{"id":1268,"date":"2025-10-15T15:57:08","date_gmt":"2025-10-15T15:57:08","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=1268"},"modified":"2025-10-31T16:30:54","modified_gmt":"2025-10-31T16:30:54","slug":"alternative-coordinate-systems-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/alternative-coordinate-systems-learn-it-4\/","title":{"raw":"Triple Integrals in Alternative Coordinate Systems: Learn It 4","rendered":"Triple Integrals in Alternative Coordinate Systems: Learn It 4"},"content":{"raw":"<h2 data-type=\"title\">Integration in Spherical Coordinates<\/h2>\r\n<p class=\"whitespace-normal break-words\">We now establish a triple integral in the spherical coordinate system, as we did before in the cylindrical coordinate system. Let the function [latex]{f}{({\\rho},{\\theta},{\\varphi})}[\/latex] be continuous in a bounded spherical box:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]{B} = {{ {({\\rho},{\\theta},{\\varphi})}{\\mid}{a} \\ {\\leq} \\ {\\rho} \\ {\\leq} \\ {{b},{\\alpha}} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {{\\beta},{\\gamma}} \\ {\\leq} \\ {\\varphi} \\ {\\leq} \\ {\\psi} }}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">We then divide each interval into [latex]l[\/latex], [latex]m[\/latex], and [latex]n[\/latex] subdivisions such that [latex]{{\\Delta}{\\rho} = {\\frac{{b}-{a}}{l}}}[\/latex], [latex]{{\\Delta}{\\theta} = {\\frac{{\\beta}-{\\alpha}}{m}}}[\/latex], and [latex]{{\\Delta}{\\varphi} = {\\frac{{\\psi}-{\\gamma}}{n}}}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">For the volume element of the subbox [latex]{\\Delta}{V}[\/latex] in spherical coordinates, we have [latex]{\\Delta}{V} = {({\\Delta}{\\rho})}{({\\rho}{\\Delta}{\\varphi})}{({\\rho} \\sin \\varphi{\\Delta}{\\theta})}[\/latex], as shown in the following figure.<\/p>\r\n\r\n\r\n[caption id=\"attachment_1398\" align=\"aligncenter\" width=\"756\"]<img class=\"size-full wp-image-1398\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26173534\/5-5-8.jpeg\" alt=\"In the spherical coordinate space, a box is projected onto the polar coordinate plane. On the polar coordinate plane, the projection has area rho sin phi Delta theta. On the z axis, a distance Delta rho is indicated, and from these boundaries, angles are made that project through the edges of the box. There is also a blown up version of the box that shows it has sides Delta rho, rho Delta phi, and rho sin phi Delta theta, with overall volume Delta V = rho squared sin phi Delta rho Delta phi Delta theta.\" width=\"756\" height=\"361\" \/> Figure 3.\u00a0The volume element of a box in spherical coordinates.[\/caption]\r\n<p class=\"whitespace-normal break-words\">Just as with cylindrical coordinates, we must account for how volume elements change in spherical coordinates. The volume element includes factors that account for the geometry of the spherical \"wedge\" shape that results from our subdivisions.<\/p>\r\n<p class=\"whitespace-normal break-words\">The triple integral in spherical coordinates is the limit of a triple Riemann sum:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\displaystyle\\lim_{l,m,n\\to\\infty}\\displaystyle\\sum^l_{i=1}\\displaystyle\\sum^m_{j=1}\\displaystyle\\sum^n_{k=1}f(\\rho^*_{ijk},\\theta^*_{ijk},\\varphi^*_{ijk})(\\rho^*_{ijk})^2\\sin\\varphi\\Delta\\rho\\Delta\\theta\\Delta\\varphi[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">provided the limit exists.<\/p>\r\n<p class=\"whitespace-normal break-words\">Notice the extra factors of [latex]\\rho^2\\sin\\varphi[\/latex] in the sum. These come from the geometry of spherical coordinates and are essential for obtaining correct results.<\/p>\r\n\r\n<div class=\"mceTemp\"><\/div>\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>converting triple integrals to spherical coordinates<\/h3>\r\n<p class=\"whitespace-normal break-words\">If [latex]g(x, y, z)[\/latex] is a continuous function in rectangular coordinates, then we can convert to spherical coordinates:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{B}{\\displaystyle\\iiint}{g}{(x,y,z)}{dV} = \\underset{B}{\\displaystyle\\iiint}{g}{({\\rho}\\sin\\varphi\\cos\\theta, {\\rho}\\sin\\varphi\\sin\\theta, {\\rho}\\cos\\varphi)}{\\rho}^{2}\\sin\\varphi \\, d{\\rho} \\, d{\\varphi} \\, d{\\theta}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">The volume element in spherical coordinates is [latex]dV = {\\rho}^{2}\\sin\\varphi \\, d{\\rho} \\, d{\\varphi} \\, d{\\theta}[\/latex].<\/p>\r\n\r\n<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\"><strong>Don't Forget [latex]\\rho^2\\sin\\varphi[\/latex]!<\/strong>\r\n<p class=\"whitespace-normal break-words\">When converting to spherical coordinates, the volume element [latex]dV[\/latex] becomes [latex]{\\rho}^{2}\\sin\\varphi , d{\\rho} , d{\\varphi} , d{\\theta}[\/latex]. Students often forget these extra factors, which leads to incorrect results. The [latex]\\rho^2[\/latex] accounts for the radial expansion, while [latex]\\sin\\varphi[\/latex] accounts for the changing circumference at different angles from the [latex]z[\/latex]-axis.<\/p>\r\n\r\n<\/section>\r\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Fubini's Theorem in Spherical Coordinates<\/h3>\r\n<p class=\"whitespace-normal break-words\">As with the other multiple integrals we have examined, all the properties work similarly for a triple integral in the spherical coordinate system, and so do the iterated integrals.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>Fubini's theorem for spherical coordinates<\/h3>\r\n<p id=\"fs-id1167793433540\">If [latex]{f}{({\\rho},{\\theta},{\\varphi})}[\/latex] is continuous on a spherical solid box [latex]{B} = {[{a},{b}]} \\ {\\times} \\ {[{\\alpha},{\\beta}]} \\ {\\times} {[{\\gamma},{\\psi}]}[\/latex], then:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\underset{B}{\\displaystyle\\iiint}{f}{({\\rho},{\\theta},{\\varphi})} \\ {\\rho}^{2}{\\sin} \\ {\\varphi} \\ {{d}{\\rho}} \\ {{d}{\\varphi}} \\ {{d}{\\theta}} = {\\displaystyle\\int^{{\\varphi}={\\psi}}_{{\\varphi}={\\gamma}}} \\ {\\displaystyle\\int^{{\\theta}={\\beta}}_{{\\theta}={\\alpha}}} \\ {\\displaystyle\\int^{{\\rho}={b}}_{{\\rho}={a}}} \\ {f}{({\\rho},{\\theta},{\\varphi})} \\ {\\rho}^{2}{\\sin} \\ {\\varphi} \\ {{d}{\\rho}} \\ {{d}{\\varphi}} \\ {{d}{\\theta}}.[\/latex]<\/p>\r\nThis iterated integral may be replaced by other iterated integrals by integrating with respect to the three variables in other orders.\r\n\r\n<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\"><strong>When to Use Spherical Coordinates<\/strong>\r\n<p class=\"whitespace-normal break-words\">Spherical coordinate systems work well for solids that are symmetric around a point, such as spheres and cones. Look for problems involving expressions like [latex]x^2 + y^2 + z^2[\/latex] or solids centered at the origin.<\/p>\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Evaluate the iterated triple integral[latex]{\\displaystyle\\int^{{\\theta}={{2}{\\pi}}}_{{\\theta}={0}}} \\ {\\displaystyle\\int^{{\\varphi}={{\\pi}\/{2}}}_{{\\varphi}={0}}} \\ {\\displaystyle\\int^{{\\rho}={1}}_{{p}={0}}} \\ {\\rho}^{2}{\\sin} \\ {\\varphi} \\ {{d}{\\rho}} \\ {{d}{\\varphi}} \\ {{d}{\\theta}}[\/latex].[reveal-answer q=\"426899382\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"426899382\"]As before, in this case the variables in the iterated integral are actually independent of each other and hence we can integrate each piece and multiply:\r\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{{2}{\\pi}}_{0}} \\ {\\displaystyle\\int^{{\\pi}\/2}_{0}} \\ {\\displaystyle\\int^{1}_{0}} \\ {\\rho}^{2}{\\sin} \\ {\\varphi} \\ {{d}{\\rho}} \\ {{d}{\\varphi}} \\ {{d}{\\theta}} = {\\displaystyle\\int^{{2}{\\pi}}_{0}} \\ {{d}{\\theta}} \\ {\\displaystyle\\int^{{\\pi}\/2}_{0}} \\ {\\sin} \\ {\\varphi} \\ {{d}{\\varphi}} \\ {\\displaystyle\\int^{1}_{0}} \\ {\\rho}^{2}{d}{\\rho} = {({2}{\\pi})}{(1)}{\\left ( {\\frac{1}{3}} \\right )} = {\\frac{{2}{\\pi}}{3}}.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Triple Integrals over General Spherical Regions<\/h3>\r\n<p id=\"fs-id1167793928675\">The concept of triple integration in spherical coordinates can be extended to integration over a general solid, using the projections onto the coordinate planes. Note that [latex]dV[\/latex] and [latex]dA[\/latex] mean the increments in volume and area, respectively. The variables [latex]V[\/latex] and [latex]A[\/latex] are used as the variables for integration to express the integrals.<\/p>\r\n<p id=\"fs-id1167793421937\">The triple integral of a continuous function [latex]{f}{({\\rho},{\\theta},{\\varphi})}[\/latex] over a general solid region:<\/p>\r\n<p style=\"text-align: center;\">[latex]{E} = {\\left \\{ {({\\rho},{\\theta},{\\varphi})}{\\mid}{({\\rho},{\\theta})} \\ {\\in} \\ {{D},{{u_1} \\ {({\\rho},{\\theta}}})} \\ {\\leq} \\ {\\varphi} \\ {\\leq} \\ {{u_2} \\ {({\\rho},{\\theta})}} \\right \\}}[\/latex]<\/p>\r\n<p id=\"fs-id1167793692948\">in [latex]{\\mathbb{R}}^{3}[\/latex], where [latex]D[\/latex] is the projection of [latex]E[\/latex] onto the [latex]{\\rho}{\\theta}[\/latex]-plane, is:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\underset{E}{\\displaystyle\\iiint}{f}{({\\rho},{\\theta},{\\varphi})}{dV} = {\\iint\\limits_{D}}{\\left [ \\ {\\int\\limits^{{u_2}{({\\rho},{\\theta})}}_{{u_1}{({\\rho},{\\theta})}}} \\ {f}{({\\rho},{\\theta},{\\varphi})}{d}{\\varphi} \\right ]}{{d}{A}}.[\/latex]<\/p>\r\nIn particular, [latex]{D} = {\\left \\{ {({\\rho},{\\theta})}{\\mid}{g_1} \\ {(\\theta)} \\ {\\leq} \\ {\\rho} \\ {\\leq} \\ {{g_2} \\ {(\\theta)},{\\alpha}} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {\\beta} \\right \\}}[\/latex], then we have:\r\n<p style=\"text-align: center;\">[latex]\\underset{E}{\\displaystyle\\iiint}{f}{({\\rho},{\\theta},{\\varphi})}{dV} = {\\displaystyle\\int^{\\beta}_{\\alpha}} \\ {\\displaystyle\\int^{{g_2}{({\\theta})}}_{{g_1}{({\\theta})}}} \\ {\\displaystyle\\int^{{u_2}{({\\rho},{\\theta})}}_{{u_1}{({\\rho},{\\theta})}}} \\ {f}{({\\rho},{\\theta},{\\varphi})}{{\\rho}^{2}}{\\sin} \\ {\\varphi} \\ {{d}{\\varphi}} \\ {{d}{\\rho}} \\ {{d}{\\theta}}.[\/latex]<\/p>\r\n<p id=\"fs-id1167793244072\">Similar formulas occur for projections onto the other coordinate planes.<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">Set up an integral for the volume of the region bounded by the cone [latex]{z} = {\\sqrt{{3}{({x^2}+{y^2})}}}[\/latex] and the hemisphere [latex]{z} = {\\sqrt{{4}-{x^2}-{y^2}}}[\/latex] (see the figure below).\r\n\r\n[caption id=\"attachment_1399\" align=\"aligncenter\" width=\"490\"]<img class=\"size-full wp-image-1399\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26173615\/5-5-9.jpeg\" alt=\"A hemisphere with equation z = the square root of (4 minus x squared minus y squared) in the upper half plane, and within it, a cone with equation z = the square root of (3 times (x squared + y squared)) that is pointing down, with vertex at the origin.\" width=\"490\" height=\"340\" \/> Figure 4.\u00a0A region bounded below by a cone and above by a hemisphere.[\/caption]\r\n\r\n[reveal-answer q=\"972450157\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"972450157\"]\r\n<p id=\"fs-id1167794160034\">Using the conversion formulas from rectangular coordinates to spherical coordinates, we have:<\/p>\r\n<p id=\"fs-id1167793384878\">For the cone: [latex]{z} = {\\sqrt{{3}{({x^2}+{y^2})}}}[\/latex] or [latex]{\\rho} \\ {\\cos} \\ {\\varphi} = {\\sqrt{3}} \\ {\\rho} \\ {\\sin} \\ {\\varphi}[\/latex] or [latex]{\\tan} \\ {\\varphi} = {\\frac{1}{\\sqrt{3}}}[\/latex] or [latex]{\\varphi} = {\\frac{\\pi}{6}}[\/latex].<\/p>\r\n<p id=\"fs-id1167793261178\">For the sphere: [latex]{z} = {\\sqrt{{4}-{x^2}-{y^2}}}[\/latex] or [latex]{z^2} + {x^2} + {y^2} = {4}[\/latex] or [latex]{{\\rho}^{2}} = {4}[\/latex] or [latex]{\\rho} = {2}[\/latex].<\/p>\r\n<p id=\"fs-id1167793433725\">Thus, the triple integral for the volume is [latex]{{V}{(E)}} = {\\displaystyle\\int^{{\\theta} = {{2}{\\pi}}}_{{\\theta} = {0}}} \\ \\ {\\displaystyle\\int^{{\\varphi} = {{\\pi}\/{6}}}_{{\\phi} = {0}}} \\ \\ {\\displaystyle\\int^{{\\rho} = {2}}_{{\\rho} = {0}}} \\ {{\\rho}^{2}}{\\sin} \\ {\\varphi} \\ {{d}{\\rho}} \\ {{d}{\\varphi}} \\ {{d}{\\theta}}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<p id=\"fs-id1167794212109\">Let [latex]E[\/latex]\u00a0<span style=\"font-size: 1rem; text-align: initial;\">be the region bounded below by the cone [latex]{z} = {\\sqrt{{x^2}+{y^2}}}[\/latex] and above by the sphere [latex]{z} = {x^2} + {y^2} + {z^2}[\/latex] (<\/span>Figure 5<span style=\"font-size: 1rem; text-align: initial;\">). Set up a triple integral in spherical coordinates and find the volume of the region using the following orders of integration:<\/span><\/p>\r\n\r\n<ol id=\"fs-id1167793400632\" type=\"a\">\r\n \t<li>[latex]{{d}{\\rho}} \\ {{d}{\\phi}} \\ {{d}{\\theta}}[\/latex],<\/li>\r\n \t<li><span data-type=\"newline\">[latex]{{d}{\\varphi}} \\ {{d}{\\rho}} \\ {{d}{\\theta}}[\/latex].\r\n<\/span><\/li>\r\n<\/ol>\r\n[caption id=\"attachment_1401\" align=\"aligncenter\" width=\"480\"]<img class=\"size-full wp-image-1401\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26173806\/5-5-10.jpeg\" alt=\"A sphere with equation z = x squared + y squared + z squared, and within it, a cone with equation z = the square root of (x squared + y squared) that is pointing down, with vertex at the origin.\" width=\"480\" height=\"341\" \/> Figure 5.\u00a0A region bounded below by a cone and above by a sphere.[\/caption]\r\n\r\n[reveal-answer q=\"178836902\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"178836902\"]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Use the conversion formulas to write the equations of the sphere and cone in spherical coordinates.<span data-type=\"newline\">\r\n<\/span>For the sphere:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned} {x^2} + {y^2} + \\ &amp; {z^2} = {z} \\\\\r\n&amp; {{\\rho}^{2}} = {\\rho} \\ {\\cos} \\ {\\varphi} \\\\\r\n&amp; {\\rho} = {\\cos} \\ {\\varphi}.\r\n\\end{aligned}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">For the cone:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{z} &amp; = {\\sqrt{{x^2}+{y^2}}} \\\\\r\n{\\rho} \\ {\\cos} \\ {\\varphi} &amp; = {\\sqrt{{{\\rho}^{2}}{{\\sin}^{2}}{\\varphi}{{\\cos}^{2}}{\\phi}+{{\\rho}^{2}}{{\\sin}^{2}}{\\varphi}{{\\sin}^{2}}{\\phi}}} \\\\\r\n{\\rho} \\ {\\cos} \\ {\\varphi} &amp; = {\\sqrt{{{\\rho}^{2}}{{\\sin}^{2}}{\\varphi}{({{\\cos}^{2}}{\\phi}+{{\\sin}^{2}}{\\phi})}}} \\\\\r\n{\\rho} \\ {\\cos} \\ {\\varphi} &amp; = {\\rho} \\ {\\sin} \\ {\\varphi} \\\\\r\n{\\cos} \\ {\\varphi} &amp; = {\\sin} \\ {\\varphi} \\\\\r\n{\\varphi} &amp; = {{\\pi}\/{4}}.\r\n\\end{aligned}[\/latex]<\/p>\r\nHence the integral for the volume of the solid region [latex]E[\/latex] becomes\r\n<p style=\"text-align: center;\">[latex]{{V}{(E)}} = {\\displaystyle\\int^{{\\theta} = {{2}{\\pi}}}_{{\\theta} = {0}}} \\ {\\displaystyle\\int^{{\\varphi} = {{\\pi}\/{4}}}_{{\\varphi} = {0}}} \\ {\\displaystyle\\int^{{\\rho} = {{\\cos}{\\varphi}}}_{{\\rho} = {0}}} \\ {{{\\rho}^{2}}{\\sin} \\ {\\varphi} \\ {{d}{\\rho}} \\ {{d}{\\varphi}} \\ {{d}{\\theta}}}.[\/latex]<\/p>\r\n<\/li>\r\n \t<li>Consider the [latex]{\\varphi}{\\rho}[\/latex]-plane. Note that the ranges for [latex]{\\varphi}[\/latex] and [latex]{\\rho}[\/latex] (from part a.) are\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{0} &amp; \\ {\\leq} \\ {\\varphi} \\ {\\leq} \\ {{\\pi}\/4} \\\\\r\n{0} &amp; \\ {\\leq} \\ {\\phi} \\ {\\leq} \\ {{\\cos}{\\varphi}}.\r\n\\end{aligned}[\/latex]<\/p>\r\nThe curve [latex]{\\rho} = {\\cos}{\\varphi}[\/latex] meets the line [latex]{\\varphi} = {{\\pi}\/{4}}[\/latex] at the point [latex]({{\\pi}\/{4}},{{\\sqrt{2}}\/{2}})[\/latex]. Thus, to change the order of integration, we need to use two pieces:\r\n\r\n[latex]\\hspace{8cm}\\begin{align}&amp;0\\leq\\rho\\leq\\sqrt2\/2 \\\\&amp;0\\leq\\varphi\\leq\\pi\/4\\end{align}[\/latex] [latex]\\text{ and }[\/latex] [latex]\\begin{align}&amp;\\sqrt2\/2\\leq\\rho\\leq1 \\\\&amp;0\\leq\\varphi\\leq\\cos^{-1}\\rho\\end{align}[\/latex]\r\n\r\nHence the integral for the volume of the solid region [latex]E[\/latex] becomes\r\n<p style=\"text-align: center;\">[latex]{{V}{(E)}} = {\\displaystyle\\int^{{\\theta} = {{2}{\\pi}}}_{{\\theta} = {0}}} \\ {\\displaystyle\\int^{{\\rho} = {{\\sqrt{2}\/{2}}}}_{{\\rho} = {0}}} \\ {\\displaystyle\\int^{{\\varphi} = {{\\pi}\/{4}}}_{{\\varphi} = {0}}} \\ {{{\\rho}^{2}}{\\sin} \\ {\\varphi} \\ {{d}{\\varphi}} \\ {{d}{\\rho}} \\ {{d}{\\theta}}} \\ + {\\displaystyle\\int^{{\\theta} = {{2}{\\pi}}}_{{\\theta} = {0}}} \\ {\\displaystyle\\int^{{\\rho} = {1}}_{{\\rho} = {{\\sqrt{2}\/{2}}}}} \\ {\\displaystyle\\int^{{\\varphi} = {{\\cos}^{-1}}{\\rho}}_{{\\varphi} = {0}}} \\ {{{\\rho}^{2}}{\\sin} \\ {\\varphi} \\ {{d}{\\varphi}} \\ {{d}{\\rho}} \\ {{d}{\\theta}}}.[\/latex]<\/p>\r\nIn each case, the integration results in [latex]{{V}{(E)}} = {\\frac{\\pi}{8}}[\/latex].<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Convert the following integral into cylindrical coordinates:\r\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{{y} = {1}}_{{y} = {-1}}}\\displaystyle\\int^{{x} = {\\sqrt{{1} - {{y}^{2}}}}}_{{x} = {0}} \\ {\\displaystyle\\int^{{z} = {\\sqrt{{x^2}+{y^2}}}}_{{z} = {{x}^{2}}+{{y}^{2}}}} \\ {{x}{y}{z}} \\ {{d}{z}} \\ {{d}{x}} \\ {{d}{y}}.[\/latex]<\/p>\r\n[reveal-answer q=\"368178320\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"368178320\"]\r\n<p id=\"fs-id1167793544553\">The ranges of the variables are<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{-1} &amp; \\ \\ \\ {\\leq} \\ \\ \\ {y}{\\leq}{1} \\\\\r\n{0} &amp; \\ \\ \\ {\\leq} \\ \\ \\ {x}{\\leq}{\\sqrt{{1} - {{y}^{2}}}} \\\\\r\n{{x}^{2}} + {{y}^{2}} &amp; \\ \\ \\ {\\leq} \\ \\ \\ {z}{\\leq}{\\sqrt{{{x}^{2}}+{{y}^{2}}}}.\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1167793464597\">The first two inequalities describe the right half of a circle of radius 1. Therefore, the ranges for [latex]{\\theta}[\/latex] and [latex]r[\/latex] are<\/p>\r\n<p style=\"text-align: center;\">[latex]{-{\\frac{\\pi}{2}}} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {\\frac{\\pi}{2}} \\ {\\text{and}} \\ {0} \\ {\\leq} \\ {r} \\ {\\leq} \\ {1}.[\/latex]<\/p>\r\n<p id=\"fs-id1167793514913\">The limits of [latex]z[\/latex] are [latex]{{r}^{2}}[\/latex], hence<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{{y} = {1}}_{{y} = {-1}}}\\displaystyle\\int^{{x} = {\\sqrt{{1} - {{y}^{2}}}}}_{{x} = {0}}\\displaystyle\\int^{{z} = {\\sqrt{{{x}^{2}} + {{y}^{2}}}}}_{{z} = {{x}^{2}} + {{y}^{2}}} \\ {{x}{y}{z}} \\ {{d}{z}} \\ {{d}{x}} \\ {{d}{y}} = {\\displaystyle\\int^{{\\theta} = {{\\pi}\/{2}}}_{{\\theta} = {-{\\pi}\/{2}}}} \\ {\\displaystyle\\int^{{r} = {1}}_{{r} = {0}}} \\ {\\displaystyle\\int^{{z} = {r}}_{{z} = {{r}^{2}}}} \\ {r}{({r}{\\cos}{\\theta})}{({r}{\\sin}{\\theta})}{z} \\ {{d}{z}} \\ {{d}{r}} \\ {{d}{\\theta}}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Convert the following integral into spherical coordinates:\r\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{{y} = {3}}_{{y} = {0}}}\\displaystyle\\int^{{x} = {\\sqrt{{9} - {{y}^{2}}}}}_{{x} = {0}}\\displaystyle\\int^{{z} = {\\sqrt{{18} - {{x}^{2}} - {{y}^{2}}}}}_{{z} = {\\sqrt{{x}^{2} + {{y}^{2}}}}} \\ {({{x}^{2}} + {{y}^{2}} + {{z}^{2}})} \\ {{d}{z}} \\ {{d}{x}} \\ {{d}{y}}.[\/latex]<\/p>\r\n[reveal-answer q=\"255795354\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"255795354\"]\r\n<p id=\"fs-id1167794052495\">The ranges of the variables are<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{0} \\ \\ \\ &amp; {\\leq} \\ \\ \\ {y} \\ {\\leq} \\ {3} \\\\\r\n{0} \\ \\ \\ &amp; {\\leq} \\ \\ \\ {x} \\ {\\leq} \\ {\\sqrt{{9} - {{y}^{2}}}} \\\\\r\n{\\sqrt{{{x}^{2}} + {{y}^{2}}}} \\ \\ \\ &amp; {\\leq} \\ \\ \\ {z} \\ {\\leq} \\ {\\sqrt{{18} - {{x}^{2}} - {{y}^{2}}}}.\r\n\\end{aligned}[\/latex]<\/p>\r\n<p id=\"fs-id1167793777086\">The first two ranges of variables describe a quarter disk in the first quadrant of the [latex]xy[\/latex]-plane. Hence the range for [latex]{\\theta}[\/latex] is [latex]{0} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {\\frac{\\pi}{2}}[\/latex].<\/p>\r\n<p id=\"fs-id1167794067621\">The lower bound [latex]{z} = {\\sqrt{{{x}^{2}} + {{y}^{2}}}}[\/latex] is the upper half of a cone and the upper bound [latex]{z} = {\\sqrt{{18} - {{x}^{2}} - {{y}^{2}}}}[\/latex] is the upper half of a sphere. Therefore, we have [latex]{0} \\ {\\leq} \\ {\\rho} \\ {\\leq} \\ {\\sqrt{18}}[\/latex], which is [latex]{0} \\ {\\leq} \\ {\\rho} \\ {\\leq} \\ {3{\\sqrt{2}}}[\/latex].<\/p>\r\n<p id=\"fs-id1167794168074\">For the ranges of [latex]{\\varphi}[\/latex], we need to find where the cone and the sphere intersect, so solve the equation<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{{{r}^{2}} + {{z}^{2}}} \\ \\ \\ &amp; = \\ \\ \\ {18} \\\\\r\n{({\\sqrt{{{x}^{2}} + {{y}^{2}}}})^2} + {{z}^{2}} \\ \\ \\ &amp; = \\ \\ \\ {18} \\\\\r\n{{z}^{2}} + {{z}^{2}} \\ \\ \\ &amp; = \\ \\ \\ {18} \\\\\r\n{{2}{{z}^{2}}} \\ \\ \\ &amp; = \\ \\ \\ {18} \\\\\r\n{{z}^{2}} \\ \\ \\ &amp; = \\ \\ \\ {9} \\\\\r\n{z} \\ \\ \\ &amp; = \\ \\ \\ {3}.\r\n\\end{aligned}[\/latex]<\/p>\r\n<p id=\"fs-id1167793610938\">This gives<\/p>\r\n[latex]\\hspace{10cm}\\begin{align}\r\n\r\n3\\sqrt2\\cos\\varphi&amp;=3 \\\\\r\n\r\n\\cos\\varphi&amp;=\\frac1{\\sqrt2} \\\\\r\n\r\n\\varphi&amp;=\\frac{\\pi}4.\r\n\r\n\\end{align}[\/latex]\r\n<p id=\"fs-id1167793924123\">Putting this together, we obtain<\/p>\r\n\r\n<center>[latex]\\displaystyle\\int_{y=0}^{y=3}\\displaystyle\\int_{x=0}^{x=\\sqrt{9-y^2}}\\displaystyle\\int_{z=\\sqrt{x^2+y^2}}^{z=\\sqrt{18-x^2-y^2}}(x^2+y^2+z^2)dz \\ dx \\ dy=\\displaystyle\\int_{\\varphi=0}^{\\varphi=\\pi\/4}\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi\/2}\\displaystyle\\int_{\\rho=0}^{\\rho=3\\sqrt2}\\rho^4\\sin\\varphi \\ d\\rho \\ d\\theta \\ d\\varphi.[\/latex]<\/center>[\/hidden-answer]\r\n\r\n<\/section>\r\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Finding Volumes Using Spherical Coordinates<\/h3>\r\n<p class=\"whitespace-normal break-words\">Now that we are familiar with the spherical coordinate system, let's find the volume of some known geometric figures, such as spheres and ellipsoids.<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">Find the volume of the spherical planetarium in l\u2019Hemisph\u00e8ric in Valencia, Spain, which is five stories tall and has a radius of approximately [latex]50[\/latex] ft, using the equation [latex]x^2+y^2+z^2=r^2[\/latex].\r\n\r\n[caption id=\"attachment_5569\" align=\"aligncenter\" width=\"325\"]<img class=\"size-full wp-image-5569\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/03022644\/5.60.jpeg\" alt=\"A picture of l\u2019Hemisph\u00e8ric, which is a giant glass structure that is in the shape of an ellipsoid.\" width=\"325\" height=\"219\" \/> Figure 7. (credit: modification of work by Javier Yaya Tur, Wikimedia Commons)[\/caption]\r\n\r\n[reveal-answer q=\"204959485\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"204959485\"]\r\n<p id=\"fs-id1167793416704\">We calculate the volume of the ball in the first octant, where [latex]x\\geq0[\/latex], [latex]y\\geq0[\/latex], and [latex]z\\geq0[\/latex], using spherical coordinates, and then multiply the result by [latex]8[\/latex] for symmetry. Since we consider the region [latex]D[\/latex] as the first octant in the integral, the ranges of the variables are<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{0\\leq\\varphi\\leq\\frac{\\pi}2, \\ 0\\leq\\rho\\leq{r}, \\ 0\\leq\\theta\\leq\\frac{\\pi}2}[\/latex]<\/p>\r\n<p id=\"fs-id1167793500489\">Therefore,<\/p>\r\n[latex]\\hspace{6cm}\\begin{align}\r\n\r\nV&amp;=\\underset{D}{\\displaystyle\\iiint}dx \\ dy \\ dz = 8\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi\/2}\\displaystyle\\int_{\\rho=0}^{\\rho=\\pi}\\displaystyle\\int_{\\varphi=0}^{\\varphi=\\pi\/2}\\rho^2\\sin\\theta \\ d\\varphi \\ d\\rho \\ d\\theta \\\\\r\n\r\n&amp;=8\\displaystyle\\int_{\\varphi=0}^{\\varphi=\\pi\/2}d\\varphi\\displaystyle\\int_{\\rho=0}^{\\rho=\\pi}\\rho^2 \\ d\\rho\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi\/2}\\sin\\theta \\ d\\theta \\\\\r\n\r\n&amp;=8\\left(\\frac{\\pi}2\\right)\\left(\\frac{r^3}3\\right)(1) \\\\\r\n\r\n&amp;=\\frac43\\pi{r}^3\r\n\r\n\\end{align}[\/latex]\r\n<p id=\"fs-id1167793522367\">This exactly matches with what we knew. So for a sphere with a radius of approximately [latex]50[\/latex] ft, the volume is [latex]\\frac43\\pi{r}^3\\approx523,000\\text{ ft.}^3[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Find the volume of the ellipsoid [latex]\\frac{x^2}{a^2}+\\frac{y^2}{b^2}+\\frac{z^2}{c^2}=1[\/latex].[reveal-answer q=\"543745872\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"543745872\"]\r\n<p id=\"fs-id1167793251274\">We again use symmetry and evaluate the volume of the ellipsoid using spherical coordinates. As before, we use the first octant [latex]x\\geq0[\/latex], [latex]y\\geq0[\/latex], and [latex]z\\geq0[\/latex] and then multiply the result by 8.<\/p>\r\n<p id=\"fs-id1167793219213\">In this case the ranges of the variables are<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{0\\leq\\varphi\\leq\\frac{\\pi}2, \\ 0\\leq\\rho\\leq\\frac{\\pi}2, \\ 0\\leq\\rho\\leq1, \\ 0\\leq\\theta\\leq\\frac{\\pi}2}[\/latex]<\/p>\r\n<p id=\"fs-id1167793540294\">Also, we need to change the rectangular to spherical coordinates in this way:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{x=a\\rho\\cos\\varphi\\sin\\theta, \\ y=b\\rho\\sin\\varphi\\sin\\theta,\\text{ and }z=c\\rho\\cos\\theta}.[\/latex]<\/p>\r\n<p id=\"fs-id1167794213144\">Then the volume of the ellipsoid becomes<\/p>\r\n[latex]\\hspace{6cm}\\begin{align}V&amp;=\\underset{D}{\\displaystyle\\iiint}dx \\ dy \\ dz \\\\&amp;=8\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi\/2}\\displaystyle\\int_{\\rho=0}^{\\rho=1}\\displaystyle\\int_{\\varphi=0}^{\\varphi=\\pi\/2}abc\\rho^2\\sin\\theta \\ d\\varphi \\ d\\rho \\ d\\theta \\\\&amp;=8abc\\displaystyle\\int_{\\varphi=0}^{\\varphi=\\pi\/2}d\\varphi\\displaystyle\\int_{\\rho=0}^{\\rho=1}\\rho^2 \\ d\\rho\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi\/2}\\sin\\theta \\ d\\theta \\\\&amp;=8abc\\left(\\frac{\\pi}2\\right)\\left(\\frac13\\right)(1) \\\\&amp;=\\frac43\\pi{abc}\\end{align}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Find the volume of the space inside the ellipsoid [latex]\\frac{x^2}{75^2}+\\frac{y^2}{80^2}+\\frac{z^2}{90^2}=1[\/latex] and outside the sphere [latex]x^2+y^2+z^2=50^2[\/latex].[reveal-answer q=\"234085739\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"234085739\"]\r\n<p id=\"fs-id1167793925860\">This problem is directly related to the l\u2019Hemisph\u00e8ric structure. The volume of space inside the ellipsoid and outside the sphere might be useful to find the expense of heating or cooling that space. We can use the preceding two examples for the volume of the sphere and ellipsoid and then subtract.<\/p>\r\n<p id=\"fs-id1167793732638\">First we find the volume of the ellipsoid using [latex]a=75[\/latex] ft, [latex]b=80[\/latex] ft, and [latex]c=90[\/latex] ft. The volume of the ellipsoid is:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{V_{\\text{ellipsoid}}=\\frac43\\pi(75)(80)(90)\\approx2,262,000\\text{ ft.}^3}[\/latex]<\/p>\r\n<p id=\"fs-id1167793452781\">We found earlier, the volume of the sphere is:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{V_{\\text{sphere}}\\approx523,600\\text{ ft.}^3.}[\/latex]<\/p>\r\n<p id=\"fs-id1167793556981\">Therefore, the volume of the space inside the ellipsoid [latex]\\frac{x^2}{75^2}+\\frac{y^2}{80^2}+\\frac{z^2}{90^2}=1[\/latex] and outside the sphere [latex]x^2+y^2+z^2=50^2[\/latex] is approximately<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{V_{\\text{Hemisferic}}=V_{\\text{ellipsoid}}-V_{\\text{sphere}}=1,738,400\\text{ ft.}^3.}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"<h2 data-type=\"title\">Integration in Spherical Coordinates<\/h2>\n<p class=\"whitespace-normal break-words\">We now establish a triple integral in the spherical coordinate system, as we did before in the cylindrical coordinate system. Let the function [latex]{f}{({\\rho},{\\theta},{\\varphi})}[\/latex] be continuous in a bounded spherical box:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]{B} = {{ {({\\rho},{\\theta},{\\varphi})}{\\mid}{a} \\ {\\leq} \\ {\\rho} \\ {\\leq} \\ {{b},{\\alpha}} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {{\\beta},{\\gamma}} \\ {\\leq} \\ {\\varphi} \\ {\\leq} \\ {\\psi} }}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">We then divide each interval into [latex]l[\/latex], [latex]m[\/latex], and [latex]n[\/latex] subdivisions such that [latex]{{\\Delta}{\\rho} = {\\frac{{b}-{a}}{l}}}[\/latex], [latex]{{\\Delta}{\\theta} = {\\frac{{\\beta}-{\\alpha}}{m}}}[\/latex], and [latex]{{\\Delta}{\\varphi} = {\\frac{{\\psi}-{\\gamma}}{n}}}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">For the volume element of the subbox [latex]{\\Delta}{V}[\/latex] in spherical coordinates, we have [latex]{\\Delta}{V} = {({\\Delta}{\\rho})}{({\\rho}{\\Delta}{\\varphi})}{({\\rho} \\sin \\varphi{\\Delta}{\\theta})}[\/latex], as shown in the following figure.<\/p>\n<figure id=\"attachment_1398\" aria-describedby=\"caption-attachment-1398\" style=\"width: 756px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1398\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26173534\/5-5-8.jpeg\" alt=\"In the spherical coordinate space, a box is projected onto the polar coordinate plane. On the polar coordinate plane, the projection has area rho sin phi Delta theta. On the z axis, a distance Delta rho is indicated, and from these boundaries, angles are made that project through the edges of the box. There is also a blown up version of the box that shows it has sides Delta rho, rho Delta phi, and rho sin phi Delta theta, with overall volume Delta V = rho squared sin phi Delta rho Delta phi Delta theta.\" width=\"756\" height=\"361\" \/><figcaption id=\"caption-attachment-1398\" class=\"wp-caption-text\">Figure 3.\u00a0The volume element of a box in spherical coordinates.<\/figcaption><\/figure>\n<p class=\"whitespace-normal break-words\">Just as with cylindrical coordinates, we must account for how volume elements change in spherical coordinates. The volume element includes factors that account for the geometry of the spherical &#8220;wedge&#8221; shape that results from our subdivisions.<\/p>\n<p class=\"whitespace-normal break-words\">The triple integral in spherical coordinates is the limit of a triple Riemann sum:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\displaystyle\\lim_{l,m,n\\to\\infty}\\displaystyle\\sum^l_{i=1}\\displaystyle\\sum^m_{j=1}\\displaystyle\\sum^n_{k=1}f(\\rho^*_{ijk},\\theta^*_{ijk},\\varphi^*_{ijk})(\\rho^*_{ijk})^2\\sin\\varphi\\Delta\\rho\\Delta\\theta\\Delta\\varphi[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">provided the limit exists.<\/p>\n<p class=\"whitespace-normal break-words\">Notice the extra factors of [latex]\\rho^2\\sin\\varphi[\/latex] in the sum. These come from the geometry of spherical coordinates and are essential for obtaining correct results.<\/p>\n<div class=\"mceTemp\"><\/div>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>converting triple integrals to spherical coordinates<\/h3>\n<p class=\"whitespace-normal break-words\">If [latex]g(x, y, z)[\/latex] is a continuous function in rectangular coordinates, then we can convert to spherical coordinates:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{B}{\\displaystyle\\iiint}{g}{(x,y,z)}{dV} = \\underset{B}{\\displaystyle\\iiint}{g}{({\\rho}\\sin\\varphi\\cos\\theta, {\\rho}\\sin\\varphi\\sin\\theta, {\\rho}\\cos\\varphi)}{\\rho}^{2}\\sin\\varphi \\, d{\\rho} \\, d{\\varphi} \\, d{\\theta}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">The volume element in spherical coordinates is [latex]dV = {\\rho}^{2}\\sin\\varphi \\, d{\\rho} \\, d{\\varphi} \\, d{\\theta}[\/latex].<\/p>\n<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\"><strong>Don&#8217;t Forget [latex]\\rho^2\\sin\\varphi[\/latex]!<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">When converting to spherical coordinates, the volume element [latex]dV[\/latex] becomes [latex]{\\rho}^{2}\\sin\\varphi , d{\\rho} , d{\\varphi} , d{\\theta}[\/latex]. Students often forget these extra factors, which leads to incorrect results. The [latex]\\rho^2[\/latex] accounts for the radial expansion, while [latex]\\sin\\varphi[\/latex] accounts for the changing circumference at different angles from the [latex]z[\/latex]-axis.<\/p>\n<\/section>\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Fubini&#8217;s Theorem in Spherical Coordinates<\/h3>\n<p class=\"whitespace-normal break-words\">As with the other multiple integrals we have examined, all the properties work similarly for a triple integral in the spherical coordinate system, and so do the iterated integrals.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>Fubini&#8217;s theorem for spherical coordinates<\/h3>\n<p id=\"fs-id1167793433540\">If [latex]{f}{({\\rho},{\\theta},{\\varphi})}[\/latex] is continuous on a spherical solid box [latex]{B} = {[{a},{b}]} \\ {\\times} \\ {[{\\alpha},{\\beta}]} \\ {\\times} {[{\\gamma},{\\psi}]}[\/latex], then:<\/p>\n<p style=\"text-align: center;\">[latex]\\underset{B}{\\displaystyle\\iiint}{f}{({\\rho},{\\theta},{\\varphi})} \\ {\\rho}^{2}{\\sin} \\ {\\varphi} \\ {{d}{\\rho}} \\ {{d}{\\varphi}} \\ {{d}{\\theta}} = {\\displaystyle\\int^{{\\varphi}={\\psi}}_{{\\varphi}={\\gamma}}} \\ {\\displaystyle\\int^{{\\theta}={\\beta}}_{{\\theta}={\\alpha}}} \\ {\\displaystyle\\int^{{\\rho}={b}}_{{\\rho}={a}}} \\ {f}{({\\rho},{\\theta},{\\varphi})} \\ {\\rho}^{2}{\\sin} \\ {\\varphi} \\ {{d}{\\rho}} \\ {{d}{\\varphi}} \\ {{d}{\\theta}}.[\/latex]<\/p>\n<p>This iterated integral may be replaced by other iterated integrals by integrating with respect to the three variables in other orders.<\/p>\n<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\"><strong>When to Use Spherical Coordinates<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Spherical coordinate systems work well for solids that are symmetric around a point, such as spheres and cones. Look for problems involving expressions like [latex]x^2 + y^2 + z^2[\/latex] or solids centered at the origin.<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Evaluate the iterated triple integral[latex]{\\displaystyle\\int^{{\\theta}={{2}{\\pi}}}_{{\\theta}={0}}} \\ {\\displaystyle\\int^{{\\varphi}={{\\pi}\/{2}}}_{{\\varphi}={0}}} \\ {\\displaystyle\\int^{{\\rho}={1}}_{{p}={0}}} \\ {\\rho}^{2}{\\sin} \\ {\\varphi} \\ {{d}{\\rho}} \\ {{d}{\\varphi}} \\ {{d}{\\theta}}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q426899382\">Show Solution<\/button><\/p>\n<div id=\"q426899382\" class=\"hidden-answer\" style=\"display: none\">As before, in this case the variables in the iterated integral are actually independent of each other and hence we can integrate each piece and multiply:<\/p>\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{{2}{\\pi}}_{0}} \\ {\\displaystyle\\int^{{\\pi}\/2}_{0}} \\ {\\displaystyle\\int^{1}_{0}} \\ {\\rho}^{2}{\\sin} \\ {\\varphi} \\ {{d}{\\rho}} \\ {{d}{\\varphi}} \\ {{d}{\\theta}} = {\\displaystyle\\int^{{2}{\\pi}}_{0}} \\ {{d}{\\theta}} \\ {\\displaystyle\\int^{{\\pi}\/2}_{0}} \\ {\\sin} \\ {\\varphi} \\ {{d}{\\varphi}} \\ {\\displaystyle\\int^{1}_{0}} \\ {\\rho}^{2}{d}{\\rho} = {({2}{\\pi})}{(1)}{\\left ( {\\frac{1}{3}} \\right )} = {\\frac{{2}{\\pi}}{3}}.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Triple Integrals over General Spherical Regions<\/h3>\n<p id=\"fs-id1167793928675\">The concept of triple integration in spherical coordinates can be extended to integration over a general solid, using the projections onto the coordinate planes. Note that [latex]dV[\/latex] and [latex]dA[\/latex] mean the increments in volume and area, respectively. The variables [latex]V[\/latex] and [latex]A[\/latex] are used as the variables for integration to express the integrals.<\/p>\n<p id=\"fs-id1167793421937\">The triple integral of a continuous function [latex]{f}{({\\rho},{\\theta},{\\varphi})}[\/latex] over a general solid region:<\/p>\n<p style=\"text-align: center;\">[latex]{E} = {\\left \\{ {({\\rho},{\\theta},{\\varphi})}{\\mid}{({\\rho},{\\theta})} \\ {\\in} \\ {{D},{{u_1} \\ {({\\rho},{\\theta}}})} \\ {\\leq} \\ {\\varphi} \\ {\\leq} \\ {{u_2} \\ {({\\rho},{\\theta})}} \\right \\}}[\/latex]<\/p>\n<p id=\"fs-id1167793692948\">in [latex]{\\mathbb{R}}^{3}[\/latex], where [latex]D[\/latex] is the projection of [latex]E[\/latex] onto the [latex]{\\rho}{\\theta}[\/latex]-plane, is:<\/p>\n<p style=\"text-align: center;\">[latex]\\underset{E}{\\displaystyle\\iiint}{f}{({\\rho},{\\theta},{\\varphi})}{dV} = {\\iint\\limits_{D}}{\\left [ \\ {\\int\\limits^{{u_2}{({\\rho},{\\theta})}}_{{u_1}{({\\rho},{\\theta})}}} \\ {f}{({\\rho},{\\theta},{\\varphi})}{d}{\\varphi} \\right ]}{{d}{A}}.[\/latex]<\/p>\n<p>In particular, [latex]{D} = {\\left \\{ {({\\rho},{\\theta})}{\\mid}{g_1} \\ {(\\theta)} \\ {\\leq} \\ {\\rho} \\ {\\leq} \\ {{g_2} \\ {(\\theta)},{\\alpha}} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {\\beta} \\right \\}}[\/latex], then we have:<\/p>\n<p style=\"text-align: center;\">[latex]\\underset{E}{\\displaystyle\\iiint}{f}{({\\rho},{\\theta},{\\varphi})}{dV} = {\\displaystyle\\int^{\\beta}_{\\alpha}} \\ {\\displaystyle\\int^{{g_2}{({\\theta})}}_{{g_1}{({\\theta})}}} \\ {\\displaystyle\\int^{{u_2}{({\\rho},{\\theta})}}_{{u_1}{({\\rho},{\\theta})}}} \\ {f}{({\\rho},{\\theta},{\\varphi})}{{\\rho}^{2}}{\\sin} \\ {\\varphi} \\ {{d}{\\varphi}} \\ {{d}{\\rho}} \\ {{d}{\\theta}}.[\/latex]<\/p>\n<p id=\"fs-id1167793244072\">Similar formulas occur for projections onto the other coordinate planes.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">Set up an integral for the volume of the region bounded by the cone [latex]{z} = {\\sqrt{{3}{({x^2}+{y^2})}}}[\/latex] and the hemisphere [latex]{z} = {\\sqrt{{4}-{x^2}-{y^2}}}[\/latex] (see the figure below).<\/p>\n<figure id=\"attachment_1399\" aria-describedby=\"caption-attachment-1399\" style=\"width: 490px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1399\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26173615\/5-5-9.jpeg\" alt=\"A hemisphere with equation z = the square root of (4 minus x squared minus y squared) in the upper half plane, and within it, a cone with equation z = the square root of (3 times (x squared + y squared)) that is pointing down, with vertex at the origin.\" width=\"490\" height=\"340\" \/><figcaption id=\"caption-attachment-1399\" class=\"wp-caption-text\">Figure 4.\u00a0A region bounded below by a cone and above by a hemisphere.<\/figcaption><\/figure>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q972450157\">Show Solution<\/button><\/p>\n<div id=\"q972450157\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794160034\">Using the conversion formulas from rectangular coordinates to spherical coordinates, we have:<\/p>\n<p id=\"fs-id1167793384878\">For the cone: [latex]{z} = {\\sqrt{{3}{({x^2}+{y^2})}}}[\/latex] or [latex]{\\rho} \\ {\\cos} \\ {\\varphi} = {\\sqrt{3}} \\ {\\rho} \\ {\\sin} \\ {\\varphi}[\/latex] or [latex]{\\tan} \\ {\\varphi} = {\\frac{1}{\\sqrt{3}}}[\/latex] or [latex]{\\varphi} = {\\frac{\\pi}{6}}[\/latex].<\/p>\n<p id=\"fs-id1167793261178\">For the sphere: [latex]{z} = {\\sqrt{{4}-{x^2}-{y^2}}}[\/latex] or [latex]{z^2} + {x^2} + {y^2} = {4}[\/latex] or [latex]{{\\rho}^{2}} = {4}[\/latex] or [latex]{\\rho} = {2}[\/latex].<\/p>\n<p id=\"fs-id1167793433725\">Thus, the triple integral for the volume is [latex]{{V}{(E)}} = {\\displaystyle\\int^{{\\theta} = {{2}{\\pi}}}_{{\\theta} = {0}}} \\ \\ {\\displaystyle\\int^{{\\varphi} = {{\\pi}\/{6}}}_{{\\phi} = {0}}} \\ \\ {\\displaystyle\\int^{{\\rho} = {2}}_{{\\rho} = {0}}} \\ {{\\rho}^{2}}{\\sin} \\ {\\varphi} \\ {{d}{\\rho}} \\ {{d}{\\varphi}} \\ {{d}{\\theta}}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p id=\"fs-id1167794212109\">Let [latex]E[\/latex]\u00a0<span style=\"font-size: 1rem; text-align: initial;\">be the region bounded below by the cone [latex]{z} = {\\sqrt{{x^2}+{y^2}}}[\/latex] and above by the sphere [latex]{z} = {x^2} + {y^2} + {z^2}[\/latex] (<\/span>Figure 5<span style=\"font-size: 1rem; text-align: initial;\">). Set up a triple integral in spherical coordinates and find the volume of the region using the following orders of integration:<\/span><\/p>\n<ol id=\"fs-id1167793400632\" type=\"a\">\n<li>[latex]{{d}{\\rho}} \\ {{d}{\\phi}} \\ {{d}{\\theta}}[\/latex],<\/li>\n<li><span data-type=\"newline\">[latex]{{d}{\\varphi}} \\ {{d}{\\rho}} \\ {{d}{\\theta}}[\/latex].<br \/>\n<\/span><\/li>\n<\/ol>\n<figure id=\"attachment_1401\" aria-describedby=\"caption-attachment-1401\" style=\"width: 480px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1401\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26173806\/5-5-10.jpeg\" alt=\"A sphere with equation z = x squared + y squared + z squared, and within it, a cone with equation z = the square root of (x squared + y squared) that is pointing down, with vertex at the origin.\" width=\"480\" height=\"341\" \/><figcaption id=\"caption-attachment-1401\" class=\"wp-caption-text\">Figure 5.\u00a0A region bounded below by a cone and above by a sphere.<\/figcaption><\/figure>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q178836902\">Show Solution<\/button><\/p>\n<div id=\"q178836902\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Use the conversion formulas to write the equations of the sphere and cone in spherical coordinates.<span data-type=\"newline\"><br \/>\n<\/span>For the sphere:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned} {x^2} + {y^2} + \\ & {z^2} = {z} \\\\  & {{\\rho}^{2}} = {\\rho} \\ {\\cos} \\ {\\varphi} \\\\  & {\\rho} = {\\cos} \\ {\\varphi}.  \\end{aligned}[\/latex]<\/p>\n<p style=\"text-align: left;\">For the cone:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {z} & = {\\sqrt{{x^2}+{y^2}}} \\\\  {\\rho} \\ {\\cos} \\ {\\varphi} & = {\\sqrt{{{\\rho}^{2}}{{\\sin}^{2}}{\\varphi}{{\\cos}^{2}}{\\phi}+{{\\rho}^{2}}{{\\sin}^{2}}{\\varphi}{{\\sin}^{2}}{\\phi}}} \\\\  {\\rho} \\ {\\cos} \\ {\\varphi} & = {\\sqrt{{{\\rho}^{2}}{{\\sin}^{2}}{\\varphi}{({{\\cos}^{2}}{\\phi}+{{\\sin}^{2}}{\\phi})}}} \\\\  {\\rho} \\ {\\cos} \\ {\\varphi} & = {\\rho} \\ {\\sin} \\ {\\varphi} \\\\  {\\cos} \\ {\\varphi} & = {\\sin} \\ {\\varphi} \\\\  {\\varphi} & = {{\\pi}\/{4}}.  \\end{aligned}[\/latex]<\/p>\n<p>Hence the integral for the volume of the solid region [latex]E[\/latex] becomes<\/p>\n<p style=\"text-align: center;\">[latex]{{V}{(E)}} = {\\displaystyle\\int^{{\\theta} = {{2}{\\pi}}}_{{\\theta} = {0}}} \\ {\\displaystyle\\int^{{\\varphi} = {{\\pi}\/{4}}}_{{\\varphi} = {0}}} \\ {\\displaystyle\\int^{{\\rho} = {{\\cos}{\\varphi}}}_{{\\rho} = {0}}} \\ {{{\\rho}^{2}}{\\sin} \\ {\\varphi} \\ {{d}{\\rho}} \\ {{d}{\\varphi}} \\ {{d}{\\theta}}}.[\/latex]<\/p>\n<\/li>\n<li>Consider the [latex]{\\varphi}{\\rho}[\/latex]-plane. Note that the ranges for [latex]{\\varphi}[\/latex] and [latex]{\\rho}[\/latex] (from part a.) are\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {0} & \\ {\\leq} \\ {\\varphi} \\ {\\leq} \\ {{\\pi}\/4} \\\\  {0} & \\ {\\leq} \\ {\\phi} \\ {\\leq} \\ {{\\cos}{\\varphi}}.  \\end{aligned}[\/latex]<\/p>\n<p>The curve [latex]{\\rho} = {\\cos}{\\varphi}[\/latex] meets the line [latex]{\\varphi} = {{\\pi}\/{4}}[\/latex] at the point [latex]({{\\pi}\/{4}},{{\\sqrt{2}}\/{2}})[\/latex]. Thus, to change the order of integration, we need to use two pieces:<\/p>\n<p>[latex]\\hspace{8cm}\\begin{align}&0\\leq\\rho\\leq\\sqrt2\/2 \\\\&0\\leq\\varphi\\leq\\pi\/4\\end{align}[\/latex] [latex]\\text{ and }[\/latex] [latex]\\begin{align}&\\sqrt2\/2\\leq\\rho\\leq1 \\\\&0\\leq\\varphi\\leq\\cos^{-1}\\rho\\end{align}[\/latex]<\/p>\n<p>Hence the integral for the volume of the solid region [latex]E[\/latex] becomes<\/p>\n<p style=\"text-align: center;\">[latex]{{V}{(E)}} = {\\displaystyle\\int^{{\\theta} = {{2}{\\pi}}}_{{\\theta} = {0}}} \\ {\\displaystyle\\int^{{\\rho} = {{\\sqrt{2}\/{2}}}}_{{\\rho} = {0}}} \\ {\\displaystyle\\int^{{\\varphi} = {{\\pi}\/{4}}}_{{\\varphi} = {0}}} \\ {{{\\rho}^{2}}{\\sin} \\ {\\varphi} \\ {{d}{\\varphi}} \\ {{d}{\\rho}} \\ {{d}{\\theta}}} \\ + {\\displaystyle\\int^{{\\theta} = {{2}{\\pi}}}_{{\\theta} = {0}}} \\ {\\displaystyle\\int^{{\\rho} = {1}}_{{\\rho} = {{\\sqrt{2}\/{2}}}}} \\ {\\displaystyle\\int^{{\\varphi} = {{\\cos}^{-1}}{\\rho}}_{{\\varphi} = {0}}} \\ {{{\\rho}^{2}}{\\sin} \\ {\\varphi} \\ {{d}{\\varphi}} \\ {{d}{\\rho}} \\ {{d}{\\theta}}}.[\/latex]<\/p>\n<p>In each case, the integration results in [latex]{{V}{(E)}} = {\\frac{\\pi}{8}}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Convert the following integral into cylindrical coordinates:<\/p>\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{{y} = {1}}_{{y} = {-1}}}\\displaystyle\\int^{{x} = {\\sqrt{{1} - {{y}^{2}}}}}_{{x} = {0}} \\ {\\displaystyle\\int^{{z} = {\\sqrt{{x^2}+{y^2}}}}_{{z} = {{x}^{2}}+{{y}^{2}}}} \\ {{x}{y}{z}} \\ {{d}{z}} \\ {{d}{x}} \\ {{d}{y}}.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q368178320\">Show Solution<\/button><\/p>\n<div id=\"q368178320\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793544553\">The ranges of the variables are<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {-1} & \\ \\ \\ {\\leq} \\ \\ \\ {y}{\\leq}{1} \\\\  {0} & \\ \\ \\ {\\leq} \\ \\ \\ {x}{\\leq}{\\sqrt{{1} - {{y}^{2}}}} \\\\  {{x}^{2}} + {{y}^{2}} & \\ \\ \\ {\\leq} \\ \\ \\ {z}{\\leq}{\\sqrt{{{x}^{2}}+{{y}^{2}}}}.  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1167793464597\">The first two inequalities describe the right half of a circle of radius 1. Therefore, the ranges for [latex]{\\theta}[\/latex] and [latex]r[\/latex] are<\/p>\n<p style=\"text-align: center;\">[latex]{-{\\frac{\\pi}{2}}} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {\\frac{\\pi}{2}} \\ {\\text{and}} \\ {0} \\ {\\leq} \\ {r} \\ {\\leq} \\ {1}.[\/latex]<\/p>\n<p id=\"fs-id1167793514913\">The limits of [latex]z[\/latex] are [latex]{{r}^{2}}[\/latex], hence<\/p>\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{{y} = {1}}_{{y} = {-1}}}\\displaystyle\\int^{{x} = {\\sqrt{{1} - {{y}^{2}}}}}_{{x} = {0}}\\displaystyle\\int^{{z} = {\\sqrt{{{x}^{2}} + {{y}^{2}}}}}_{{z} = {{x}^{2}} + {{y}^{2}}} \\ {{x}{y}{z}} \\ {{d}{z}} \\ {{d}{x}} \\ {{d}{y}} = {\\displaystyle\\int^{{\\theta} = {{\\pi}\/{2}}}_{{\\theta} = {-{\\pi}\/{2}}}} \\ {\\displaystyle\\int^{{r} = {1}}_{{r} = {0}}} \\ {\\displaystyle\\int^{{z} = {r}}_{{z} = {{r}^{2}}}} \\ {r}{({r}{\\cos}{\\theta})}{({r}{\\sin}{\\theta})}{z} \\ {{d}{z}} \\ {{d}{r}} \\ {{d}{\\theta}}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Convert the following integral into spherical coordinates:<\/p>\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{{y} = {3}}_{{y} = {0}}}\\displaystyle\\int^{{x} = {\\sqrt{{9} - {{y}^{2}}}}}_{{x} = {0}}\\displaystyle\\int^{{z} = {\\sqrt{{18} - {{x}^{2}} - {{y}^{2}}}}}_{{z} = {\\sqrt{{x}^{2} + {{y}^{2}}}}} \\ {({{x}^{2}} + {{y}^{2}} + {{z}^{2}})} \\ {{d}{z}} \\ {{d}{x}} \\ {{d}{y}}.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q255795354\">Show Solution<\/button><\/p>\n<div id=\"q255795354\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794052495\">The ranges of the variables are<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {0} \\ \\ \\ & {\\leq} \\ \\ \\ {y} \\ {\\leq} \\ {3} \\\\  {0} \\ \\ \\ & {\\leq} \\ \\ \\ {x} \\ {\\leq} \\ {\\sqrt{{9} - {{y}^{2}}}} \\\\  {\\sqrt{{{x}^{2}} + {{y}^{2}}}} \\ \\ \\ & {\\leq} \\ \\ \\ {z} \\ {\\leq} \\ {\\sqrt{{18} - {{x}^{2}} - {{y}^{2}}}}.  \\end{aligned}[\/latex]<\/p>\n<p id=\"fs-id1167793777086\">The first two ranges of variables describe a quarter disk in the first quadrant of the [latex]xy[\/latex]-plane. Hence the range for [latex]{\\theta}[\/latex] is [latex]{0} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {\\frac{\\pi}{2}}[\/latex].<\/p>\n<p id=\"fs-id1167794067621\">The lower bound [latex]{z} = {\\sqrt{{{x}^{2}} + {{y}^{2}}}}[\/latex] is the upper half of a cone and the upper bound [latex]{z} = {\\sqrt{{18} - {{x}^{2}} - {{y}^{2}}}}[\/latex] is the upper half of a sphere. Therefore, we have [latex]{0} \\ {\\leq} \\ {\\rho} \\ {\\leq} \\ {\\sqrt{18}}[\/latex], which is [latex]{0} \\ {\\leq} \\ {\\rho} \\ {\\leq} \\ {3{\\sqrt{2}}}[\/latex].<\/p>\n<p id=\"fs-id1167794168074\">For the ranges of [latex]{\\varphi}[\/latex], we need to find where the cone and the sphere intersect, so solve the equation<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {{{r}^{2}} + {{z}^{2}}} \\ \\ \\ & = \\ \\ \\ {18} \\\\  {({\\sqrt{{{x}^{2}} + {{y}^{2}}}})^2} + {{z}^{2}} \\ \\ \\ & = \\ \\ \\ {18} \\\\  {{z}^{2}} + {{z}^{2}} \\ \\ \\ & = \\ \\ \\ {18} \\\\  {{2}{{z}^{2}}} \\ \\ \\ & = \\ \\ \\ {18} \\\\  {{z}^{2}} \\ \\ \\ & = \\ \\ \\ {9} \\\\  {z} \\ \\ \\ & = \\ \\ \\ {3}.  \\end{aligned}[\/latex]<\/p>\n<p id=\"fs-id1167793610938\">This gives<\/p>\n<p>[latex]\\hspace{10cm}\\begin{align}    3\\sqrt2\\cos\\varphi&=3 \\\\    \\cos\\varphi&=\\frac1{\\sqrt2} \\\\    \\varphi&=\\frac{\\pi}4.    \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1167793924123\">Putting this together, we obtain<\/p>\n<div style=\"text-align: center;\">[latex]\\displaystyle\\int_{y=0}^{y=3}\\displaystyle\\int_{x=0}^{x=\\sqrt{9-y^2}}\\displaystyle\\int_{z=\\sqrt{x^2+y^2}}^{z=\\sqrt{18-x^2-y^2}}(x^2+y^2+z^2)dz \\ dx \\ dy=\\displaystyle\\int_{\\varphi=0}^{\\varphi=\\pi\/4}\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi\/2}\\displaystyle\\int_{\\rho=0}^{\\rho=3\\sqrt2}\\rho^4\\sin\\varphi \\ d\\rho \\ d\\theta \\ d\\varphi.[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Finding Volumes Using Spherical Coordinates<\/h3>\n<p class=\"whitespace-normal break-words\">Now that we are familiar with the spherical coordinate system, let&#8217;s find the volume of some known geometric figures, such as spheres and ellipsoids.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">Find the volume of the spherical planetarium in l\u2019Hemisph\u00e8ric in Valencia, Spain, which is five stories tall and has a radius of approximately [latex]50[\/latex] ft, using the equation [latex]x^2+y^2+z^2=r^2[\/latex].<\/p>\n<figure id=\"attachment_5569\" aria-describedby=\"caption-attachment-5569\" style=\"width: 325px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-5569\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/03022644\/5.60.jpeg\" alt=\"A picture of l\u2019Hemisph\u00e8ric, which is a giant glass structure that is in the shape of an ellipsoid.\" width=\"325\" height=\"219\" \/><figcaption id=\"caption-attachment-5569\" class=\"wp-caption-text\">Figure 7. (credit: modification of work by Javier Yaya Tur, Wikimedia Commons)<\/figcaption><\/figure>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q204959485\">Show Solution<\/button><\/p>\n<div id=\"q204959485\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793416704\">We calculate the volume of the ball in the first octant, where [latex]x\\geq0[\/latex], [latex]y\\geq0[\/latex], and [latex]z\\geq0[\/latex], using spherical coordinates, and then multiply the result by [latex]8[\/latex] for symmetry. Since we consider the region [latex]D[\/latex] as the first octant in the integral, the ranges of the variables are<\/p>\n<p style=\"text-align: center;\">[latex]\\large{0\\leq\\varphi\\leq\\frac{\\pi}2, \\ 0\\leq\\rho\\leq{r}, \\ 0\\leq\\theta\\leq\\frac{\\pi}2}[\/latex]<\/p>\n<p id=\"fs-id1167793500489\">Therefore,<\/p>\n<p>[latex]\\hspace{6cm}\\begin{align}    V&=\\underset{D}{\\displaystyle\\iiint}dx \\ dy \\ dz = 8\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi\/2}\\displaystyle\\int_{\\rho=0}^{\\rho=\\pi}\\displaystyle\\int_{\\varphi=0}^{\\varphi=\\pi\/2}\\rho^2\\sin\\theta \\ d\\varphi \\ d\\rho \\ d\\theta \\\\    &=8\\displaystyle\\int_{\\varphi=0}^{\\varphi=\\pi\/2}d\\varphi\\displaystyle\\int_{\\rho=0}^{\\rho=\\pi}\\rho^2 \\ d\\rho\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi\/2}\\sin\\theta \\ d\\theta \\\\    &=8\\left(\\frac{\\pi}2\\right)\\left(\\frac{r^3}3\\right)(1) \\\\    &=\\frac43\\pi{r}^3    \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1167793522367\">This exactly matches with what we knew. So for a sphere with a radius of approximately [latex]50[\/latex] ft, the volume is [latex]\\frac43\\pi{r}^3\\approx523,000\\text{ ft.}^3[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Find the volume of the ellipsoid [latex]\\frac{x^2}{a^2}+\\frac{y^2}{b^2}+\\frac{z^2}{c^2}=1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q543745872\">Show Solution<\/button><\/p>\n<div id=\"q543745872\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793251274\">We again use symmetry and evaluate the volume of the ellipsoid using spherical coordinates. As before, we use the first octant [latex]x\\geq0[\/latex], [latex]y\\geq0[\/latex], and [latex]z\\geq0[\/latex] and then multiply the result by 8.<\/p>\n<p id=\"fs-id1167793219213\">In this case the ranges of the variables are<\/p>\n<p style=\"text-align: center;\">[latex]\\large{0\\leq\\varphi\\leq\\frac{\\pi}2, \\ 0\\leq\\rho\\leq\\frac{\\pi}2, \\ 0\\leq\\rho\\leq1, \\ 0\\leq\\theta\\leq\\frac{\\pi}2}[\/latex]<\/p>\n<p id=\"fs-id1167793540294\">Also, we need to change the rectangular to spherical coordinates in this way:<\/p>\n<p style=\"text-align: center;\">[latex]\\large{x=a\\rho\\cos\\varphi\\sin\\theta, \\ y=b\\rho\\sin\\varphi\\sin\\theta,\\text{ and }z=c\\rho\\cos\\theta}.[\/latex]<\/p>\n<p id=\"fs-id1167794213144\">Then the volume of the ellipsoid becomes<\/p>\n<p>[latex]\\hspace{6cm}\\begin{align}V&=\\underset{D}{\\displaystyle\\iiint}dx \\ dy \\ dz \\\\&=8\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi\/2}\\displaystyle\\int_{\\rho=0}^{\\rho=1}\\displaystyle\\int_{\\varphi=0}^{\\varphi=\\pi\/2}abc\\rho^2\\sin\\theta \\ d\\varphi \\ d\\rho \\ d\\theta \\\\&=8abc\\displaystyle\\int_{\\varphi=0}^{\\varphi=\\pi\/2}d\\varphi\\displaystyle\\int_{\\rho=0}^{\\rho=1}\\rho^2 \\ d\\rho\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi\/2}\\sin\\theta \\ d\\theta \\\\&=8abc\\left(\\frac{\\pi}2\\right)\\left(\\frac13\\right)(1) \\\\&=\\frac43\\pi{abc}\\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Find the volume of the space inside the ellipsoid [latex]\\frac{x^2}{75^2}+\\frac{y^2}{80^2}+\\frac{z^2}{90^2}=1[\/latex] and outside the sphere [latex]x^2+y^2+z^2=50^2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q234085739\">Show Solution<\/button><\/p>\n<div id=\"q234085739\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793925860\">This problem is directly related to the l\u2019Hemisph\u00e8ric structure. The volume of space inside the ellipsoid and outside the sphere might be useful to find the expense of heating or cooling that space. We can use the preceding two examples for the volume of the sphere and ellipsoid and then subtract.<\/p>\n<p id=\"fs-id1167793732638\">First we find the volume of the ellipsoid using [latex]a=75[\/latex] ft, [latex]b=80[\/latex] ft, and [latex]c=90[\/latex] ft. The volume of the ellipsoid is:<\/p>\n<p style=\"text-align: center;\">[latex]\\large{V_{\\text{ellipsoid}}=\\frac43\\pi(75)(80)(90)\\approx2,262,000\\text{ ft.}^3}[\/latex]<\/p>\n<p id=\"fs-id1167793452781\">We found earlier, the volume of the sphere is:<\/p>\n<p style=\"text-align: center;\">[latex]\\large{V_{\\text{sphere}}\\approx523,600\\text{ ft.}^3.}[\/latex]<\/p>\n<p id=\"fs-id1167793556981\">Therefore, the volume of the space inside the ellipsoid [latex]\\frac{x^2}{75^2}+\\frac{y^2}{80^2}+\\frac{z^2}{90^2}=1[\/latex] and outside the sphere [latex]x^2+y^2+z^2=50^2[\/latex] is approximately<\/p>\n<p style=\"text-align: center;\">[latex]\\large{V_{\\text{Hemisferic}}=V_{\\text{ellipsoid}}-V_{\\text{sphere}}=1,738,400\\text{ ft.}^3.}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":14,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1268","chapter","type-chapter","status-publish","hentry"],"part":434,"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\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1268","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":8,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1268\/revisions"}],"predecessor-version":[{"id":1944,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1268\/revisions\/1944"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/434"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1268\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=1268"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=1268"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=1268"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=1268"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}