195\">Using the changes of variables from rectangular coordinates to polar coordinates, we have<\/p>\r\n[latex]\\hspace{3cm}\\begin{align}\r\n\r\n\\underset{R^2}{\\displaystyle\\iint}e^{-10(x^2+y^2)}dx \\ dy&amp;=\\displaystyle\\int_{\\theta=0}^{\\theta=2\\pi}\\displaystyle\\int_{r=0}^{r=\\infty}e^{-10r^2}r \\ dr \\ d\\theta=\\displaystyle\\int_{\\theta=0}^{\\theta=2\\pi}\\left(\\displaystyle\\lim_{a\\to\\infty}\\displaystyle\\int_{r=0}^{r=a}e^{-10r^2}r \\ dr\\right)d\\theta \\\\\r\n\r\n&amp;=\\left(\\displaystyle\\int_{\\theta=0}^{\\theta=2\\pi}d\\theta\\right)\\left(\\displaystyle\\lim_{a\\to\\infty}\\displaystyle\\int_{r=0}^{r=a}e^{-10r^2}r \\ dr\\right) \\\\\r\n\r\n&amp;=2\\pi\\left(\\displaystyle\\lim_{a\\to\\infty}\\displaystyle\\int_{r=0}^{r=a}e^{-10r^2}r \\ dr\\right) \\\\\r\n\r\n&amp;=2\\pi\\displaystyle\\lim_{a\\to\\infty}\\left(-\\frac1{20}\\right)\\left(e^{-10r^2}\\bigg|_0^1\\right)\\\\\r\n\r\n&amp;=2\\pi\\left(-\\frac1{20}\\right)\\displaystyle\\lim_{a\\to\\infty}\\left(e^{-10a^2}-1\\right)\\\\\r\n\r\n&amp;=\\frac{\\pi}{10}\r\n\r\n\\end{align}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nEvaluate the integral\u00a0[latex]\\underset{R^2}{\\displaystyle\\iint}{e^{{-4}{(x^2+y^2)}}}{dx^{}}{dy}[\/latex].\r\n\r\n[reveal-answer q=\"967834547\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"967834547\"]\r\n\r\n[latex]\\frac{\\pi}4[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167793952731\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\"><\/div>\r\n&nbsp;\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=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\"><\/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>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul class=\"os-abstract\">\n<li><span class=\"os-abstract-content\">Use double integrals in polar coordinates to calculate areas and volumes.<\/span><\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1167793277568\">As in rectangular coordinates, if a solid [latex]S[\/latex] is bounded by the surface [latex]{z} = {f}{(r,{\\theta})}[\/latex], as well as by the surfaces [latex]{r} = {a}, {r} = {b}, {\\theta} = {\\alpha}[\/latex], and [latex]{\\theta} = {\\beta}[\/latex] we can find the volume [latex]V[\/latex] of [latex]S[\/latex] by double integration, as<\/p>\n<p style=\"text-align: center;\">[latex]\\large{{V} = \\underset{R}{\\displaystyle\\iint}{f}{(r,{\\theta})} \\ {r} \\ {dr} \\ {d{\\theta}} = \\displaystyle\\int_{\\theta=\\alpha}^{\\theta=\\beta}\\displaystyle\\int_{r=a}^{r=b}f(r,\\theta)r \\ dr \\ d\\theta}[\/latex].<\/p>\n<p>If the base of the solid can be described as [latex]{D} = {\\left \\{{(r,{\\theta})}{\\mid}{\\alpha} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {{\\beta},{h_1}}{(\\theta)} \\ {\\leq} \\ {r} \\ {\\leq} \\ {h_2{(\\theta)}} \\right \\}}[\/latex], then the double integral for the volume becomes<\/p>\n<p style=\"text-align: center;\">[latex]\\large{{V} = \\underset{D}{\\displaystyle\\iint}{f}{(r,{\\theta})} \\ {r} \\ {dr} \\ {d{\\theta}} = \\displaystyle\\int_{\\theta=\\alpha}^{\\theta=\\beta}\\displaystyle\\int_{r=h_1(\\theta)}^{r=h_2(\\theta)}f(r,\\theta)r \\ dr \\ d\\theta}[\/latex]<\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\">We illustrate this idea with some examples.<\/span><\/p>\n<div id=\"fs-id1167793949568\" class=\"ui-has-child-title\" data-type=\"example\">\n<div class=\"textbox exercises\">\n<h3>Example: finding a volume using a double integral<\/h3>\n<p>Find the volume of the solid that lies under the paraboloid [latex]z=1-x^{2}-y^{2}[\/latex] and above the unit circle on the [latex]xy[\/latex]-plane (see the following figure).<\/p>\n<div id=\"attachment_1364\" style=\"width: 466px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1364\" class=\"size-full wp-image-1364\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25143225\/5-3-7.jpeg\" alt=\"The paraboloid z = 1 minus x squared minus y squared is shown, which in this graph looks like a sheet with the middle gently puffed up and the corners anchored.\" width=\"456\" height=\"371\" \/><\/p>\n<p id=\"caption-attachment-1364\" class=\"wp-caption-text\">Figure 1. The paraboloid\u00a0[latex]z=1-x^{2}-y^{2}[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q613772890\">Show Solution<\/span><\/p>\n<div id=\"q613772890\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793416705\">By the method of double integration, we can see that the volume is the iterated integral of the form [latex]\\underset{R}{\\displaystyle\\iint}{(1-{x^2}-{y^2})}{dA}[\/latex] where [latex]{R} = {\\left \\{{(r,{\\theta})}{\\mid}{0} \\ {\\leq} \\ {r} \\ {\\leq} \\ {{1},{0}} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {2{\\pi}} \\right \\}}[\/latex].<\/p>\n<p id=\"fs-id1167793936934\">This integration was shown before in\u00a0Example &#8220;Evaluating a Double Integral by Converting from Rectangular Coordinates&#8221;, so the volume is [latex]{\\frac{\\pi}{2}}[\/latex] cubic units.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: finding a volume using double integration<\/h3>\n<p>Find the volume of the solid that lies under the paraboloid [latex]{z} = {4} - {x^2} - {y^2}[\/latex] and above the disk [latex]{(x-1)^2} + {y^2} = {1}[\/latex] on the [latex]xy[\/latex]-plane. See the paraboloid in\u00a0Figure 2\u00a0intersecting the cylinder [latex]{(x-1)^2} + {y^2} = {1}[\/latex] above the [latex]xy[\/latex]-plane.<\/p>\n<div id=\"attachment_1365\" style=\"width: 429px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1365\" class=\"size-full wp-image-1365\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25143256\/5-3-8.jpeg\" alt=\"A paraboloid with equation z = 4 minus x squared minus y squared is intersected by a cylinder with equation (x minus 1) squared + y squared = 1.\" width=\"419\" height=\"500\" \/><\/p>\n<p id=\"caption-attachment-1365\" class=\"wp-caption-text\">Figure 2.\u00a0Finding the volume of a solid with a paraboloid cap and a circular base.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q143368392\">Show Solution<\/span><\/p>\n<div id=\"q143368392\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793938250\">First change the disk [latex]{(x-1)^2} + {y^2} = {1}[\/latex] to polar coordinates. Expanding the square term, we have [latex]{x^2} - {2x} + {1} + {y^2} = {1}[\/latex]. Then simplify to get [latex]{x^2} + {y^2} = {2x}[\/latex], which in polar coordinates becomes [latex]{r^2} = {2r}{\\cos}{\\theta}[\/latex] and then either [latex]{r} = {0}[\/latex] or [latex]{r} = {2}{\\cos}{\\theta}[\/latex]. Similarly, the equation of the paraboloid changes to [latex]{z} = {4} - {r^2}[\/latex]. Therefore we can describe the disk [latex]{(x-1)^2} + {y^2} = {1}[\/latex] on the [latex]xy[\/latex]-plane as the region<\/p>\n<p style=\"text-align: center;\">[latex]{D} = {\\left \\{{(r,{\\theta})}{\\mid}{0} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {{\\pi},{0}} \\ {\\leq} \\ {r} \\ {\\leq} \\ {2{\\cos}{\\theta}} \\right \\}}[\/latex].<\/p>\n<p id=\"fs-id1167793547432\">Hence the volume of the solid bounded above by the paraboloid [latex]{z} = {4} - {x^2} - {y^2}[\/latex] and below by [latex]{r} = {2}{\\cos}{\\theta}[\/latex] is<\/p>\n<p style=\"text-align: left;\">[latex]\\hspace{3cm}\\begin{align}  V&=\\underset{D}{\\displaystyle\\iint}f(r,\\theta)r \\ dr \\ d\\theta =\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi}\\displaystyle\\int_{r=0}^{r=2\\cos\\theta}(4-r^2)r \\ dr \\ d\\theta \\\\  &=\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi}\\left[4\\frac{r^2}2-\\frac{r^4}4\\Bigg|_{r=0}^{r=2\\cos\\theta}\\right]d\\theta \\\\  &=\\displaystyle\\int_{0}^{\\pi}[8\\cos^2\\theta-4\\cos^2\\theta]d\\theta=\\left[\\frac52\\theta+\\frac52\\sin\\theta\\cos\\theta-\\sin\\theta\\cos^3\\theta\\right]_0^{\\pi}=\\frac52\\pi  \\end{align}[\/latex].\n<\/div>\n<\/div>\n<\/div>\n<p>Notice in the next example that integration is not always easy with polar coordinates. Complexity of integration depends on the function and also on the region over which we need to perform the integration. If the region has a more natural expression in polar coordinates or if [latex]f[\/latex] has a simpler antiderivative in polar coordinates, then the change in polar coordinates is appropriate; otherwise, use rectangular coordinates.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: finding a volume using a double integral<\/h3>\n<p>Find the volume of the region that lies under the paraboloid [latex]z=x^{2}+y^{2}[\/latex] and above the triangle enclosed by the lines [latex]y=x[\/latex], [latex]x=0[\/latex] and [latex]x+y=2[\/latex] in the [latex]xy[\/latex]-plane (Figure 3).<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q300824617\">Show Solution<\/span><\/p>\n<div id=\"q300824617\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793948459\">First examine the region over which we need to set up the double integral and the accompanying paraboloid.<\/p>\n<div id=\"attachment_1366\" style=\"width: 858px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1366\" class=\"size-full wp-image-1366\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25143350\/5-3-9.jpeg\" alt=\"This figure consists of three figures. The first is simply a paraboloid that opens up. The second shows the region D bounded by x = 0, y = x, and x + y = 2 with a vertical double-sided arrow within the region. The second shows the same region but in polar coordinates, so the lines bounding D are theta = pi\/2, r = 2\/(cos theta + sin theta), and theta = pi\/4, with a double-sided arrow that has one side pointed at the origin.\" width=\"848\" height=\"305\" \/><\/p>\n<p id=\"caption-attachment-1366\" class=\"wp-caption-text\">Figure 3.\u00a0Finding the volume of a solid under a paraboloid and above a given triangle.<\/p>\n<\/div>\n<p>The region [latex]D[\/latex] is [latex]{\\left \\{{(x,y)}{\\mid}{0} \\ {\\leq} \\ {x} \\ {\\leq} \\ {1,x} \\ {\\leq} \\ {y} \\ {\\leq} \\ {2-x} \\right \\}}[\/latex]. Converting the lines [latex]y=x[\/latex], [latex]x=0[\/latex], and [latex]x+y=2[\/latex] in the [latex]xy[\/latex]-plane to functions of [latex]r[\/latex] and [latex]\\theta[\/latex], we have [latex]{{\\theta} = {\\pi}\/{4}}, \\ {{\\theta} = {\\pi}\/{2}},[\/latex] and [latex]{r} = {2}\/{({cos}{\\theta}+{sin}{\\theta})}[\/latex], respectively. Graphing the region on the [latex]xy[\/latex]-plane, we see that it looks like [latex]{D}= {\\left \\{{(r,{\\theta})}{\\mid}{{\\pi}\/{4}} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {{\\pi}\/{2},0} \\ {\\leq} \\ {r} \\ {\\leq} \\ {2\/{({\\cos}{\\theta}+{\\sin}{\\theta})}} \\right \\}}[\/latex]. Now converting the equation of the surface gives [latex]{z} = {x^2} + {y^2} = {r^2}[\/latex]. Therefore, the volume of the solid is given by the double integral<br \/>\n[latex]\\hspace{2cm}\\begin{align}    V&=\\underset{D}{\\displaystyle\\iint}f(r,\\theta)r \\ dr \\ d\\theta=\\displaystyle\\int_{\\theta=\\pi\/4}^{\\theta=\\pi\/2}\\displaystyle\\int_{r=0}^{r=2\/(\\cos\\theta+\\sin\\theta}r^2r \\ dr \\ d\\theta=\\displaystyle\\int_{\\pi\/4}^{\\pi\/2}\\left[\\frac{r^4}4\\right]_{0}^{2\/(\\cos\\theta+\\sin\\theta)} \\ d\\theta \\\\    &=\\frac14\\displaystyle\\int_{\\pi\/4}^{\\pi\/2}\\left(\\frac2{(\\cos\\theta+\\sin\\theta)}\\right)^4 \\ d\\theta = \\frac{16}4\\displaystyle\\int_{\\pi\/4}^{\\pi\/2}\\left(\\frac1{(\\cos\\theta+\\sin\\theta)}\\right)^4 \\ d\\theta =4\\displaystyle\\int_{\\pi\/4}^{\\pi\/2}\\left(\\frac1{(\\cos\\theta+\\sin\\theta)}\\right)^4 \\ d\\theta.    \\end{align}[\/latex]<br \/>\nAs you can see, this integral is very complicated. So, we can instead evaluate this double integral in rectangular coordinates as<\/p>\n<p style=\"text-align: center;\">[latex]\\large{V=\\displaystyle\\int_0^1\\displaystyle\\int_x^{2-x}(x^2+y^2)dydx}[\/latex].<\/p>\n<p>Evaluating gives<br \/>\n[latex]\\hspace{6cm}\\begin{align}    V&=\\displaystyle\\int_0^1\\displaystyle\\int_x^{2-x}(x^2+y^2)dydx=\\displaystyle\\int_0^1\\left[x^2y+\\frac{y^3}3\\right]\\Bigg|_x^{2-x}dx \\\\    &=\\displaystyle\\int_0^1\\frac83-4x+4x^2-\\frac{8x^3}3dx \\\\    &=\\left[\\frac{8x}3-2x^2+\\frac{4x^3}3-\\frac{2x^4}3\\right]\\bigg|_0^1=\\frac43.    \\end{align}[\/latex]\n<\/p><\/div>\n<\/div>\n<\/div>\n<p>To answer the question of how the formulas for the volumes of different standard solids such as a sphere, a cone, or a cylinder are found, we want to demonstrate an example and find the volume of an arbitrary cone.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: finding a volume using a double integral<\/h3>\n<p>Use polar coordinates to find the volume inside the cone [latex]{z} = {2} - {\\sqrt{{x^2}+{y^2}}}[\/latex] and above the [latex]xy[\/latex]&#8211;<span class=\"os-math-in-para\"><span id=\"MathJax-Element-550-Frame\" class=\"MathJax\" style=\"box-sizing: border-box; overflow: initial; display: inline-table; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;\" tabindex=\"0\" role=\"presentation\" data-mathml=\"&lt;math xmlns=&quot;http:\/\/www.w3.org\/1998\/Math\/MathML&quot; display=&quot;inline&quot;&gt;&lt;semantics&gt;&lt;mrow&gt;&lt;mrow&gt;&lt;mi&gt;x&lt;\/mi&gt;&lt;mi&gt;y&lt;\/mi&gt;&lt;mtext&gt;-plane&lt;\/mtext&gt;&lt;mtext&gt;.&lt;\/mtext&gt;&lt;\/mrow&gt;&lt;\/mrow&gt;&lt;annotation-xml encoding=&quot;MathML-Content&quot;&gt;&lt;mrow&gt;&lt;mi&gt;x&lt;\/mi&gt;&lt;mi&gt;y&lt;\/mi&gt;&lt;mtext&gt;-plane&lt;\/mtext&gt;&lt;mtext&gt;.&lt;\/mtext&gt;&lt;\/mrow&gt;&lt;\/annotation-xml&gt;&lt;\/semantics&gt;&lt;\/math&gt;\"><span id=\"MathJax-Span-13451\" class=\"math\"><span id=\"MathJax-Span-13452\" class=\"mrow\"><span id=\"MathJax-Span-13453\" class=\"semantics\"><span id=\"MathJax-Span-13454\" class=\"mrow\"><span id=\"MathJax-Span-13455\" class=\"mrow\"><span id=\"MathJax-Span-13458\" class=\"mtext\">plane<\/span><span id=\"MathJax-Span-13459\" class=\"mtext\">.<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q135668234\">Show Solution<\/span><\/p>\n<div id=\"q135668234\" class=\"hidden-answer\" style=\"display: none\">\n<h2 data-type=\"solution-title\"><span class=\"os-title-label\">Solution<\/span><\/h2>\n<div class=\"os-solution-container\">\n<p id=\"fs-id1167793887539\">The region [latex]D[\/latex] for the integration is the base of the cone, which appears to be a circle on the [latex]xy[\/latex]-plane (see the following figure).<\/p>\n<div id=\"attachment_1367\" style=\"width: 566px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1367\" class=\"size-full wp-image-1367\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25143428\/5-3-10.jpeg\" alt=\"A cone given by z = 2 minus the square root of (x squared plus y squared) and a circle given by x squared plus y squared = 4. The cone is above the circle in xyz space.\" width=\"556\" height=\"533\" \/><\/p>\n<p id=\"caption-attachment-1367\" class=\"wp-caption-text\">Figure 4.\u00a0Finding the volume of a solid inside the cone and above the [latex]xy[\/latex]-plane.<\/p>\n<\/div>\n<div id=\"fs-id1167793958280\" class=\"ui-solution-visible\" data-type=\"solution\" aria-label=\"hide solution\" aria-expanded=\"true\">\n<section class=\"ui-body\" role=\"alert\">\n<div class=\"os-solution-container\">\n<p id=\"fs-id1167794049092\">We find the equation of the circle by setting [latex]z=0[\/latex]:<\/p>\n<p>[latex]\\hspace{10cm}\\large{\\begin{align}  {0} &= {2} - {\\sqrt{{x^2}+{y^2}}} \\\\  {2} &= {\\sqrt{{x^2}+{y^2}}} \\\\  {x^2} + {y^2} &= {4}  \\end{align}}[\/latex]<\/p>\n<p id=\"fs-id1167794123037\">This means the radius of the circle is 2, so for the integration we have [latex]{0} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {2{\\pi}}[\/latex] and [latex]{0} \\ {\\leq} \\ {r} \\ {\\leq} \\ {2}[\/latex]. Substituting [latex]{x} = {r}{cos}{\\theta}[\/latex] and [latex]{y} = {r}{sin}{\\theta}[\/latex] in the equation [latex]{z} = {2} - {\\sqrt{{x^2}+{y^2}}}[\/latex] we have [latex]z=2-r[\/latex]. Therefore, the volume of the cone is<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\int_{\\theta=0}^{\\theta=2\\pi}\\displaystyle\\int_{r=0}^{r=2}(2-r)r \\ dr \\ d\\theta = 2\\pi\\frac43=\\frac{8\\pi}3}[\/latex] cubic units.<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1167794172290\" data-type=\"commentary\">\n<h2 id=\"26\" data-type=\"commentary-title\"><span class=\"os-title-label\">Analysis<\/span><\/h2>\n<p id=\"fs-id1167794172296\">Note that if we were to find the volume of an arbitrary cone with radius [latex]a[\/latex] and height [latex]h[\/latex] units, then the equation of the cone would be [latex]{z} = {h} - {\\frac{h}{a}}{\\sqrt{{x^2}+{y^2}}}[\/latex].<\/p>\n<p id=\"fs-id1167793895360\">We can still use\u00a0Figure 4\u00a0and set up the integral as [latex]\\displaystyle\\int_{\\theta=0}^{\\theta=2\\pi}\\displaystyle\\int_{r=0}^{r=a}\\left(h-\\frac{h}{a}\\right)r \\ dr \\ d\\theta[\/latex].<\/p>\n<p id=\"fs-id1167793419012\">Evaluating the integral, we get [latex]{\\frac{1}{3}}{\\pi}{a^2}{h}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q512348890\">Show Solution<\/span><\/p>\n<div id=\"q512348890\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]V=\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^{2\\sqrt2}(16-2r^2)r \\ dr \\ d\\theta=64\\pi[\/latex] cubic units.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167793605536\">As with rectangular coordinates, we can also use polar coordinates to find areas of certain regions using a double integral. As before, we need to understand the region whose area we want to compute. Sketching a graph and identifying the region can be helpful to realize the limits of integration. Generally, the area formula in double integration will look like<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\text{Area }A = \\displaystyle\\int_\\alpha^\\beta\\displaystyle\\int_{h_1(\\theta)}^{h_2(\\theta)}{1r} \\ {dr} \\ {d{\\theta}}}[\/latex].<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: finding an area using a double integral in polar coordinates<\/h3>\n<p>Evaluate the area bounded by the curve\u00a0[latex]{r} = {\\cos}{4}{\\theta}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q411567920\">Show Solution<\/span><\/p>\n<div id=\"q411567920\" class=\"hidden-answer\" style=\"display: none\">\n<p>Sketching the graph of the function [latex]{r} = {\\cos}{4}{\\theta}[\/latex] reveals that it is a polar rose with eight petals (see the following figure).<\/p>\n<div id=\"attachment_1368\" style=\"width: 370px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1368\" class=\"size-full wp-image-1368\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25143517\/5-3-11.jpeg\" alt=\"A rose with eight petals given by r = cos (4 theta).\" width=\"360\" height=\"360\" \/><\/p>\n<p id=\"caption-attachment-1368\" class=\"wp-caption-text\">Figure 5.\u00a0Finding the area of a polar rose with eight petals.<\/p>\n<\/div>\n<p>Using\u00a0<span id=\"82852cef-a933-495b-9be8-6b3aa0676026_term218\" class=\"no-emphasis\" data-type=\"term\">symmetry<\/span>, we can see that we need to find the area of one petal and then multiply it by 8. Notice that the values of [latex]\\theta[\/latex] for which the graph passes through the origin are the zeros of the function [latex]{\\cos}{4}{\\theta}[\/latex], and these are odd multiples of [latex]{\\pi}\/{8}[\/latex]. Thus, one of the petals corresponds to the values of [latex]\\theta[\/latex] in the interval [latex][{{-\\pi}\/{8}},{{\\pi}\/{8}}][\/latex]. Therefore, the area bounded by the curve [latex]{r} = {\\cos}{4}{\\theta}[\/latex] is<\/p>\n<div style=\"text-align: center;\">\n[latex]\\begin{align}<\/div>\n<p>A&=8\\displaystyle\\int_{\\theta=-\\pi\/8}^{\\theta=\\pi\/8}\\displaystyle\\int_{r=0}^{r=\\cos4\\theta}1 r \\ dr \\ d\\theta \\\\    &=8\\displaystyle\\int_{-\\pi\/8}^{\\pi\/8}\\left[\\frac12r^2\\bigg|_{0}^{\\cos4\\theta}\\right]d\\theta=8\\displaystyle\\int_{-\\pi\/8}^{\\pi\/8}\\frac12\\cos^2{4}\\theta \\ d\\theta=8\\left[\\frac14\\theta+\\frac1{16}\\sin4\\theta\\cos4\\theta\\bigg|_{-\\pi\/8}^{\\pi\/8}\\right]=8[\\frac{\\pi}{16}]=\\frac{\\pi}2.    \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: finding area between two polar curves<\/h3>\n<p>Find the area enclosed by the circle [latex]{r} = {3}{\\cos}{\\theta}[\/latex] and the cardioid [latex]{r} = {1} + {\\cos}{\\theta}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q351268027\">Show Solution<\/span><\/p>\n<div id=\"q351268027\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793544550\">First and foremost, sketch the graphs of the region (Figure 6).<\/p>\n<div id=\"attachment_1369\" style=\"width: 667px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1369\" class=\"size-full wp-image-1369\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25143555\/5-3-12.jpeg\" alt=\"A cardioid with equation 1 + cos theta is shown overlapping a circle given by r = 3 cos theta, which is a circle of radius 3 with center (1.5, 0). The area bounded by the x axis, the cardioid, and the dashed line connecting the origin to the intersection of the cardioid and circle on the r = 2 line is shaded.\" width=\"657\" height=\"644\" \/><\/p>\n<p id=\"caption-attachment-1369\" class=\"wp-caption-text\">Figure 6.\u00a0Finding the area enclosed by both a circle and a cardioid.<\/p>\n<\/div>\n<p id=\"fs-id1167793589743\">We can from see the symmetry of the graph that we need to find the points of intersection. Setting the two equations equal to each other gives<\/p>\n<p style=\"text-align: center;\">[latex]{3}{cos}{\\theta} = {1} + {\\cos}{\\theta}[\/latex].<\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\">One of the points of intersection is [latex]{\\theta} = {{\\pi}\/{3}}[\/latex]. The area above the polar axis consists of two parts, with one part defined by the cardioid from [latex]{\\theta} = {0}[\/latex] to [latex]{\\theta} = {{\\pi}\/{3}}[\/latex] and the other part defined by the circle from [latex]{\\theta} = {{\\pi}\/{3}}[\/latex] to [latex]{\\theta} = {{\\pi}\/{2}}[\/latex]. By symmetry, the total area is twice the area above the polar axis. Thus, we have<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\large{A=2\\left[\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi\/3}\\displaystyle\\int_{r=0}^{r=1+\\cos\\theta}1r \\ dr \\ d\\theta+\\displaystyle\\int_{\\theta=\\pi\/3}^{\\theta=\\pi\/2}\\displaystyle\\int_{r=0}^{r=3\\cos\\theta}1r \\ dr \\ d\\theta\\right]}[\/latex].<\/p>\n<p>Evaluating each piece separately, we find that the area is<\/p>\n<p style=\"text-align: center;\">[latex]{A} = {2}{\\left ({{\\frac{1}{4}}{\\pi}} + {{\\frac{9}{16}}{\\sqrt{3}}} + {{\\frac{3}{8}}{\\pi}} - {{\\frac{9}{16}}{\\sqrt{3}}} \\right)} = {2}{\\left ({\\frac{5}{8}}{\\pi} \\right )} = {{\\frac{5}{4}}{\\pi}}[\/latex] square units.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q579921064\">Show Solution<\/span><\/p>\n<div id=\"q579921064\" class=\"hidden-answer\" style=\"display: none\">\n<p>[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]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: evaluating an improper double integral in polar coordinates<\/h3>\n<p>Evaluate the integral\u00a0[latex]\\underset{R^2}{\\displaystyle\\iint}e^{-10(x^2+y^2)}dx \\ dy[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q271367529\">Show Solution<\/span><\/p>\n<div id=\"q271367529\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793444452\">This is an improper integral because we are integrating over an unbounded region [latex]R^2[\/latex]. In polar coordinates, the entire plane [latex]R^2[\/latex] can be seen as [latex]{0} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {{{2}{\\pi}},0} \\ {\\leq} \\ {r} \\ {\\leq} \\ {\\infty}[\/latex].<\/p>\n<p id=\"fs-id1167793387195\">Using the changes of variables from rectangular coordinates to polar coordinates, we have<\/p>\n<p>[latex]\\hspace{3cm}\\begin{align}    \\underset{R^2}{\\displaystyle\\iint}e^{-10(x^2+y^2)}dx \\ dy&=\\displaystyle\\int_{\\theta=0}^{\\theta=2\\pi}\\displaystyle\\int_{r=0}^{r=\\infty}e^{-10r^2}r \\ dr \\ d\\theta=\\displaystyle\\int_{\\theta=0}^{\\theta=2\\pi}\\left(\\displaystyle\\lim_{a\\to\\infty}\\displaystyle\\int_{r=0}^{r=a}e^{-10r^2}r \\ dr\\right)d\\theta \\\\    &=\\left(\\displaystyle\\int_{\\theta=0}^{\\theta=2\\pi}d\\theta\\right)\\left(\\displaystyle\\lim_{a\\to\\infty}\\displaystyle\\int_{r=0}^{r=a}e^{-10r^2}r \\ dr\\right) \\\\    &=2\\pi\\left(\\displaystyle\\lim_{a\\to\\infty}\\displaystyle\\int_{r=0}^{r=a}e^{-10r^2}r \\ dr\\right) \\\\    &=2\\pi\\displaystyle\\lim_{a\\to\\infty}\\left(-\\frac1{20}\\right)\\left(e^{-10r^2}\\bigg|_0^1\\right)\\\\    &=2\\pi\\left(-\\frac1{20}\\right)\\displaystyle\\lim_{a\\to\\infty}\\left(e^{-10a^2}-1\\right)\\\\    &=\\frac{\\pi}{10}    \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Evaluate the integral\u00a0[latex]\\underset{R^2}{\\displaystyle\\iint}{e^{{-4}{(x^2+y^2)}}}{dx^{}}{dy}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q967834547\">Show Solution<\/span><\/p>\n<div id=\"q967834547\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{\\pi}4[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167793952731\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\"><\/div>\n<p>&nbsp;<\/p>\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=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\"><\/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\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1066\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>CP 5.22. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 3. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":428269,"menu_order":14,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 3\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"CP 5.22\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1066","chapter","type-chapter","status-publish","hentry"],"part":23,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1066","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/428269"}],"version-history":[{"count":119,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1066\/revisions"}],"predecessor-version":[{"id":6454,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1066\/revisions\/6454"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/23"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1066\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=1066"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=1066"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=1066"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=1066"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}