{"id":438,"date":"2025-02-13T19:44:55","date_gmt":"2025-02-13T19:44:55","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/volumes-of-revolution-cylindrical-shells-learn-it-2\/"},"modified":"2025-02-13T19:44:55","modified_gmt":"2025-02-13T19:44:55","slug":"volumes-of-revolution-cylindrical-shells-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/volumes-of-revolution-cylindrical-shells-learn-it-2\/","title":{"raw":"Volumes of Revolution: Cylindrical Shells: Learn It 2","rendered":"Volumes of Revolution: Cylindrical Shells: Learn It 2"},"content":{"raw":"\n<h2>Cylindrical Shells Method Cont.<\/h2>\n<p id=\"fs-id1167793419292\">As with the disk method and the washer method, we can use the method of cylindrical shells with solids of revolution, revolved around the [latex]x\\text{-axis},[\/latex] when we want to integrate with respect to [latex]y.[\/latex]<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>the method of cylindrical shells for solids of revolution around the [latex]x[\/latex]-axis<\/h3>\n<p id=\"fs-id1167793514545\">Let [latex]g(y)[\/latex] be continuous and nonnegative.<\/p>\n<p>&nbsp;<\/p>\n<p>Define [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]x\\text{-axis}[\/latex] is given by:<\/p>\n<div id=\"fs-id1167793559101\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V={\\displaystyle\\int }_{c}^{d}(2\\pi yg(y))dy[\/latex]<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>Define [latex]Q[\/latex] as the region bounded on the right by the graph of [latex]g(y)=2\\sqrt{y}[\/latex] and on the left by the [latex]y\\text{-axis}[\/latex] for [latex]y\\in \\left[0,4\\right].[\/latex] Find the volume of the solid of revolution formed by revolving [latex]Q[\/latex] around the [latex]x[\/latex]-axis.<\/p>\n\n[reveal-answer q=\"fs-id1167793480237\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1167793480237\"]\n\n<p id=\"fs-id1167793480237\">First, we need to graph the region [latex]Q[\/latex] and the associated solid of revolution, as shown in the following figure.<\/p>\n\n[caption id=\"\" align=\"aligncenter\" width=\"634\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213017\/CNX_Calc_Figure_06_03_006.jpg\" alt=\"This figure has two graphs. The first graph is labeled \u201ca\u201d and is the curve g(y)=2squareroot(y). It is an increasing curve in the first quadrant beginning at the origin. Between the y-axis and the curve, there is a shaded region labeled \u201cQ\u201d. The shaded region is bounded above by the line y=4. The second graph is the same curve in \u201ca\u201d and labeled \u201cb\u201d. It also has a solid region that has been formed by rotating the curve in \u201ca\u201d about the x-axis. The solid starts at the y-axis and stops at x=4.\" width=\"634\" height=\"461\"> Figure 7. (a) The region [latex]Q[\/latex] to the left of the function [latex]g(y)[\/latex] over the interval [latex]\\left[0,4\\right].[\/latex] (b) The solid of revolution generated by revolving [latex]Q[\/latex] around the [latex]x\\text{-axis}.[\/latex][\/caption]\n\n<p id=\"fs-id1167793931828\">Label the shaded region [latex]Q.[\/latex] Then the volume of the solid is given by:<\/p>\n<div id=\"fs-id1167793931838\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill V&amp; ={\\displaystyle\\int }_{c}^{d}(2\\pi yg(y))dy\\hfill \\\\ &amp; ={\\displaystyle\\int }_{0}^{4}(2\\pi y(2\\sqrt{y}))dy=4\\pi {\\displaystyle\\int }_{0}^{4}{y}^{3\\text{\/}2}dy\\hfill \\\\ &amp; ={4\\pi \\left[\\frac{2{y}^{5\\text{\/}2}}{5}\\right]|}_{0}^{4}=\\frac{256\\pi }{5}{\\text{units}}^{3}\\text{.}\\hfill \\end{array}[\/latex][\/hidden-answer]<\/div>\n<\/section>\n<p>For the next example, we look at a solid of revolution for which the graph of a function is revolved around a line other than one of the two coordinate axes. To set this up, we need to revisit the development of the method of cylindrical shells.<\/p>\n<p>Recall that we found the volume of one of the shells to be given by:<\/p>\n<div id=\"fs-id1167794146622\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill {V}_{\\text{shell}}&amp; =f({x}_{i}^{*})(\\pi {x}_{i}^{2}-\\pi {x}_{i-1}^{2})\\hfill \\\\ &amp; =\\pi f({x}_{i}^{*})({x}_{i}^{2}-{x}_{i-1}^{2})\\hfill \\\\ &amp; =\\pi f({x}_{i}^{*})({x}_{i}+{x}_{i-1})({x}_{i}-{x}_{i-1})\\hfill \\\\ &amp; =2\\pi f({x}_{i}^{*})(\\frac{{x}_{i}+{x}_{i-1}}{2})({x}_{i}-{x}_{i-1}).\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167793931535\">This was based on a shell with an outer radius of [latex]{x}_{i}[\/latex] and an inner radius of [latex]{x}_{i-1}.[\/latex] If, however, we rotate the region around a line other than the [latex]y\\text{-axis},[\/latex] we have a different outer and inner radius. <br>\n<br>\nSuppose, for example, that we rotate the region around the line [latex]x=\\text{\u2212}k,[\/latex] where [latex]k[\/latex] is some positive constant. Then, the outer radius of the shell is [latex]{x}_{i}+k[\/latex] and the inner radius of the shell is [latex]{x}_{i-1}+k.[\/latex] <br>\n<br>\nSubstituting these terms into the expression for volume, we see that when a plane region is rotated around the line [latex]x=\\text{\u2212}k,[\/latex] the volume of a shell is given by:<\/p>\n<div id=\"fs-id1167793421191\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill {V}_{\\text{shell}}&amp; =2\\pi f({x}_{i}^{*})(\\frac{({x}_{i}+k)+({x}_{i-1}+k)}{2})(({x}_{i}+k)-({x}_{i-1}+k))\\hfill \\\\ &amp; =2\\pi f({x}_{i}^{*})((\\frac{{x}_{i}+{x}_{i-2}}{2})+k)\\text{\u0394}x.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167793478786\">As before, we notice that [latex]\\frac{{x}_{i}+{x}_{i-1}}{2}[\/latex] is the midpoint of the interval [latex]\\left[{x}_{i-1},{x}_{i}\\right][\/latex] and can be approximated by [latex]{x}_{i}^{*}.[\/latex] Then, the approximate volume of the shell is:<\/p>\n<div id=\"fs-id1167793355111\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{V}_{\\text{shell}}\\approx 2\\pi ({x}_{i}^{*}+k)f({x}_{i}^{*})\\text{\u0394}x[\/latex]<\/div>\n<p id=\"fs-id1167793729500\">The remainder of the development proceeds as before, and we see that:<\/p>\n<div id=\"fs-id1167793729503\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V={\\displaystyle\\int }_{a}^{b}(2\\pi (x+k)f(x))dx[\/latex]<\/div>\n<p id=\"fs-id1167794326027\">We could also rotate the region around other horizontal or vertical lines, such as a vertical line in the right half plane. In each case, the volume formula must be adjusted accordingly. Specifically, the [latex]x\\text{-term}[\/latex] in the integral must be replaced with an expression representing the radius of a shell. To see how this works, consider the following example.<\/p>\n<section class=\"textbox example\">\n<p>Define [latex]R[\/latex] as the region bounded above by the graph of [latex]f(x)=x[\/latex] and below by the [latex]x\\text{-axis}[\/latex] over the interval [latex]\\left[1,2\\right].[\/latex] Find the volume of the solid of revolution formed by revolving [latex]R[\/latex] around the line [latex]x=-1.[\/latex]<\/p>\n\n[reveal-answer q=\"fs-id1167793940512\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1167793940512\"]\n\n<p id=\"fs-id1167793940512\">First, graph the region [latex]R[\/latex] and the associated solid of revolution, as shown in the following figure.<\/p>\n\n[caption id=\"\" align=\"aligncenter\" width=\"859\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213021\/CNX_Calc_Figure_06_03_007.jpg\" alt=\"This figure has two graphs. The first graph is labeled \u201ca\u201d and is the line f(x)=x, a diagonal line through the origin. There is a shaded region above the x-axis under the line labeled \u201cR\u201d. This region is bounded to the left by the line x=1 and to the right by the line x=2. There is also the vertical line x=-1 on the graph. The second figure has the same graphs as \u201ca\u201d and is labeled \u201cb\u201d. Also on the graph is a solid formed by rotating the region \u201cR\u201d from the first graph about the line x=-1.\" width=\"859\" height=\"461\"> Figure 8. (a) The region [latex]R[\/latex] between the graph of [latex]f(x)[\/latex] and the [latex]x\\text{-axis}[\/latex] over the interval [latex]\\left[1,2\\right].[\/latex] (b) The solid of revolution generated by revolving [latex]R[\/latex] around the line [latex]x=-1.[\/latex][\/caption]\n\n<p id=\"fs-id1167794095303\">Note that the radius of a shell is given by [latex]x+1.[\/latex] Then the volume of the solid is given by:<\/p>\n<div id=\"fs-id1167794095318\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill V&amp; ={\\displaystyle\\int }_{1}^{2}(2\\pi (x+1)f(x))dx\\hfill \\\\ &amp; ={\\displaystyle\\int }_{1}^{2}(2\\pi (x+1)x)dx=2\\pi {\\displaystyle\\int }_{1}^{2}({x}^{2}+x)dx\\hfill \\\\ &amp; ={2\\pi \\left[\\frac{{x}^{3}}{3}+\\frac{{x}^{2}}{2}\\right]|}_{1}^{2}=\\frac{23\\pi }{3}{\\text{units}}^{3}\\text{.}\\hfill \\end{array}[\/latex][\/hidden-answer]<\/div>\n<\/section>\n<p id=\"fs-id1167791543268\">For our final example, let\u2019s look at the volume of a solid of revolution for which the region of revolution is bounded by the graphs of two functions.<\/p>\n<section class=\"textbox example\">\n<p>Define [latex]R[\/latex] as the region bounded above by the graph of the function [latex]f(x)=\\sqrt{x}[\/latex] and below by the graph of the function [latex]g(x)=\\frac{1}{x}[\/latex] over the interval [latex]\\left[1,4\\right].[\/latex] Find the volume of the solid of revolution generated by revolving [latex]R[\/latex] around the [latex]y\\text{-axis}.[\/latex]<\/p>\n\n[reveal-answer q=\"fs-id1167793604193\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1167793604193\"]\n\n<p id=\"fs-id1167793604193\">First, graph the region [latex]R[\/latex] and the associated solid of revolution, as shown in the following figure.<\/p>\n\n[caption id=\"\" align=\"aligncenter\" width=\"859\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213024\/CNX_Calc_Figure_06_03_008.jpg\" alt=\"This figure has two graphs. The first graph is labeled \u201ca\u201d and has two curves. The curves are the graphs of f(x)=squareroot(x) and g(x)=1\/x. In the first quadrant the curves intersect at (1,1). In between the curves in the first quadrant there is a shaded region labeled \u201cR\u201d, bounded to the right by the line x=4. The second graph is labeled \u201cb\u201d and is the same as the graphs in \u201ca\u201d. Also on this graph is a solid that has been formed by rotating the region \u201cR\u201d from the figure \u201ca\u201d about the y-axis.\" width=\"859\" height=\"461\"> Figure 9. (a) The region [latex]R[\/latex] between the graph of [latex]f(x)[\/latex] and the graph of [latex]g(x)[\/latex] over the interval [latex]\\left[1,4\\right].[\/latex] (b) The solid of revolution generated by revolving [latex]R[\/latex] around the [latex]y\\text{-axis}.[\/latex][\/caption]\n\n<p id=\"fs-id1167793521467\">Note that the axis of revolution is the [latex]y\\text{-axis},[\/latex] so the radius of a shell is given simply by [latex]x.[\/latex] We don\u2019t need to make any adjustments to the [latex]x[\/latex]-term of our integrand. The height of a shell, though, is given by [latex]f(x)-g(x),[\/latex] so in this case we need to adjust the [latex]f(x)[\/latex] term of the integrand.<\/p>\n<p>Then the volume of the solid is given by:<\/p>\n<div id=\"fs-id1167793590358\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill V&amp; ={\\displaystyle\\int }_{1}^{4}(2\\pi x(f(x)-g(x)))dx\\hfill \\\\ &amp; ={\\displaystyle\\int }_{1}^{4}(2\\pi x(\\sqrt{x}-\\frac{1}{x}))dx=2\\pi {\\displaystyle\\int }_{1}^{4}({x}^{3\\text{\/}2}-1)dx\\hfill \\\\ &amp; ={2\\pi \\left[\\frac{2{x}^{5\\text{\/}2}}{5}-x\\right]|}_{1}^{4}=\\frac{94\\pi }{5}{\\text{units}}^{3}.\\hfill \\end{array}[\/latex][\/hidden-answer]<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<p>[ohm_question hide_question_numbers=1]288441[\/ohm_question]<\/p>\n<\/section>\n","rendered":"<h2>Cylindrical Shells Method Cont.<\/h2>\n<p id=\"fs-id1167793419292\">As with the disk method and the washer method, we can use the method of cylindrical shells with solids of revolution, revolved around the [latex]x\\text{-axis},[\/latex] when we want to integrate with respect to [latex]y.[\/latex]<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>the method of cylindrical shells for solids of revolution around the [latex]x[\/latex]-axis<\/h3>\n<p id=\"fs-id1167793514545\">Let [latex]g(y)[\/latex] be continuous and nonnegative.<\/p>\n<p>&nbsp;<\/p>\n<p>Define [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]x\\text{-axis}[\/latex] is given by:<\/p>\n<div id=\"fs-id1167793559101\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V={\\displaystyle\\int }_{c}^{d}(2\\pi yg(y))dy[\/latex]<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>Define [latex]Q[\/latex] as the region bounded on the right by the graph of [latex]g(y)=2\\sqrt{y}[\/latex] and on the left by the [latex]y\\text{-axis}[\/latex] for [latex]y\\in \\left[0,4\\right].[\/latex] Find the volume of the solid of revolution formed by revolving [latex]Q[\/latex] around the [latex]x[\/latex]-axis.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167793480237\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167793480237\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793480237\">First, we need to graph the region [latex]Q[\/latex] and the associated solid of revolution, as shown in the following figure.<\/p>\n<figure style=\"width: 634px\" 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\/11213017\/CNX_Calc_Figure_06_03_006.jpg\" alt=\"This figure has two graphs. The first graph is labeled \u201ca\u201d and is the curve g(y)=2squareroot(y). It is an increasing curve in the first quadrant beginning at the origin. Between the y-axis and the curve, there is a shaded region labeled \u201cQ\u201d. The shaded region is bounded above by the line y=4. The second graph is the same curve in \u201ca\u201d and labeled \u201cb\u201d. It also has a solid region that has been formed by rotating the curve in \u201ca\u201d about the x-axis. The solid starts at the y-axis and stops at x=4.\" width=\"634\" height=\"461\" \/><figcaption class=\"wp-caption-text\">Figure 7. (a) The region [latex]Q[\/latex] to the left of the function [latex]g(y)[\/latex] over the interval [latex]\\left[0,4\\right].[\/latex] (b) The solid of revolution generated by revolving [latex]Q[\/latex] around the [latex]x\\text{-axis}.[\/latex]<\/figcaption><\/figure>\n<p id=\"fs-id1167793931828\">Label the shaded region [latex]Q.[\/latex] Then the volume of the solid is given by:<\/p>\n<div id=\"fs-id1167793931838\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill V& ={\\displaystyle\\int }_{c}^{d}(2\\pi yg(y))dy\\hfill \\\\ & ={\\displaystyle\\int }_{0}^{4}(2\\pi y(2\\sqrt{y}))dy=4\\pi {\\displaystyle\\int }_{0}^{4}{y}^{3\\text{\/}2}dy\\hfill \\\\ & ={4\\pi \\left[\\frac{2{y}^{5\\text{\/}2}}{5}\\right]|}_{0}^{4}=\\frac{256\\pi }{5}{\\text{units}}^{3}\\text{.}\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<p>For the next example, we look at a solid of revolution for which the graph of a function is revolved around a line other than one of the two coordinate axes. To set this up, we need to revisit the development of the method of cylindrical shells.<\/p>\n<p>Recall that we found the volume of one of the shells to be given by:<\/p>\n<div id=\"fs-id1167794146622\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill {V}_{\\text{shell}}& =f({x}_{i}^{*})(\\pi {x}_{i}^{2}-\\pi {x}_{i-1}^{2})\\hfill \\\\ & =\\pi f({x}_{i}^{*})({x}_{i}^{2}-{x}_{i-1}^{2})\\hfill \\\\ & =\\pi f({x}_{i}^{*})({x}_{i}+{x}_{i-1})({x}_{i}-{x}_{i-1})\\hfill \\\\ & =2\\pi f({x}_{i}^{*})(\\frac{{x}_{i}+{x}_{i-1}}{2})({x}_{i}-{x}_{i-1}).\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167793931535\">This was based on a shell with an outer radius of [latex]{x}_{i}[\/latex] and an inner radius of [latex]{x}_{i-1}.[\/latex] If, however, we rotate the region around a line other than the [latex]y\\text{-axis},[\/latex] we have a different outer and inner radius. <\/p>\n<p>Suppose, for example, that we rotate the region around the line [latex]x=\\text{\u2212}k,[\/latex] where [latex]k[\/latex] is some positive constant. Then, the outer radius of the shell is [latex]{x}_{i}+k[\/latex] and the inner radius of the shell is [latex]{x}_{i-1}+k.[\/latex] <\/p>\n<p>Substituting these terms into the expression for volume, we see that when a plane region is rotated around the line [latex]x=\\text{\u2212}k,[\/latex] the volume of a shell is given by:<\/p>\n<div id=\"fs-id1167793421191\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill {V}_{\\text{shell}}& =2\\pi f({x}_{i}^{*})(\\frac{({x}_{i}+k)+({x}_{i-1}+k)}{2})(({x}_{i}+k)-({x}_{i-1}+k))\\hfill \\\\ & =2\\pi f({x}_{i}^{*})((\\frac{{x}_{i}+{x}_{i-2}}{2})+k)\\text{\u0394}x.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167793478786\">As before, we notice that [latex]\\frac{{x}_{i}+{x}_{i-1}}{2}[\/latex] is the midpoint of the interval [latex]\\left[{x}_{i-1},{x}_{i}\\right][\/latex] and can be approximated by [latex]{x}_{i}^{*}.[\/latex] Then, the approximate volume of the shell is:<\/p>\n<div id=\"fs-id1167793355111\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{V}_{\\text{shell}}\\approx 2\\pi ({x}_{i}^{*}+k)f({x}_{i}^{*})\\text{\u0394}x[\/latex]<\/div>\n<p id=\"fs-id1167793729500\">The remainder of the development proceeds as before, and we see that:<\/p>\n<div id=\"fs-id1167793729503\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V={\\displaystyle\\int }_{a}^{b}(2\\pi (x+k)f(x))dx[\/latex]<\/div>\n<p id=\"fs-id1167794326027\">We could also rotate the region around other horizontal or vertical lines, such as a vertical line in the right half plane. In each case, the volume formula must be adjusted accordingly. Specifically, the [latex]x\\text{-term}[\/latex] in the integral must be replaced with an expression representing the radius of a shell. To see how this works, consider the following example.<\/p>\n<section class=\"textbox example\">\n<p>Define [latex]R[\/latex] as the region bounded above by the graph of [latex]f(x)=x[\/latex] and below by the [latex]x\\text{-axis}[\/latex] over the interval [latex]\\left[1,2\\right].[\/latex] Find the volume of the solid of revolution formed by revolving [latex]R[\/latex] around the line [latex]x=-1.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167793940512\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167793940512\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793940512\">First, graph the region [latex]R[\/latex] and the associated solid of revolution, as shown in the following figure.<\/p>\n<figure style=\"width: 859px\" 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\/11213021\/CNX_Calc_Figure_06_03_007.jpg\" alt=\"This figure has two graphs. The first graph is labeled \u201ca\u201d and is the line f(x)=x, a diagonal line through the origin. There is a shaded region above the x-axis under the line labeled \u201cR\u201d. This region is bounded to the left by the line x=1 and to the right by the line x=2. There is also the vertical line x=-1 on the graph. The second figure has the same graphs as \u201ca\u201d and is labeled \u201cb\u201d. Also on the graph is a solid formed by rotating the region \u201cR\u201d from the first graph about the line x=-1.\" width=\"859\" height=\"461\" \/><figcaption class=\"wp-caption-text\">Figure 8. (a) The region [latex]R[\/latex] between the graph of [latex]f(x)[\/latex] and the [latex]x\\text{-axis}[\/latex] over the interval [latex]\\left[1,2\\right].[\/latex] (b) The solid of revolution generated by revolving [latex]R[\/latex] around the line [latex]x=-1.[\/latex]<\/figcaption><\/figure>\n<p id=\"fs-id1167794095303\">Note that the radius of a shell is given by [latex]x+1.[\/latex] Then the volume of the solid is given by:<\/p>\n<div id=\"fs-id1167794095318\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill V& ={\\displaystyle\\int }_{1}^{2}(2\\pi (x+1)f(x))dx\\hfill \\\\ & ={\\displaystyle\\int }_{1}^{2}(2\\pi (x+1)x)dx=2\\pi {\\displaystyle\\int }_{1}^{2}({x}^{2}+x)dx\\hfill \\\\ & ={2\\pi \\left[\\frac{{x}^{3}}{3}+\\frac{{x}^{2}}{2}\\right]|}_{1}^{2}=\\frac{23\\pi }{3}{\\text{units}}^{3}\\text{.}\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<p id=\"fs-id1167791543268\">For our final example, let\u2019s look at the volume of a solid of revolution for which the region of revolution is bounded by the graphs of two functions.<\/p>\n<section class=\"textbox example\">\n<p>Define [latex]R[\/latex] as the region bounded above by the graph of the function [latex]f(x)=\\sqrt{x}[\/latex] and below by the graph of the function [latex]g(x)=\\frac{1}{x}[\/latex] over the interval [latex]\\left[1,4\\right].[\/latex] Find the volume of the solid of revolution generated by revolving [latex]R[\/latex] around the [latex]y\\text{-axis}.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167793604193\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167793604193\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793604193\">First, graph the region [latex]R[\/latex] and the associated solid of revolution, as shown in the following figure.<\/p>\n<figure style=\"width: 859px\" 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\/11213024\/CNX_Calc_Figure_06_03_008.jpg\" alt=\"This figure has two graphs. The first graph is labeled \u201ca\u201d and has two curves. The curves are the graphs of f(x)=squareroot(x) and g(x)=1\/x. In the first quadrant the curves intersect at (1,1). In between the curves in the first quadrant there is a shaded region labeled \u201cR\u201d, bounded to the right by the line x=4. The second graph is labeled \u201cb\u201d and is the same as the graphs in \u201ca\u201d. Also on this graph is a solid that has been formed by rotating the region \u201cR\u201d from the figure \u201ca\u201d about the y-axis.\" width=\"859\" height=\"461\" \/><figcaption class=\"wp-caption-text\">Figure 9. (a) The region [latex]R[\/latex] between the graph of [latex]f(x)[\/latex] and the graph of [latex]g(x)[\/latex] over the interval [latex]\\left[1,4\\right].[\/latex] (b) The solid of revolution generated by revolving [latex]R[\/latex] around the [latex]y\\text{-axis}.[\/latex]<\/figcaption><\/figure>\n<p id=\"fs-id1167793521467\">Note that the axis of revolution is the [latex]y\\text{-axis},[\/latex] so the radius of a shell is given simply by [latex]x.[\/latex] We don\u2019t need to make any adjustments to the [latex]x[\/latex]-term of our integrand. The height of a shell, though, is given by [latex]f(x)-g(x),[\/latex] so in this case we need to adjust the [latex]f(x)[\/latex] term of the integrand.<\/p>\n<p>Then the volume of the solid is given by:<\/p>\n<div id=\"fs-id1167793590358\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill V& ={\\displaystyle\\int }_{1}^{4}(2\\pi x(f(x)-g(x)))dx\\hfill \\\\ & ={\\displaystyle\\int }_{1}^{4}(2\\pi x(\\sqrt{x}-\\frac{1}{x}))dx=2\\pi {\\displaystyle\\int }_{1}^{4}({x}^{3\\text{\/}2}-1)dx\\hfill \\\\ & ={2\\pi \\left[\\frac{2{x}^{5\\text{\/}2}}{5}-x\\right]|}_{1}^{4}=\\frac{94\\pi }{5}{\\text{units}}^{3}.\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm288441\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=288441&theme=lumen&iframe_resize_id=ohm288441&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n","protected":false},"author":6,"menu_order":17,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":421,"module-header":"","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/438"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":0,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/438\/revisions"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/421"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/438\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=438"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=438"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=438"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=438"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}