{"id":2654,"date":"2025-11-20T16:29:22","date_gmt":"2025-11-20T16:29:22","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=2654"},"modified":"2026-05-26T14:12:04","modified_gmt":"2026-05-26T14:12:04","slug":"double-integration-and-applications-background-youll-need-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/double-integration-and-applications-background-youll-need-4\/","title":{"raw":"Double Integration and Applications: Background You\u2019ll Need 4","rendered":"Double Integration and Applications: Background You\u2019ll Need 4"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li><span data-sheets-root=\"1\">Find the volume of a solid by using the disk method<\/span><\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>The Disk Method<\/h2>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">The disk method is a technique for finding volumes of solids of revolution. When you revolve a region around an axis, the cross-sections perpendicular to that axis are circular disks. This method uses integration to sum the volumes of these infinitely thin disks.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">Consider revolving the region between the graph of [latex]f(x)={(x-1)}^{2}+1[\/latex] and the [latex]x[\/latex]-axis over the interval [latex][-1,3][\/latex] around the [latex]x[\/latex]-axis. The process creates a three-dimensional solid, and each cross-section perpendicular to the [latex]x[\/latex]-axis is a circular disk.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"899\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11212809\/CNX_Calc_Figure_06_02_009.jpg\" alt=\"This figure has four graphs. The first graph, labeled \u201ca\u201d is a parabola f(x)=(x-1)^2+1. The curve is above the x-axis and intersects the y-axis at y=2. Under the curve in the first quadrant is a vertical rectangle starting at the x-axis and stopping at the curve. The second graph, labeled \u201cb\u201d is the same parabola as in the first graph. The rectangle under the parabola from the first graph has been rotated around the x-axis forming a solid disk. The third graph labeled \u201cc\u201d is the same parabola as the first graph. There is a shaded region bounded above by the parabola, to the left by the line x=-1 and to the right by the line x=3, and below by the x-axis. The fourth graph labeled \u201cd\u201d is the same parabola as the first graph. The region from the third graph has been revolved around the x-axis to form a solid.\" width=\"899\" height=\"978\" \/> Figure 9. (a) A thin rectangle for approximating the area under a curve. (b) A representative disk formed by revolving the rectangle about the [latex]x\\text{-axis}\\text{.}[\/latex] (c) The region under the curve is revolved about the [latex]x\\text{-axis},[\/latex] resulting in (d) the solid of revolution.[\/caption]\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">The disk method builds on the general slicing method for volumes. You already know that volume can be found using:<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\" style=\"text-align: center;\">[latex]V=\\int_{a}^{b}A(x)dx[\/latex]<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">The key difference with the disk method is that the cross-sectional area [latex]A(x)[\/latex] is always the area of a circle: [latex]A(x)=\\pi[r(x)]^{2}[\/latex], where [latex]r(x)[\/latex] is the radius of the disk at position [latex]x[\/latex].<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>the disk method<\/h3>\r\n<p id=\"fs-id1167793895393\">Let [latex]f(x)[\/latex] be continuous and nonnegative.<\/p>\r\nDefine [latex]R[\/latex] as the region bounded above by the graph of [latex]f(x),[\/latex] below by the [latex]x\\text{-axis,}[\/latex] on the left by the line [latex]x=a,[\/latex] and on the right by the line [latex]x=b.[\/latex]\r\n\r\n&nbsp;\r\n\r\nThen, the volume of the solid of revolution formed by revolving [latex]R[\/latex] around the [latex]x\\text{-axis}[\/latex] is given by:\r\n<div id=\"fs-id1167794099899\" class=\"equation\" style=\"text-align: center;\">[latex]V={\\displaystyle\\int }_{a}^{b}\\pi {\\left[f(x)\\right]}^{2}dx.[\/latex]<\/div>\r\n<\/section><section class=\"textbox example\">Use the disk method to find the volume of the solid of revolution generated by rotating the region between the graph of [latex]f(x)=\\sqrt{x}[\/latex] and the [latex]x\\text{-axis}[\/latex] over the interval [latex]\\left[1,4\\right][\/latex] around the [latex]x\\text{-axis}\\text{.}[\/latex][reveal-answer q=\"fs-id1167793419916\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793419916\"]\r\n<p id=\"fs-id1167793419916\">The graphs of the function and the solid of revolution are shown in the following figure.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"740\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11212813\/CNX_Calc_Figure_06_02_010.jpg\" alt=\"This figure has two graphs. The first graph labeled \u201ca\u201d is the curve f(x) = squareroot(x). It is an increasing curve above the x-axis. The curve is in the first quadrant. Under the curve is a region bounded by x=1 and x=4. The bottom of the region is the x-axis. The second graph labeled \u201cb\u201d is the same curve as the first graph. The solid region from the first graph has been rotated around the x-axis to form a solid region.\" width=\"740\" height=\"386\" \/> Figure 10. (a) The function [latex]f(x)=\\sqrt{x}[\/latex] over the interval [latex]\\left[1,4\\right].[\/latex] (b) The solid of revolution obtained by revolving the region under the graph of [latex]f(x)[\/latex] about the [latex]x\\text{-axis}.[\/latex][\/caption]\r\n<p id=\"fs-id1167793389815\">We have<\/p>\r\n\r\n<div id=\"fs-id1167793287538\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill V&amp; ={\\displaystyle\\int }_{a}^{b}\\pi {\\left[f(x)\\right]}^{2}dx\\hfill \\\\ &amp; ={\\displaystyle\\int }_{1}^{4}\\pi {\\left[\\sqrt{x}\\right]}^{2}dx=\\pi {\\displaystyle\\int }_{1}^{4}xdx\\hfill \\\\ &amp; ={\\frac{\\pi }{2}{x}^{2}|}_{1}^{4}=\\frac{15\\pi }{2}.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1167794047779\">The volume is [latex]\\frac{(15\\pi )}{2}[\/latex] units<sup>3<\/sup>.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<h3 id=\"fs-id1167793951744\">Revolving Around the [latex]y[\/latex]-axis<\/h3>\r\nYou can revolve a region around any horizontal or vertical line. When revolving around the [latex]y[\/latex]-axis, the mechanics are similar, but you express the function in terms of [latex]y[\/latex] and integrate with respect to [latex]y[\/latex].\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>the disk method for solids of revolution around the [latex]y[\/latex]-axis<\/h3>\r\n<p id=\"fs-id1167794006397\">Let [latex]g(y)[\/latex] be continuous and nonnegative.\u00a0Define [latex]Q[\/latex] as the region bounded on the right by the graph of [latex]g(y),[\/latex] on the left by the [latex]y\\text{-axis,}[\/latex] below by the line [latex]y=c,[\/latex] and above by the line [latex]y=d.[\/latex]<\/p>\r\n&nbsp;\r\n\r\nThen, the volume of the solid of revolution formed by revolving [latex]Q[\/latex] around the [latex]y\\text{-axis}[\/latex] is given by:\r\n<div id=\"fs-id1167793499149\" class=\"equation\" style=\"text-align: center;\">[latex]V={\\displaystyle\\int }_{c}^{d}\\pi {\\left[g(y)\\right]}^{2}dy.[\/latex]<\/div>\r\n<\/section><section class=\"textbox example\">Let [latex]R[\/latex] be the region bounded by the graph of [latex]g(y)=\\sqrt{4-y}[\/latex] and the [latex]y\\text{-axis}[\/latex] over the [latex]y\\text{-axis}[\/latex] interval [latex]\\left[0,4\\right].[\/latex] Use the disk method to find the volume of the solid of revolution generated by rotating [latex]R[\/latex] around the [latex]y\\text{-axis}\\text{.}[\/latex][reveal-answer q=\"fs-id1167793368748\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793368748\"]\r\n<p id=\"fs-id1167793368748\">Figure 11 shows the function and a representative disk that can be used to estimate the volume. Notice that since we are revolving the function around the [latex]y\\text{-axis,}[\/latex] the disks are horizontal, rather than vertical.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"740\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11212816\/CNX_Calc_Figure_06_02_011.jpg\" alt=\"This figure has two graphs. The first graph labeled \u201ca\u201d is the curve g(y) = squareroot(4-y). It is a decreasing curve starting on the y-axis at y=4. Between the curve and the y-axis is a horizontal rectangle. The rectangle starts at the y-axis and stops at the curve. The second graph labeled \u201cb\u201d is the same curve as the first graph. The rectangle from the first graph has been rotated around the y-axis to form a horizontal disk.\" width=\"740\" height=\"386\" \/> Figure 11. (a) Shown is a thin rectangle between the curve of the function [latex]g(y)=\\sqrt{4-y}[\/latex] and the [latex]y\\text{-axis}\\text{.}[\/latex] (b) The rectangle forms a representative disk after revolution around the [latex]y\\text{-axis}\\text{.}[\/latex][\/caption]\r\n<p id=\"fs-id1167794142277\">The region to be revolved and the full solid of revolution are depicted in the following figure.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"740\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11212820\/CNX_Calc_Figure_06_02_012.jpg\" alt=\"This figure has two graphs. The first graph labeled \u201ca\u201d is the curve g(y) = squareroot(4-y). It is a decreasing curve starting on the y-axis at y=4. The region formed by the x-axis, the y-axis, and the curve is shaded. This region is in the first quadrant. The second graph labeled \u201cb\u201d is the same curve as the first graph. The region from the first graph has been rotated around the y-axis to form a solid.\" width=\"740\" height=\"386\" \/> Figure 12. (a) The region to the left of the function [latex]g(y)=\\sqrt{4-y}[\/latex] over the [latex]y\\text{-axis}[\/latex] interval [latex]\\left[0,4\\right].[\/latex] (b) The solid of revolution formed by revolving the region about the [latex]y\\text{-axis}\\text{.}[\/latex][\/caption]\r\n<p id=\"fs-id1167794331230\">To find the volume, we integrate with respect to [latex]y.[\/latex] We obtain<\/p>\r\n\r\n<div id=\"fs-id1167793498482\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill V&amp; ={\\displaystyle\\int }_{c}^{d}\\pi {\\left[g(y)\\right]}^{2}dy\\hfill \\\\ &amp; ={\\displaystyle\\int }_{0}^{4}\\pi {\\left[\\sqrt{4-y}\\right]}^{2}dy=\\pi {\\displaystyle\\int }_{0}^{4}(4-y)dy\\hfill \\\\ &amp; ={\\pi \\left[4y-\\frac{{y}^{2}}{2}\\right]|}_{0}^{4}=8\\pi .\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793259617\">The volume is [latex]8\\pi [\/latex] units<sup>3<\/sup>.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\">[ohm_question hide_question_numbers=1]325821[\/ohm_question]<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li><span data-sheets-root=\"1\">Find the volume of a solid by using the disk method<\/span><\/li>\n<\/ul>\n<\/section>\n<h2>The Disk Method<\/h2>\n<p class=\"font-claude-response-body whitespace-normal break-words\">The disk method is a technique for finding volumes of solids of revolution. When you revolve a region around an axis, the cross-sections perpendicular to that axis are circular disks. This method uses integration to sum the volumes of these infinitely thin disks.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">Consider revolving the region between the graph of [latex]f(x)={(x-1)}^{2}+1[\/latex] and the [latex]x[\/latex]-axis over the interval [latex][-1,3][\/latex] around the [latex]x[\/latex]-axis. The process creates a three-dimensional solid, and each cross-section perpendicular to the [latex]x[\/latex]-axis is a circular disk.<\/p>\n<figure style=\"width: 899px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11212809\/CNX_Calc_Figure_06_02_009.jpg\" alt=\"This figure has four graphs. The first graph, labeled \u201ca\u201d is a parabola f(x)=(x-1)^2+1. The curve is above the x-axis and intersects the y-axis at y=2. Under the curve in the first quadrant is a vertical rectangle starting at the x-axis and stopping at the curve. The second graph, labeled \u201cb\u201d is the same parabola as in the first graph. The rectangle under the parabola from the first graph has been rotated around the x-axis forming a solid disk. The third graph labeled \u201cc\u201d is the same parabola as the first graph. There is a shaded region bounded above by the parabola, to the left by the line x=-1 and to the right by the line x=3, and below by the x-axis. The fourth graph labeled \u201cd\u201d is the same parabola as the first graph. The region from the third graph has been revolved around the x-axis to form a solid.\" width=\"899\" height=\"978\" \/><figcaption class=\"wp-caption-text\">Figure 9. (a) A thin rectangle for approximating the area under a curve. (b) A representative disk formed by revolving the rectangle about the [latex]x\\text{-axis}\\text{.}[\/latex] (c) The region under the curve is revolved about the [latex]x\\text{-axis},[\/latex] resulting in (d) the solid of revolution.<\/figcaption><\/figure>\n<p class=\"font-claude-response-body whitespace-normal break-words\">The disk method builds on the general slicing method for volumes. You already know that volume can be found using:<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\" style=\"text-align: center;\">[latex]V=\\int_{a}^{b}A(x)dx[\/latex]<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">The key difference with the disk method is that the cross-sectional area [latex]A(x)[\/latex] is always the area of a circle: [latex]A(x)=\\pi[r(x)]^{2}[\/latex], where [latex]r(x)[\/latex] is the radius of the disk at position [latex]x[\/latex].<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>the disk method<\/h3>\n<p id=\"fs-id1167793895393\">Let [latex]f(x)[\/latex] be continuous and nonnegative.<\/p>\n<p>Define [latex]R[\/latex] as the region bounded above by the graph of [latex]f(x),[\/latex] below by the [latex]x\\text{-axis,}[\/latex] on the left by the line [latex]x=a,[\/latex] and on the right by the line [latex]x=b.[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>Then, the volume of the solid of revolution formed by revolving [latex]R[\/latex] around the [latex]x\\text{-axis}[\/latex] is given by:<\/p>\n<div id=\"fs-id1167794099899\" class=\"equation\" style=\"text-align: center;\">[latex]V={\\displaystyle\\int }_{a}^{b}\\pi {\\left[f(x)\\right]}^{2}dx.[\/latex]<\/div>\n<\/section>\n<section class=\"textbox example\">Use the disk method to find the volume of the solid of revolution generated by rotating the region between the graph of [latex]f(x)=\\sqrt{x}[\/latex] and the [latex]x\\text{-axis}[\/latex] over the interval [latex]\\left[1,4\\right][\/latex] around the [latex]x\\text{-axis}\\text{.}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167793419916\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167793419916\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793419916\">The graphs of the function and the solid of revolution are shown in the following figure.<\/p>\n<figure style=\"width: 740px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11212813\/CNX_Calc_Figure_06_02_010.jpg\" alt=\"This figure has two graphs. The first graph labeled \u201ca\u201d is the curve f(x) = squareroot(x). It is an increasing curve above the x-axis. The curve is in the first quadrant. Under the curve is a region bounded by x=1 and x=4. The bottom of the region is the x-axis. The second graph labeled \u201cb\u201d is the same curve as the first graph. The solid region from the first graph has been rotated around the x-axis to form a solid region.\" width=\"740\" height=\"386\" \/><figcaption class=\"wp-caption-text\">Figure 10. (a) The function [latex]f(x)=\\sqrt{x}[\/latex] over the interval [latex]\\left[1,4\\right].[\/latex] (b) The solid of revolution obtained by revolving the region under the graph of [latex]f(x)[\/latex] about the [latex]x\\text{-axis}.[\/latex]<\/figcaption><\/figure>\n<p id=\"fs-id1167793389815\">We have<\/p>\n<div id=\"fs-id1167793287538\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill V& ={\\displaystyle\\int }_{a}^{b}\\pi {\\left[f(x)\\right]}^{2}dx\\hfill \\\\ & ={\\displaystyle\\int }_{1}^{4}\\pi {\\left[\\sqrt{x}\\right]}^{2}dx=\\pi {\\displaystyle\\int }_{1}^{4}xdx\\hfill \\\\ & ={\\frac{\\pi }{2}{x}^{2}|}_{1}^{4}=\\frac{15\\pi }{2}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167794047779\">The volume is [latex]\\frac{(15\\pi )}{2}[\/latex] units<sup>3<\/sup>.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<h3 id=\"fs-id1167793951744\">Revolving Around the [latex]y[\/latex]-axis<\/h3>\n<p>You can revolve a region around any horizontal or vertical line. When revolving around the [latex]y[\/latex]-axis, the mechanics are similar, but you express the function in terms of [latex]y[\/latex] and integrate with respect to [latex]y[\/latex].<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>the disk method for solids of revolution around the [latex]y[\/latex]-axis<\/h3>\n<p id=\"fs-id1167794006397\">Let [latex]g(y)[\/latex] be continuous and nonnegative.\u00a0Define [latex]Q[\/latex] as the region bounded on the right by the graph of [latex]g(y),[\/latex] on the left by the [latex]y\\text{-axis,}[\/latex] below by the line [latex]y=c,[\/latex] and above by the line [latex]y=d.[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>Then, the volume of the solid of revolution formed by revolving [latex]Q[\/latex] around the [latex]y\\text{-axis}[\/latex] is given by:<\/p>\n<div id=\"fs-id1167793499149\" class=\"equation\" style=\"text-align: center;\">[latex]V={\\displaystyle\\int }_{c}^{d}\\pi {\\left[g(y)\\right]}^{2}dy.[\/latex]<\/div>\n<\/section>\n<section class=\"textbox example\">Let [latex]R[\/latex] be the region bounded by the graph of [latex]g(y)=\\sqrt{4-y}[\/latex] and the [latex]y\\text{-axis}[\/latex] over the [latex]y\\text{-axis}[\/latex] interval [latex]\\left[0,4\\right].[\/latex] Use the disk method to find the volume of the solid of revolution generated by rotating [latex]R[\/latex] around the [latex]y\\text{-axis}\\text{.}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167793368748\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167793368748\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793368748\">Figure 11 shows the function and a representative disk that can be used to estimate the volume. Notice that since we are revolving the function around the [latex]y\\text{-axis,}[\/latex] the disks are horizontal, rather than vertical.<\/p>\n<figure style=\"width: 740px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11212816\/CNX_Calc_Figure_06_02_011.jpg\" alt=\"This figure has two graphs. The first graph labeled \u201ca\u201d is the curve g(y) = squareroot(4-y). It is a decreasing curve starting on the y-axis at y=4. Between the curve and the y-axis is a horizontal rectangle. The rectangle starts at the y-axis and stops at the curve. The second graph labeled \u201cb\u201d is the same curve as the first graph. The rectangle from the first graph has been rotated around the y-axis to form a horizontal disk.\" width=\"740\" height=\"386\" \/><figcaption class=\"wp-caption-text\">Figure 11. (a) Shown is a thin rectangle between the curve of the function [latex]g(y)=\\sqrt{4-y}[\/latex] and the [latex]y\\text{-axis}\\text{.}[\/latex] (b) The rectangle forms a representative disk after revolution around the [latex]y\\text{-axis}\\text{.}[\/latex]<\/figcaption><\/figure>\n<p id=\"fs-id1167794142277\">The region to be revolved and the full solid of revolution are depicted in the following figure.<\/p>\n<figure style=\"width: 740px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11212820\/CNX_Calc_Figure_06_02_012.jpg\" alt=\"This figure has two graphs. The first graph labeled \u201ca\u201d is the curve g(y) = squareroot(4-y). It is a decreasing curve starting on the y-axis at y=4. The region formed by the x-axis, the y-axis, and the curve is shaded. This region is in the first quadrant. The second graph labeled \u201cb\u201d is the same curve as the first graph. The region from the first graph has been rotated around the y-axis to form a solid.\" width=\"740\" height=\"386\" \/><figcaption class=\"wp-caption-text\">Figure 12. (a) The region to the left of the function [latex]g(y)=\\sqrt{4-y}[\/latex] over the [latex]y\\text{-axis}[\/latex] interval [latex]\\left[0,4\\right].[\/latex] (b) The solid of revolution formed by revolving the region about the [latex]y\\text{-axis}\\text{.}[\/latex]<\/figcaption><\/figure>\n<p id=\"fs-id1167794331230\">To find the volume, we integrate with respect to [latex]y.[\/latex] We obtain<\/p>\n<div id=\"fs-id1167793498482\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill V& ={\\displaystyle\\int }_{c}^{d}\\pi {\\left[g(y)\\right]}^{2}dy\\hfill \\\\ & ={\\displaystyle\\int }_{0}^{4}\\pi {\\left[\\sqrt{4-y}\\right]}^{2}dy=\\pi {\\displaystyle\\int }_{0}^{4}(4-y)dy\\hfill \\\\ & ={\\pi \\left[4y-\\frac{{y}^{2}}{2}\\right]|}_{0}^{4}=8\\pi .\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793259617\">The volume is [latex]8\\pi[\/latex] units<sup>3<\/sup>.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm325821\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=325821&theme=lumen&iframe_resize_id=ohm325821&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":15,"menu_order":5,"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-2654","chapter","type-chapter","status-publish","hentry"],"part":432,"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\/2654","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":4,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/2654\/revisions"}],"predecessor-version":[{"id":2875,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/2654\/revisions\/2875"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/432"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/2654\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=2654"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=2654"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=2654"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=2654"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}