{"id":434,"date":"2025-02-13T19:44:53","date_gmt":"2025-02-13T19:44:53","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/determining-volumes-by-slicing-learn-it-4\/"},"modified":"2025-02-13T19:44:53","modified_gmt":"2025-02-13T19:44:53","slug":"determining-volumes-by-slicing-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/determining-volumes-by-slicing-learn-it-4\/","title":{"raw":"Determining Volumes by Slicing: Learn It 4","rendered":"Determining Volumes by Slicing: Learn It 4"},"content":{"raw":"\n<h2>The Washer Method<\/h2>\n<p id=\"fs-id1167793478980\">Some solids of revolution have cavities in the middle; they are not solid all the way to the axis of revolution. This can happen because of the shape of the region of revolution with respect to the axis. Cavities also arise when the region of revolution is defined between the graphs of two functions or [latex]x[\/latex]-axis or [latex]y[\/latex]-axis is selected for revolution.<\/p>\n<p>When the solid of revolution has a cavity, the slices used to approximate the volume are not disks but washers (disks with holes in the center).<\/p>\n<p id=\"fs-id1167793831962\">For example, consider 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)=1[\/latex] over the interval [latex]\\left[1,4\\right].[\/latex] When this region is revolved around the [latex]x\\text{-axis,}[\/latex] the result is a solid with a cavity in the middle, and the slices are washers.&nbsp;<\/p>\n\n\n[caption id=\"\" align=\"aligncenter\" width=\"669\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11212825\/CNX_Calc_Figure_06_02_013.jpg\" alt=\"This figure has four graphs. The first graph is labeled \u201ca\u201d and has the two functions f(x)=squareroot(x) and g(x)=1 graphed in the first quadrant. f(x) is an increasing curve starting at the origin and g(x) is a horizontal line at y=1. The curves intersect at the ordered pair (1,1). In between the curves is a shaded rectangle with the bottom on g(x) and the top at f(x). The second graph labeled \u201cb\u201d is the same two curves as the first graph. The shaded rectangle between the curves from the first graph has been rotated around the x-axis to form an open disk or washer. The third graph labeled \u201ca\u201d has the same two curves as the first graph. There is a shaded region between the two curves between where they intersect and a line at x=4. The fourth graph is the same two curves as the first with the region from the third graph rotated around the x-axis forming a solid region with a hollow center. The hollow center is represented on the graph with broken horizontal lines at y=1 and y=-1.\" width=\"669\" height=\"860\"> Figure 13. (a) A thin rectangle in the region between two curves. (b) A representative disk formed by revolving the rectangle about the [latex]x\\text{-axis}.[\/latex] (c) The region between the curves over the given interval. (d) The resulting solid of revolution.[\/caption]\n\n\n<p id=\"fs-id1167794003626\">The cross-sectional area, then, is the area of the outer circle less the area of the inner circle. In this case,<\/p>\n<div id=\"fs-id1167791543230\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A(x)=\\pi {(\\sqrt{x})}^{2}-\\pi {(1)}^{2}=\\pi (x-1).[\/latex]<\/div>\n<p id=\"fs-id1167793630184\">Then the volume of the solid is:<\/p>\n<div id=\"fs-id1167794100454\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill V&amp; ={\\displaystyle\\int }_{a}^{b}A(x)dx\\hfill \\\\ &amp; ={\\displaystyle\\int }_{1}^{4}\\pi (x-1)dx={\\pi \\left[\\frac{{x}^{2}}{2}-x\\right]|}_{1}^{4}=\\frac{9}{2}\\pi {\\text{units}}^{3}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167793465550\">Generalizing this process gives the <strong>washer method<\/strong>.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>the washer method<\/h3>\n<p id=\"fs-id1167794334065\">Suppose [latex]f(x)[\/latex] and [latex]g(x)[\/latex] are continuous, nonnegative functions such that [latex]f(x)\\ge g(x)[\/latex] over [latex]\\left[a,b\\right].[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>Let [latex]R[\/latex] denote the region bounded above by the graph of [latex]f(x),[\/latex] below by the graph of [latex]g(x),[\/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-id1167793503097\" class=\"equation\" style=\"text-align: center;\">[latex]V={\\displaystyle\\int }_{a}^{b}\\pi \\left[{(f(x))}^{2}-{(g(x))}^{2}\\right]dx.[\/latex]<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>Find the volume of a solid of revolution formed by revolving the region bounded above by the graph of [latex]f(x)=x[\/latex] and below by the graph of [latex]g(x)=\\frac{1}{x}[\/latex] over the interval [latex]\\left[1,4\\right][\/latex] around the [latex]x\\text{-axis}\\text{.}[\/latex]<\/p>\n\n\n[reveal-answer q=\"fs-id1167794211097\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1167794211097\"]\n\n\n<p id=\"fs-id1167794211097\">The graphs of the functions and the solid of revolution are shown in the following figure.<\/p>\n\n\n[caption id=\"\" align=\"aligncenter\" width=\"690\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11212828\/CNX_Calc_Figure_06_02_014.jpg\" alt=\"This figure has two graphs. The first graph is labeled \u201ca\u201d and has the two curves f(x)=x and g(x)=1\/x. They are graphed only in the first quadrant. f(x) is a diagonal line starting at the origin and g(x) is a decreasing curve with the y-axis as a vertical asymptote and the x-axis as a horizontal asymptote. The graphs intersect at (1,1). There is a shaded region between the graphs, bounded to the right by a line at x=4. The second graph is the same two curves. There is a solid formed by rotating the shaded region from the first graph around the x-axis.\" width=\"690\" height=\"386\"> Figure 14. (a) The region between the graphs of the functions [latex]f(x)=x[\/latex] and [latex]g(x)=\\frac{1}{x}[\/latex] over the interval [latex]\\left[1,4\\right].[\/latex] (b) Revolving the region about the [latex]x\\text{-axis}[\/latex] generates a solid of revolution with a cavity in the middle.[\/caption]\n\n\n<p id=\"fs-id1167793420906\">We have:<\/p>\n<div id=\"fs-id1167794213421\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill V&amp; ={\\displaystyle\\int }_{a}^{b}\\pi \\left[{(f(x))}^{2}-{(g(x))}^{2}\\right]dx\\hfill \\\\ &amp; =\\pi {\\displaystyle\\int }_{1}^{4}\\left[{x}^{2}-{(\\frac{1}{x})}^{2}\\right]dx\\text{}={\\pi \\left[\\frac{{x}^{3}}{3}+\\frac{1}{x}\\right]|}_{1}^{4}=\\frac{81\\pi }{4}{\\text{units}}^{3}.\\hfill \\end{array}[\/latex][\/hidden-answer]<\/div>\n<\/section>\n<p>As with the disk method, we can also apply the washer method to solids of revolution that result from revolving a region around the [latex]y[\/latex]-axis. In this case, the following rule applies.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>the washer method for solids of revolution around the [latex]y[\/latex]-axis<\/h3>\n<p id=\"fs-id1167793628604\">Suppose [latex]u(y)[\/latex] and [latex]v(y)[\/latex] are continuous, nonnegative functions such that [latex]v(y)\\le u(y)[\/latex] for [latex]y\\in \\left[c,d\\right].[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>Let [latex]Q[\/latex] denote the region bounded on the right by the graph of [latex]u(y),[\/latex] on the left by the graph of [latex]v(y),[\/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-id1167793953572\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V={\\displaystyle\\int }_{c}^{d}\\pi \\left[{(u(y))}^{2}-{(v(y))}^{2}\\right]dy[\/latex]<\/div>\n<\/section>\n<p id=\"fs-id1167794067848\">Rather than looking at an example of the washer method with the [latex]y\\text{-axis}[\/latex] as the axis of revolution, we now consider an example in which the axis of revolution is a line other than one of the two coordinate axes. The same general method applies, but you may have to visualize just how to describe the cross-sectional area of the volume.<\/p>\n<section class=\"textbox proTip\">\n<p>An important thing to remember is that for both the disk and washer method, the rectangles (the radii of the cross-sectional circles) are always perpendicular to the axis of revolution.<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p>Find the volume of a solid of revolution formed by revolving the region bounded above by [latex]f(x)=4-x[\/latex] and below by the [latex]x\\text{-axis}[\/latex] over the interval [latex]\\left[0,4\\right][\/latex] around the line [latex]y=-2.[\/latex]<\/p>\n\n\n[reveal-answer q=\"fs-id1167794218804\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1167794218804\"]\n\n\n<p id=\"fs-id1167794218804\">The graph of the region and the solid of revolution are shown in the following figure.<\/p>\n\n\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\/11212832\/CNX_Calc_Figure_06_02_015.jpg\" alt=\"This figure has two graphs. The first graph is labeled \u201ca\u201d and has the two curves f(x)=4-x and -2. There is a shaded region making a triangle bounded by the decreasing line f(x), the y-axis and the x-axis. The second graph is the same two curves. There is a solid formed by rotating the shaded region from the first graph around the line y=-2. There is a hollow cylinder inside of the solid represented by the lines y=-2 and y=-4.\" width=\"740\" height=\"536\"> Figure 15. (a) The region between the graph of the function [latex]f(x)=4-x[\/latex] and the [latex]x\\text{-axis}[\/latex] over the interval [latex]\\left[0,4\\right].[\/latex] (b) Revolving the region about the line [latex]y=-2[\/latex] generates a solid of revolution with a cylindrical hole through its middle.[\/caption]\n\n\n<p id=\"fs-id1167793501976\">We can\u2019t apply the volume formula to this problem directly because the axis of revolution is not one of the coordinate axes. However, we still know that the area of the cross-section is the area of the outer circle less the area of the inner circle.<\/p>\n<p>Looking at the graph of the function, we see the radius of the outer circle is given by [latex]f(x)+2,[\/latex] which simplifies to:<\/p>\n<div id=\"fs-id1167793619980\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)+2=(4-x)+2=6-x.[\/latex]<\/div>\n<p id=\"fs-id1167794200068\">The radius of the inner circle is [latex]g(x)=2.[\/latex] Therefore, we have:<\/p>\n<div id=\"fs-id1167794296437\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill V&amp; ={\\displaystyle\\int }_{0}^{4}\\pi \\left[{(6-x)}^{2}-{(2)}^{2}\\right]dx\\hfill \\\\ &amp; =\\pi {\\displaystyle\\int }_{0}^{4}({x}^{2}-12x+32)dx\\text{}={\\pi \\left[\\frac{{x}^{3}}{3}-6{x}^{2}+32x\\right]|}_{0}^{4}=\\frac{160\\pi }{3}{\\text{units}}^{3}.\\hfill \\end{array}[\/latex][\/hidden-answer]<\/div>\n<\/section>\n","rendered":"<h2>The Washer Method<\/h2>\n<p id=\"fs-id1167793478980\">Some solids of revolution have cavities in the middle; they are not solid all the way to the axis of revolution. This can happen because of the shape of the region of revolution with respect to the axis. Cavities also arise when the region of revolution is defined between the graphs of two functions or [latex]x[\/latex]-axis or [latex]y[\/latex]-axis is selected for revolution.<\/p>\n<p>When the solid of revolution has a cavity, the slices used to approximate the volume are not disks but washers (disks with holes in the center).<\/p>\n<p id=\"fs-id1167793831962\">For example, consider 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)=1[\/latex] over the interval [latex]\\left[1,4\\right].[\/latex] When this region is revolved around the [latex]x\\text{-axis,}[\/latex] the result is a solid with a cavity in the middle, and the slices are washers.&nbsp;<\/p>\n<figure style=\"width: 669px\" 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\/11212825\/CNX_Calc_Figure_06_02_013.jpg\" alt=\"This figure has four graphs. The first graph is labeled \u201ca\u201d and has the two functions f(x)=squareroot(x) and g(x)=1 graphed in the first quadrant. f(x) is an increasing curve starting at the origin and g(x) is a horizontal line at y=1. The curves intersect at the ordered pair (1,1). In between the curves is a shaded rectangle with the bottom on g(x) and the top at f(x). The second graph labeled \u201cb\u201d is the same two curves as the first graph. The shaded rectangle between the curves from the first graph has been rotated around the x-axis to form an open disk or washer. The third graph labeled \u201ca\u201d has the same two curves as the first graph. There is a shaded region between the two curves between where they intersect and a line at x=4. The fourth graph is the same two curves as the first with the region from the third graph rotated around the x-axis forming a solid region with a hollow center. The hollow center is represented on the graph with broken horizontal lines at y=1 and y=-1.\" width=\"669\" height=\"860\" \/><figcaption class=\"wp-caption-text\">Figure 13. (a) A thin rectangle in the region between two curves. (b) A representative disk formed by revolving the rectangle about the [latex]x\\text{-axis}.[\/latex] (c) The region between the curves over the given interval. (d) The resulting solid of revolution.<\/figcaption><\/figure>\n<p id=\"fs-id1167794003626\">The cross-sectional area, then, is the area of the outer circle less the area of the inner circle. In this case,<\/p>\n<div id=\"fs-id1167791543230\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A(x)=\\pi {(\\sqrt{x})}^{2}-\\pi {(1)}^{2}=\\pi (x-1).[\/latex]<\/div>\n<p id=\"fs-id1167793630184\">Then the volume of the solid is:<\/p>\n<div id=\"fs-id1167794100454\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill V& ={\\displaystyle\\int }_{a}^{b}A(x)dx\\hfill \\\\ & ={\\displaystyle\\int }_{1}^{4}\\pi (x-1)dx={\\pi \\left[\\frac{{x}^{2}}{2}-x\\right]|}_{1}^{4}=\\frac{9}{2}\\pi {\\text{units}}^{3}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167793465550\">Generalizing this process gives the <strong>washer method<\/strong>.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>the washer method<\/h3>\n<p id=\"fs-id1167794334065\">Suppose [latex]f(x)[\/latex] and [latex]g(x)[\/latex] are continuous, nonnegative functions such that [latex]f(x)\\ge g(x)[\/latex] over [latex]\\left[a,b\\right].[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>Let [latex]R[\/latex] denote the region bounded above by the graph of [latex]f(x),[\/latex] below by the graph of [latex]g(x),[\/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-id1167793503097\" class=\"equation\" style=\"text-align: center;\">[latex]V={\\displaystyle\\int }_{a}^{b}\\pi \\left[{(f(x))}^{2}-{(g(x))}^{2}\\right]dx.[\/latex]<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>Find the volume of a solid of revolution formed by revolving the region bounded above by the graph of [latex]f(x)=x[\/latex] and below by the graph of [latex]g(x)=\\frac{1}{x}[\/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-id1167794211097\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167794211097\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794211097\">The graphs of the functions and the solid of revolution are shown in the following figure.<\/p>\n<figure style=\"width: 690px\" 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\/11212828\/CNX_Calc_Figure_06_02_014.jpg\" alt=\"This figure has two graphs. The first graph is labeled \u201ca\u201d and has the two curves f(x)=x and g(x)=1\/x. They are graphed only in the first quadrant. f(x) is a diagonal line starting at the origin and g(x) is a decreasing curve with the y-axis as a vertical asymptote and the x-axis as a horizontal asymptote. The graphs intersect at (1,1). There is a shaded region between the graphs, bounded to the right by a line at x=4. The second graph is the same two curves. There is a solid formed by rotating the shaded region from the first graph around the x-axis.\" width=\"690\" height=\"386\" \/><figcaption class=\"wp-caption-text\">Figure 14. (a) The region between the graphs of the functions [latex]f(x)=x[\/latex] and [latex]g(x)=\\frac{1}{x}[\/latex] over the interval [latex]\\left[1,4\\right].[\/latex] (b) Revolving the region about the [latex]x\\text{-axis}[\/latex] generates a solid of revolution with a cavity in the middle.<\/figcaption><\/figure>\n<p id=\"fs-id1167793420906\">We have:<\/p>\n<div id=\"fs-id1167794213421\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill V& ={\\displaystyle\\int }_{a}^{b}\\pi \\left[{(f(x))}^{2}-{(g(x))}^{2}\\right]dx\\hfill \\\\ & =\\pi {\\displaystyle\\int }_{1}^{4}\\left[{x}^{2}-{(\\frac{1}{x})}^{2}\\right]dx\\text{}={\\pi \\left[\\frac{{x}^{3}}{3}+\\frac{1}{x}\\right]|}_{1}^{4}=\\frac{81\\pi }{4}{\\text{units}}^{3}.\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<p>As with the disk method, we can also apply the washer method to solids of revolution that result from revolving a region around the [latex]y[\/latex]-axis. In this case, the following rule applies.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>the washer method for solids of revolution around the [latex]y[\/latex]-axis<\/h3>\n<p id=\"fs-id1167793628604\">Suppose [latex]u(y)[\/latex] and [latex]v(y)[\/latex] are continuous, nonnegative functions such that [latex]v(y)\\le u(y)[\/latex] for [latex]y\\in \\left[c,d\\right].[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>Let [latex]Q[\/latex] denote the region bounded on the right by the graph of [latex]u(y),[\/latex] on the left by the graph of [latex]v(y),[\/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-id1167793953572\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V={\\displaystyle\\int }_{c}^{d}\\pi \\left[{(u(y))}^{2}-{(v(y))}^{2}\\right]dy[\/latex]<\/div>\n<\/section>\n<p id=\"fs-id1167794067848\">Rather than looking at an example of the washer method with the [latex]y\\text{-axis}[\/latex] as the axis of revolution, we now consider an example in which the axis of revolution is a line other than one of the two coordinate axes. The same general method applies, but you may have to visualize just how to describe the cross-sectional area of the volume.<\/p>\n<section class=\"textbox proTip\">\n<p>An important thing to remember is that for both the disk and washer method, the rectangles (the radii of the cross-sectional circles) are always perpendicular to the axis of revolution.<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p>Find the volume of a solid of revolution formed by revolving the region bounded above by [latex]f(x)=4-x[\/latex] and below by the [latex]x\\text{-axis}[\/latex] over the interval [latex]\\left[0,4\\right][\/latex] around the line [latex]y=-2.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167794218804\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167794218804\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794218804\">The graph of the region 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\/11212832\/CNX_Calc_Figure_06_02_015.jpg\" alt=\"This figure has two graphs. The first graph is labeled \u201ca\u201d and has the two curves f(x)=4-x and -2. There is a shaded region making a triangle bounded by the decreasing line f(x), the y-axis and the x-axis. The second graph is the same two curves. There is a solid formed by rotating the shaded region from the first graph around the line y=-2. There is a hollow cylinder inside of the solid represented by the lines y=-2 and y=-4.\" width=\"740\" height=\"536\" \/><figcaption class=\"wp-caption-text\">Figure 15. (a) The region between the graph of the function [latex]f(x)=4-x[\/latex] and the [latex]x\\text{-axis}[\/latex] over the interval [latex]\\left[0,4\\right].[\/latex] (b) Revolving the region about the line [latex]y=-2[\/latex] generates a solid of revolution with a cylindrical hole through its middle.<\/figcaption><\/figure>\n<p id=\"fs-id1167793501976\">We can\u2019t apply the volume formula to this problem directly because the axis of revolution is not one of the coordinate axes. However, we still know that the area of the cross-section is the area of the outer circle less the area of the inner circle.<\/p>\n<p>Looking at the graph of the function, we see the radius of the outer circle is given by [latex]f(x)+2,[\/latex] which simplifies to:<\/p>\n<div id=\"fs-id1167793619980\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)+2=(4-x)+2=6-x.[\/latex]<\/div>\n<p id=\"fs-id1167794200068\">The radius of the inner circle is [latex]g(x)=2.[\/latex] Therefore, we have:<\/p>\n<div id=\"fs-id1167794296437\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill V& ={\\displaystyle\\int }_{0}^{4}\\pi \\left[{(6-x)}^{2}-{(2)}^{2}\\right]dx\\hfill \\\\ & =\\pi {\\displaystyle\\int }_{0}^{4}({x}^{2}-12x+32)dx\\text{}={\\pi \\left[\\frac{{x}^{3}}{3}-6{x}^{2}+32x\\right]|}_{0}^{4}=\\frac{160\\pi }{3}{\\text{units}}^{3}.\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":6,"menu_order":13,"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\/434"}],"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\/434\/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\/434\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=434"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=434"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=434"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=434"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}