{"id":718,"date":"2025-10-09T15:34:21","date_gmt":"2025-10-09T15:34:21","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=718"},"modified":"2025-11-13T15:16:19","modified_gmt":"2025-11-13T15:16:19","slug":"alternative-coordinate-systems-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/alternative-coordinate-systems-fresh-take\/","title":{"raw":"Triple Integrals in Alternative Coordinate Systems: Fresh Take","rendered":"Triple Integrals in Alternative Coordinate Systems: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Evaluate a triple integral by changing to cylindrical coordinates<\/li>\r\n \t<li>Evaluate a triple integral by changing to spherical coordinates<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2 data-type=\"title\">Integration in Cylindrical Coordinates<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">When setting up triple integrals in cylindrical coordinates, the critical element is the volume element [latex]dV = r \\, dr \\, d\\theta \\, dz[\/latex]. That extra factor of [latex]r[\/latex] is essential\u2014it comes from the geometry of how cylindrical \"wedges\" expand as you move away from the [latex]z[\/latex]-axis.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Converting to Cylindrical Coordinates<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">To convert a triple integral from rectangular to cylindrical coordinates, substitute [latex]x = r\\cos\\theta[\/latex], [latex]y = r\\sin\\theta[\/latex], and [latex]z = z[\/latex] into the function. The integral becomes:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{B}{\\iiint} g(x,y,z) , dV = \\underset{B}{\\iiint} g(r\\cos\\theta, r\\sin\\theta, z) \\, r \\, dr \\, d\\theta \\, dz[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Don't forget that extra [latex]r[\/latex]! It's not part of the function\u2014it's part of the volume element itself.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Common Surface Equations<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Cylindrical coordinates simplify many surface equations. A circular cylinder [latex]x^2 + y^2 = c^2[\/latex] becomes simply [latex]r = c[\/latex]. A cone [latex]z^2 = c^2(x^2 + y^2)[\/latex] becomes [latex]z = cr[\/latex]. A paraboloid [latex]z = c(x^2 + y^2)[\/latex] becomes [latex]z = cr^2[\/latex]. A sphere [latex]x^2 + y^2 + z^2 = c^2[\/latex] becomes [latex]r^2 + z^2 = c^2[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Setting Up Bounds for General Regions<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">For a general region [latex]E[\/latex] bounded between two surfaces, project onto the [latex]r\\theta[\/latex]-plane to find [latex]D[\/latex]. If [latex]z[\/latex] varies between [latex]u_1(r,\\theta)[\/latex] and [latex]u_2(r,\\theta)[\/latex], the integral becomes:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{E}{\\iiint} f(r,\\theta,z) \\, r \\, dr \\, d\\theta \\, dz = \\underset{D}{\\iint} \\left[\\int_{u_1(r,\\theta)}^{u_2(r,\\theta)} f(r,\\theta,z) , dz\\right] r \\, dr \\, d\\theta [\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Identify surfaces with circular symmetry or expressions containing [latex]x^2 + y^2[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Convert surface equations to cylindrical form<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Determine bounds by examining projections and intersections<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Set up the integral with the volume element [latex]r \\, dr \\, d\\theta \\, dz[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Integrate in the order that simplifies computation\u2014the order can be changed just as with rectangular coordinates<\/li>\r\n<\/ol>\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Evaluate the triple integral\u00a0[latex]{\\displaystyle\\int^{{\\theta} = {\\pi}}_{{\\theta} = {0}}} \\ {\\displaystyle\\int^{r = 1}_{r = 0}} \\ {\\displaystyle\\int^{z = 4}_{z = 0}} \\ {rz} \\ {\\sin} \\ {{\\theta}{r}} \\ {dz} \\ {{d}{\\theta}}[\/latex].[reveal-answer q=\"470317685\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"470317685\"][latex]\\frac{16}{3}[\/latex][\/hidden-answer]<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8197108&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=jqgmsgHqSCY&amp;video_target=tpm-plugin-ryjsuejd-jqgmsgHqSCY\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP5.27_transcript.html\">transcript for \u201cCP 5.27\u201d here (opens in new window).<\/a><\/center><\/section><section class=\"textbox example\" aria-label=\"Example\">Consider the region [latex]E[\/latex] inside the right circular cylinder with equation [latex]{r} = {2}{\\sin} \\ {\\theta}[\/latex],\u00a0 bounded below by the [latex]{r} \\ {\\theta}[\/latex]-plane and bounded above by [latex]z=4-y[\/latex]. Set up a triple integral with a function [latex]{f}{({r},{\\theta},{z})}[\/latex] in cylindrical coordinates.[reveal-answer q=\"512447813\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"512447813\"][latex]\\underset{E}{\\displaystyle\\iiint}f(r,\\theta,z)r \\ dz \\ d\\theta=\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi}\\displaystyle\\int_{r=0}^{r=2\\sin\\theta}\\displaystyle\\int_{z=0}^{z=4-r\\sin\\theta}f(r,\\theta,z)r \\ dz \\ dr \\ d\\theta.[\/latex][\/hidden-answer]<\/section>\r\n<h2 data-type=\"title\">Integration in Spherical Coordinates<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Spherical coordinates shine when dealing with regions that have symmetry around a point, especially spheres, cones, and other shapes centered at the origin. The critical element is the volume element [latex]dV = \\rho^2 \\sin\\varphi , d\\rho , d\\varphi , d\\theta[\/latex]. Those extra factors [latex]\\rho^2 \\sin\\varphi[\/latex] account for how spherical \"wedges\" change size based on distance from the origin and angle from the [latex]z[\/latex]-axis.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Converting to Spherical Coordinates<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">To convert a triple integral from rectangular to spherical coordinates, substitute [latex]x = \\rho\\sin\\varphi\\cos\\theta[\/latex], [latex]y = \\rho\\sin\\varphi\\sin\\theta[\/latex], and [latex]z = \\rho\\cos\\varphi[\/latex] into the function. The integral becomes:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{B}{\\iiint} g(x,y,z) \\, dV = \\underset{B}{\\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 [latex]\\rho^2[\/latex] accounts for radial expansion (things get bigger farther from the origin), while [latex]\\sin\\varphi[\/latex] accounts for how the circumference at angle [latex]\\varphi[\/latex] from the [latex]z[\/latex]-axis changes.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>When to Use Spherical Coordinates<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Look for expressions like [latex]x^2 + y^2 + z^2[\/latex], which becomes simply [latex]\\rho^2[\/latex] in spherical coordinates. Spherical coordinates are ideal for spheres, cones pointing along the [latex]z[\/latex]-axis, and regions centered at the origin. A sphere [latex]x^2 + y^2 + z^2 = c^2[\/latex] becomes [latex]\\rho = c[\/latex]. A cone [latex]z = \\sqrt{x^2 + y^2}[\/latex] becomes [latex]\\varphi = \\frac{\\pi}{4}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Setting Up Bounds<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">For a spherical box, ranges are straightforward: [latex]a \\leq \\rho \\leq b[\/latex], [latex]\\alpha \\leq \\theta \\leq \\beta[\/latex], and [latex]\\gamma \\leq \\varphi \\leq \\psi[\/latex]. For general regions, project onto the [latex]\\rho\\theta[\/latex]-plane and determine how [latex]\\varphi[\/latex] varies. Find intersections by converting surface equations to spherical form and solving for the variables.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Convert surface equations to spherical coordinates<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Identify geometric boundaries (spheres become constants for [latex]\\rho[\/latex], cones become constants for [latex]\\varphi[\/latex])<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Find intersection points to determine ranges<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Set up the integral with the volume element [latex]\\rho^2 \\sin\\varphi , d\\rho , d\\varphi , d\\theta[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Use symmetry when possible to simplify calculations<\/li>\r\n<\/ol>\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Set up a triple integral for the volume of the solid region bounded above by the sphere [latex]{\\rho} = {2}[\/latex] and bounded below by the cone [latex]{\\varphi} = {{\\pi}\/{3}}[\/latex].[reveal-answer q=\"422572964\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"422572964\"][latex]V(E)=\\displaystyle\\int_{\\theta=0}^{\\theta=2\\pi}\\displaystyle\\int_{\\phi=0}^{\\varphi=\\pi\/3}\\displaystyle\\int_{\\rho=0}^{\\rho=2}\\rho^2\\sin\\varphi \\ d\\rho \\ d\\varphi \\ d\\theta.[\/latex][\/hidden-answer]<\/section><section class=\"textbox example\" aria-label=\"Example\">Use rectangular, cylindrical, and spherical coordinates to set up triple integrals for finding the volume of the region inside the sphere [latex]x^2+y^2+z^2=4[\/latex] but outside the cylinder\u00a0[latex]x^2+y^2=1[\/latex].\r\n\r\n[caption id=\"attachment_1402\" align=\"aligncenter\" width=\"493\"]<img class=\"size-full wp-image-1402\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26173912\/5-5-11.jpeg\" alt=\"A sphere with equation x squared + y squared + z squared = 4, and within it, a cylinder with equation x squared + y squared = 1.\" width=\"493\" height=\"402\" \/> Figure 6.[\/caption]\r\n\r\n[reveal-answer q=\"842692876\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"842692876\"]\r\n\r\nRectangular:\u00a0[latex]\\displaystyle\\int_{x=-2}^{x=2}\\displaystyle\\int_{y=-\\sqrt{4-x^2}}^{y=\\sqrt{4-x^2}}\\displaystyle\\int_{z=-\\sqrt{4-x^2-y^2}}^{z=\\sqrt{4-x^2-y^2}}dz \\ dy \\ dx - \\displaystyle\\int_{x=-1}^{x=1}\\displaystyle\\int_{y=-\\sqrt{1-x^2}}^{y=\\sqrt{1-x^2}}\\displaystyle\\int_{z=-\\sqrt{4-x^2-y^2}}^{z=\\sqrt{4-x^2-y^2}}dz \\ dy \\ dx.[\/latex]\r\n\r\nCylindrical:\u00a0[latex]\\displaystyle\\int_{\\theta=0}^{\\theta=2\\pi}\\displaystyle\\int_{r=1}^{r=2}\\displaystyle\\int_{z=-\\sqrt{4-r^2}}^{z=\\sqrt{4-r^2}}r \\ dz \\ dr \\ d\\theta.[\/latex]\r\n\r\nSpherical:\u00a0[latex]\\displaystyle\\int_{\\varphi=\\pi\/6}^{\\varphi=5\\pi\/6}\\displaystyle\\int_{\\theta=0}^{\\theta=2\\pi}\\displaystyle\\int_{\\rho=\\csc\\varphi}^{\\rho=2} \\rho^2 \\sin{\\varphi} d\\rho \\ d\\theta \\ d\\varphi[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8197109&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=7iLbCVweWdI&amp;video_target=tpm-plugin-9elr1s1g-7iLbCVweWdI\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP5.32_transcript.html\">transcript for \u201cCP 5.32\u201d here (opens in new window).<\/a><\/center><\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Evaluate a triple integral by changing to cylindrical coordinates<\/li>\n<li>Evaluate a triple integral by changing to spherical coordinates<\/li>\n<\/ul>\n<\/section>\n<h2 data-type=\"title\">Integration in Cylindrical Coordinates<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">When setting up triple integrals in cylindrical coordinates, the critical element is the volume element [latex]dV = r \\, dr \\, d\\theta \\, dz[\/latex]. That extra factor of [latex]r[\/latex] is essential\u2014it comes from the geometry of how cylindrical &#8220;wedges&#8221; expand as you move away from the [latex]z[\/latex]-axis.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Converting to Cylindrical Coordinates<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">To convert a triple integral from rectangular to cylindrical coordinates, substitute [latex]x = r\\cos\\theta[\/latex], [latex]y = r\\sin\\theta[\/latex], and [latex]z = z[\/latex] into the function. The integral becomes:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{B}{\\iiint} g(x,y,z) , dV = \\underset{B}{\\iiint} g(r\\cos\\theta, r\\sin\\theta, z) \\, r \\, dr \\, d\\theta \\, dz[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Don&#8217;t forget that extra [latex]r[\/latex]! It&#8217;s not part of the function\u2014it&#8217;s part of the volume element itself.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Common Surface Equations<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Cylindrical coordinates simplify many surface equations. A circular cylinder [latex]x^2 + y^2 = c^2[\/latex] becomes simply [latex]r = c[\/latex]. A cone [latex]z^2 = c^2(x^2 + y^2)[\/latex] becomes [latex]z = cr[\/latex]. A paraboloid [latex]z = c(x^2 + y^2)[\/latex] becomes [latex]z = cr^2[\/latex]. A sphere [latex]x^2 + y^2 + z^2 = c^2[\/latex] becomes [latex]r^2 + z^2 = c^2[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Setting Up Bounds for General Regions<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">For a general region [latex]E[\/latex] bounded between two surfaces, project onto the [latex]r\\theta[\/latex]-plane to find [latex]D[\/latex]. If [latex]z[\/latex] varies between [latex]u_1(r,\\theta)[\/latex] and [latex]u_2(r,\\theta)[\/latex], the integral becomes:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{E}{\\iiint} f(r,\\theta,z) \\, r \\, dr \\, d\\theta \\, dz = \\underset{D}{\\iint} \\left[\\int_{u_1(r,\\theta)}^{u_2(r,\\theta)} f(r,\\theta,z) , dz\\right] r \\, dr \\, d\\theta[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Identify surfaces with circular symmetry or expressions containing [latex]x^2 + y^2[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Convert surface equations to cylindrical form<\/li>\n<li class=\"whitespace-normal break-words\">Determine bounds by examining projections and intersections<\/li>\n<li class=\"whitespace-normal break-words\">Set up the integral with the volume element [latex]r \\, dr \\, d\\theta \\, dz[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Integrate in the order that simplifies computation\u2014the order can be changed just as with rectangular coordinates<\/li>\n<\/ol>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Evaluate the triple integral\u00a0[latex]{\\displaystyle\\int^{{\\theta} = {\\pi}}_{{\\theta} = {0}}} \\ {\\displaystyle\\int^{r = 1}_{r = 0}} \\ {\\displaystyle\\int^{z = 4}_{z = 0}} \\ {rz} \\ {\\sin} \\ {{\\theta}{r}} \\ {dz} \\ {{d}{\\theta}}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q470317685\">Show Solution<\/button><\/p>\n<div id=\"q470317685\" class=\"hidden-answer\" style=\"display: none\">[latex]\\frac{16}{3}[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8197108&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=jqgmsgHqSCY&amp;video_target=tpm-plugin-ryjsuejd-jqgmsgHqSCY\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP5.27_transcript.html\">transcript for \u201cCP 5.27\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Consider the region [latex]E[\/latex] inside the right circular cylinder with equation [latex]{r} = {2}{\\sin} \\ {\\theta}[\/latex],\u00a0 bounded below by the [latex]{r} \\ {\\theta}[\/latex]-plane and bounded above by [latex]z=4-y[\/latex]. Set up a triple integral with a function [latex]{f}{({r},{\\theta},{z})}[\/latex] in cylindrical coordinates.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q512447813\">Show Solution<\/button><\/p>\n<div id=\"q512447813\" class=\"hidden-answer\" style=\"display: none\">[latex]\\underset{E}{\\displaystyle\\iiint}f(r,\\theta,z)r \\ dz \\ d\\theta=\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi}\\displaystyle\\int_{r=0}^{r=2\\sin\\theta}\\displaystyle\\int_{z=0}^{z=4-r\\sin\\theta}f(r,\\theta,z)r \\ dz \\ dr \\ d\\theta.[\/latex]<\/div>\n<\/div>\n<\/section>\n<h2 data-type=\"title\">Integration in Spherical Coordinates<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Spherical coordinates shine when dealing with regions that have symmetry around a point, especially spheres, cones, and other shapes centered at the origin. The critical element is the volume element [latex]dV = \\rho^2 \\sin\\varphi , d\\rho , d\\varphi , d\\theta[\/latex]. Those extra factors [latex]\\rho^2 \\sin\\varphi[\/latex] account for how spherical &#8220;wedges&#8221; change size based on distance from the origin and angle from the [latex]z[\/latex]-axis.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Converting to Spherical Coordinates<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">To convert a triple integral from rectangular to spherical coordinates, substitute [latex]x = \\rho\\sin\\varphi\\cos\\theta[\/latex], [latex]y = \\rho\\sin\\varphi\\sin\\theta[\/latex], and [latex]z = \\rho\\cos\\varphi[\/latex] into the function. The integral becomes:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{B}{\\iiint} g(x,y,z) \\, dV = \\underset{B}{\\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 [latex]\\rho^2[\/latex] accounts for radial expansion (things get bigger farther from the origin), while [latex]\\sin\\varphi[\/latex] accounts for how the circumference at angle [latex]\\varphi[\/latex] from the [latex]z[\/latex]-axis changes.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>When to Use Spherical Coordinates<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Look for expressions like [latex]x^2 + y^2 + z^2[\/latex], which becomes simply [latex]\\rho^2[\/latex] in spherical coordinates. Spherical coordinates are ideal for spheres, cones pointing along the [latex]z[\/latex]-axis, and regions centered at the origin. A sphere [latex]x^2 + y^2 + z^2 = c^2[\/latex] becomes [latex]\\rho = c[\/latex]. A cone [latex]z = \\sqrt{x^2 + y^2}[\/latex] becomes [latex]\\varphi = \\frac{\\pi}{4}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Setting Up Bounds<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">For a spherical box, ranges are straightforward: [latex]a \\leq \\rho \\leq b[\/latex], [latex]\\alpha \\leq \\theta \\leq \\beta[\/latex], and [latex]\\gamma \\leq \\varphi \\leq \\psi[\/latex]. For general regions, project onto the [latex]\\rho\\theta[\/latex]-plane and determine how [latex]\\varphi[\/latex] varies. Find intersections by converting surface equations to spherical form and solving for the variables.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Convert surface equations to spherical coordinates<\/li>\n<li class=\"whitespace-normal break-words\">Identify geometric boundaries (spheres become constants for [latex]\\rho[\/latex], cones become constants for [latex]\\varphi[\/latex])<\/li>\n<li class=\"whitespace-normal break-words\">Find intersection points to determine ranges<\/li>\n<li class=\"whitespace-normal break-words\">Set up the integral with the volume element [latex]\\rho^2 \\sin\\varphi , d\\rho , d\\varphi , d\\theta[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Use symmetry when possible to simplify calculations<\/li>\n<\/ol>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Set up a triple integral for the volume of the solid region bounded above by the sphere [latex]{\\rho} = {2}[\/latex] and bounded below by the cone [latex]{\\varphi} = {{\\pi}\/{3}}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q422572964\">Show Solution<\/button><\/p>\n<div id=\"q422572964\" class=\"hidden-answer\" style=\"display: none\">[latex]V(E)=\\displaystyle\\int_{\\theta=0}^{\\theta=2\\pi}\\displaystyle\\int_{\\phi=0}^{\\varphi=\\pi\/3}\\displaystyle\\int_{\\rho=0}^{\\rho=2}\\rho^2\\sin\\varphi \\ d\\rho \\ d\\varphi \\ d\\theta.[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Use rectangular, cylindrical, and spherical coordinates to set up triple integrals for finding the volume of the region inside the sphere [latex]x^2+y^2+z^2=4[\/latex] but outside the cylinder\u00a0[latex]x^2+y^2=1[\/latex].<\/p>\n<figure id=\"attachment_1402\" aria-describedby=\"caption-attachment-1402\" style=\"width: 493px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1402\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26173912\/5-5-11.jpeg\" alt=\"A sphere with equation x squared + y squared + z squared = 4, and within it, a cylinder with equation x squared + y squared = 1.\" width=\"493\" height=\"402\" \/><figcaption id=\"caption-attachment-1402\" class=\"wp-caption-text\">Figure 6.<\/figcaption><\/figure>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q842692876\">Show Solution<\/button><\/p>\n<div id=\"q842692876\" class=\"hidden-answer\" style=\"display: none\">\n<p>Rectangular:\u00a0[latex]\\displaystyle\\int_{x=-2}^{x=2}\\displaystyle\\int_{y=-\\sqrt{4-x^2}}^{y=\\sqrt{4-x^2}}\\displaystyle\\int_{z=-\\sqrt{4-x^2-y^2}}^{z=\\sqrt{4-x^2-y^2}}dz \\ dy \\ dx - \\displaystyle\\int_{x=-1}^{x=1}\\displaystyle\\int_{y=-\\sqrt{1-x^2}}^{y=\\sqrt{1-x^2}}\\displaystyle\\int_{z=-\\sqrt{4-x^2-y^2}}^{z=\\sqrt{4-x^2-y^2}}dz \\ dy \\ dx.[\/latex]<\/p>\n<p>Cylindrical:\u00a0[latex]\\displaystyle\\int_{\\theta=0}^{\\theta=2\\pi}\\displaystyle\\int_{r=1}^{r=2}\\displaystyle\\int_{z=-\\sqrt{4-r^2}}^{z=\\sqrt{4-r^2}}r \\ dz \\ dr \\ d\\theta.[\/latex]<\/p>\n<p>Spherical:\u00a0[latex]\\displaystyle\\int_{\\varphi=\\pi\/6}^{\\varphi=5\\pi\/6}\\displaystyle\\int_{\\theta=0}^{\\theta=2\\pi}\\displaystyle\\int_{\\rho=\\csc\\varphi}^{\\rho=2} \\rho^2 \\sin{\\varphi} d\\rho \\ d\\theta \\ d\\varphi[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8197109&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=7iLbCVweWdI&amp;video_target=tpm-plugin-9elr1s1g-7iLbCVweWdI\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP5.32_transcript.html\">transcript for \u201cCP 5.32\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":16,"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-718","chapter","type-chapter","status-publish","hentry"],"part":434,"module-header":"fresh_take","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/718","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":12,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/718\/revisions"}],"predecessor-version":[{"id":2205,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/718\/revisions\/2205"}],"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\/718\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=718"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=718"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=718"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=718"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}