{"id":1168,"date":"2021-06-30T17:02:03","date_gmt":"2021-06-30T17:02:03","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/disk-and-washer-methods\/"},"modified":"2022-03-19T03:33:01","modified_gmt":"2022-03-19T03:33:01","slug":"disk-and-washer-methods","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/disk-and-washer-methods\/","title":{"raw":"Disk and Washer Methods","rendered":"Disk and Washer Methods"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Find the volume of a solid of revolution using the disk method<\/li>\r\n \t<li>Find the volume of a solid of revolution with a cavity using the washer method<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>The Disk Method<\/h2>\r\n<p id=\"fs-id1167793777573\">When we use the slicing method with solids of revolution, it is often called the <strong>disk method<\/strong> because, for solids of revolution, the slices used to over approximate the volume of the solid are disks. To see this, consider the solid of revolution generated by revolving the region between the graph of the function [latex]f(x)={(x-1)}^{2}+1[\/latex] and the [latex]x\\text{-axis}[\/latex] over the interval [latex]\\left[-1,3\\right][\/latex] around the [latex]x\\text{-axis}\\text{.}[\/latex] The graph of the function and a representative disk are shown in Figure 9(a) and (b). The region of revolution and the resulting solid are shown in Figure 9(c) and (d).<\/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 id=\"fs-id1167793551150\">We already used the formal Riemann sum development of the volume formula when we developed the slicing method. We know that<\/p>\r\n\r\n<div id=\"fs-id1167793443010\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V={\\displaystyle\\int }_{a}^{b}A(x)dx[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794333239\">The only difference with the disk method is that we know the formula for the cross-sectional area ahead of time; it is the area of a circle. This gives the following rule.<\/p>\r\n\r\n<div id=\"fs-id1167793370844\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">The Disk Method<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167793895393\">Let [latex]f(x)[\/latex] be continuous and nonnegative. 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] 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>\r\n\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<\/div>\r\n<p id=\"fs-id1167793497293\">The volume of the solid we have been studying (Figure 9) is given by<\/p>\r\n\r\n<div id=\"fs-id1167794040142\" 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}^{3}\\pi {\\left[{(x-1)}^{2}+1\\right]}^{2}dx=\\pi {\\displaystyle\\int }_{-1}^{3}{\\left[{(x-1)}^{4}+2{(x-1)}^{2}+1\\right]}^{2}dx\\hfill \\\\ &amp; =\\pi {\\left[\\frac{1}{5}{(x-1)}^{5}+\\frac{2}{3}{(x-1)}^{3}+x\\right]|}_{-1}^{3}=\\pi \\left[(\\frac{32}{5}+\\frac{16}{3}+3)-(-\\frac{32}{5}-\\frac{16}{3}-1)\\right]=\\frac{412\\pi }{15}{\\text{units}}^{3}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794329525\">Let\u2019s look at some examples.<\/p>\r\n\r\n<div id=\"fs-id1167793956465\" class=\"textbook exercises\">\r\n<h3>Example: Using the Disk Method to Find the Volume of a Solid of Revolution 1<\/h3>\r\nUse 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]\r\n<div id=\"fs-id1167793618941\" class=\"exercise\">[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<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167794037149\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nUse 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{4-x}[\/latex] and the [latex]x\\text{-axis}[\/latex] over the interval [latex]\\left[0,4\\right][\/latex] around the [latex]x\\text{-axis}\\text{.}[\/latex]\r\n<div id=\"fs-id1167793730425\" class=\"exercise\">[reveal-answer q=\"fs-id1167794039169\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794039169\"]\r\n<p id=\"fs-id1167794039169\">[latex]8\\pi [\/latex] units<sup>3<\/sup><\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try It.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/7dG21yfKXHg?controls=0&amp;start=755&amp;end=847&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266834\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.2DeterminingVolumesBySlicing755to847_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"2.2 Determining Volumes by Slicing\" here (opens in new window)<\/a>.[hidden-answer]\r\n<p id=\"fs-id1167793951744\">So far, our examples have all concerned regions revolved around the [latex]x\\text{-axis,}[\/latex] but we can generate a solid of revolution by revolving a plane region around any horizontal or vertical line. In the next example, we look at a solid of revolution that has been generated by revolving a region around the [latex]y\\text{-axis}\\text{.}[\/latex] The mechanics of the disk method are nearly the same as when the [latex]x\\text{-axis}[\/latex] is the axis of revolution, but we express the function in terms of [latex]y[\/latex] and we integrate with respect to [latex]y[\/latex] as well. This is summarized in the following rule.<\/p>\r\n\r\n<div id=\"fs-id1167793571437\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">The Disk Method for Solids of Revolution around the [latex]y[\/latex]-axis<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167794006397\">Let [latex]g(y)[\/latex] be continuous and nonnegative. 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] 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>\r\n\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&nbsp;\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793377534\">The next example shows how this rule works in practice.<\/p>\r\n\r\n<div id=\"fs-id1167793271607\" class=\"textbook exercises\">\r\n<h3>example: Using the Disk Method to Find the Volume of a Solid of Revolution 2<\/h3>\r\nLet [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]\r\n<div id=\"fs-id1167793271609\" class=\"exercise\">[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<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167793455063\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nUse the disk method to find the volume of the solid of revolution generated by rotating the region between the graph of [latex]g(y)=y[\/latex] and the [latex]y\\text{-axis}[\/latex] over the interval [latex]\\left[1,4\\right][\/latex] around the [latex]y\\text{-axis}\\text{.}[\/latex]\r\n<div>[reveal-answer q=\"476774\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"476774\"]Use the procedure from the last example.[\/hidden-answer]<\/div>\r\n<div><\/div>\r\n<div id=\"fs-id1167793455066\" class=\"exercise\">[reveal-answer q=\"fs-id1167793960078\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793960078\"]\r\n<p id=\"fs-id1167793960078\">[latex]21\\pi [\/latex] units<sup>3<\/sup><\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]20075[\/ohm_question]\r\n\r\n<\/div>\r\n<div id=\"fs-id1167794172336\" class=\"bc-section section\">\r\n<h2>The Washer Method<\/h2>\r\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. Sometimes, this is just a result of the way the region of revolution is shaped with respect to the axis of revolution. In other cases, cavities arise when the region of revolution is defined as the region between the graphs of two functions. A third way this can happen is when an axis of revolution other than the [latex]x\\text{-axis}[\/latex] or [latex]y\\text{-axis}[\/latex] is selected.<\/p>\r\n<p id=\"fs-id1167793831962\">When the solid of revolution has a cavity in the middle, the slices used to approximate the volume are not disks, but washers (disks with holes in the center). 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. The graph of the function and a representative washer are shown in Figure 13(a) and (b). The region of revolution and the resulting solid are shown in Figure 13(c) and (d).<\/p>\r\n\r\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]\r\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>\r\n\r\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>\r\n&nbsp;\r\n<p id=\"fs-id1167793630184\">Then the volume of the solid is<\/p>\r\n\r\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>\r\n&nbsp;\r\n<p id=\"fs-id1167793465550\">Generalizing this process gives the <strong>washer method<\/strong>.<\/p>\r\n\r\n<div id=\"fs-id1167793397841\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">The Washer Method<\/h3>\r\n\r\n<hr \/>\r\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] 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] 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>\r\n\r\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>\r\n&nbsp;\r\n\r\n<\/div>\r\n<div id=\"fs-id1167794209450\" class=\"textbook exercises\">\r\n<h3>Example: Using the Washer Method<\/h3>\r\nFind 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]\r\n<div id=\"fs-id1167794209452\" class=\"exercise\">[reveal-answer q=\"fs-id1167794211097\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794211097\"]\r\n<p id=\"fs-id1167794211097\">The graphs of the functions and the solid of revolution are shown in the following figure.<\/p>\r\n\r\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]\r\n<p id=\"fs-id1167793420906\">We have<\/p>\r\n\r\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>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167794324562\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind the volume of a solid of revolution formed by revolving the region bounded by the graphs of [latex]f(x)=\\sqrt{x}[\/latex] and [latex]g(x)=\\frac{1}{x}[\/latex] over the interval [latex]\\left[1,3\\right][\/latex] around the [latex]x\\text{-axis}\\text{.}[\/latex]\r\n<div>[reveal-answer q=\"888359\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"888359\"]Graph the functions to determine which graph forms the upper bound and which graph forms the lower bound, then use the procedure from the previous example.[\/hidden-answer]<\/div>\r\n<div><\/div>\r\n<div id=\"fs-id1167794324566\" class=\"exercise\">[reveal-answer q=\"fs-id1167794030044\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794030044\"]\r\n<p id=\"fs-id1167794030044\">[latex]\\frac{10\\pi }{3}[\/latex] units<sup>3<\/sup><\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try It.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/7dG21yfKXHg?controls=0&amp;start=1315&amp;end=1476&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266833\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266833\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.2DeterminingVolumesBySlicing1315to1476_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"2.2 Determining Volumes by Slicing\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<p id=\"fs-id1167793940198\">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>\r\n\r\n<div id=\"fs-id1167793565052\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">The Washer Method for Solids of Revolution around the [latex]y[\/latex]-axis<\/h3>\r\n\r\n<hr \/>\r\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] 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] 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>\r\n\r\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>\r\n<\/div>\r\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>\r\nAn 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.\r\n<div id=\"fs-id1167793912719\" class=\"textbook exercises\">\r\n<h3>Example: The Washer Method with a Different Axis of Revolution<\/h3>\r\nFind 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]\r\n<div id=\"fs-id1167793912721\" class=\"exercise\">[reveal-answer q=\"fs-id1167794218804\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794218804\"]\r\n<p id=\"fs-id1167794218804\">The graph of the region 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\/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]\r\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. 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>\r\n\r\n<div id=\"fs-id1167793619980\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)+2=(4-x)+2=6-x.[\/latex]<\/div>\r\n<p id=\"fs-id1167794200068\">The radius of the inner circle is [latex]g(x)=2.[\/latex] Therefore, we have<\/p>\r\n\r\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>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167793571148\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind the volume of a solid of revolution formed by revolving the region bounded above by the graph of [latex]f(x)=x+2[\/latex] and below by the [latex]x\\text{-axis}[\/latex] over the interval [latex]\\left[0,3\\right][\/latex] around the line [latex]y=-1.[\/latex]\r\n<div>[reveal-answer q=\"541009\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"541009\"]Use the procedure from the last example.[\/hidden-answer]<\/div>\r\n<div><\/div>\r\n<div id=\"fs-id1167793571152\" class=\"exercise\">[reveal-answer q=\"fs-id1167793638825\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793638825\"]\r\n<p id=\"fs-id1167793638825\">[latex]60\\pi [\/latex] units<sup>3<\/sup><\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Find the volume of a solid of revolution using the disk method<\/li>\n<li>Find the volume of a solid of revolution with a cavity using the washer method<\/li>\n<\/ul>\n<\/div>\n<h2>The Disk Method<\/h2>\n<p id=\"fs-id1167793777573\">When we use the slicing method with solids of revolution, it is often called the <strong>disk method<\/strong> because, for solids of revolution, the slices used to over approximate the volume of the solid are disks. To see this, consider the solid of revolution generated by revolving the region between the graph of the function [latex]f(x)={(x-1)}^{2}+1[\/latex] and the [latex]x\\text{-axis}[\/latex] over the interval [latex]\\left[-1,3\\right][\/latex] around the [latex]x\\text{-axis}\\text{.}[\/latex] The graph of the function and a representative disk are shown in Figure 9(a) and (b). The region of revolution and the resulting solid are shown in Figure 9(c) and (d).<\/p>\n<div style=\"width: 909px\" 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\" \/><\/p>\n<p 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.<\/p>\n<\/div>\n<p id=\"fs-id1167793551150\">We already used the formal Riemann sum development of the volume formula when we developed the slicing method. We know that<\/p>\n<div id=\"fs-id1167793443010\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V={\\displaystyle\\int }_{a}^{b}A(x)dx[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794333239\">The only difference with the disk method is that we know the formula for the cross-sectional area ahead of time; it is the area of a circle. This gives the following rule.<\/p>\n<div id=\"fs-id1167793370844\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">The Disk Method<\/h3>\n<hr \/>\n<p id=\"fs-id1167793895393\">Let [latex]f(x)[\/latex] be continuous and nonnegative. 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] 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<\/div>\n<p id=\"fs-id1167793497293\">The volume of the solid we have been studying (Figure 9) is given by<\/p>\n<div id=\"fs-id1167794040142\" 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}^{3}\\pi {\\left[{(x-1)}^{2}+1\\right]}^{2}dx=\\pi {\\displaystyle\\int }_{-1}^{3}{\\left[{(x-1)}^{4}+2{(x-1)}^{2}+1\\right]}^{2}dx\\hfill \\\\ & =\\pi {\\left[\\frac{1}{5}{(x-1)}^{5}+\\frac{2}{3}{(x-1)}^{3}+x\\right]|}_{-1}^{3}=\\pi \\left[(\\frac{32}{5}+\\frac{16}{3}+3)-(-\\frac{32}{5}-\\frac{16}{3}-1)\\right]=\\frac{412\\pi }{15}{\\text{units}}^{3}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794329525\">Let\u2019s look at some examples.<\/p>\n<div id=\"fs-id1167793956465\" class=\"textbook exercises\">\n<h3>Example: Using the Disk Method to Find the Volume of a Solid of Revolution 1<\/h3>\n<p>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 id=\"fs-id1167793618941\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793419916\">Show Solution<\/span><\/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<div style=\"width: 750px\" 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\" \/><\/p>\n<p 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]<\/p>\n<\/div>\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<\/div>\n<\/div>\n<div id=\"fs-id1167794037149\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>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{4-x}[\/latex] and the [latex]x\\text{-axis}[\/latex] over the interval [latex]\\left[0,4\\right][\/latex] around the [latex]x\\text{-axis}\\text{.}[\/latex]<\/p>\n<div id=\"fs-id1167793730425\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794039169\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167794039169\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794039169\">[latex]8\\pi[\/latex] units<sup>3<\/sup><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/7dG21yfKXHg?controls=0&amp;start=755&amp;end=847&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266834\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.2DeterminingVolumesBySlicing755to847_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;2.2 Determining Volumes by Slicing&#8221; here (opens in new window)<\/a>.<\/p>\n<div id=\"qdefault 1\" class=\"hidden-answer\" style=\"display: none\"><\/div>\n<\/div>\n<p id=\"fs-id1167793951744\">So far, our examples have all concerned regions revolved around the [latex]x\\text{-axis,}[\/latex] but we can generate a solid of revolution by revolving a plane region around any horizontal or vertical line. In the next example, we look at a solid of revolution that has been generated by revolving a region around the [latex]y\\text{-axis}\\text{.}[\/latex] The mechanics of the disk method are nearly the same as when the [latex]x\\text{-axis}[\/latex] is the axis of revolution, but we express the function in terms of [latex]y[\/latex] and we integrate with respect to [latex]y[\/latex] as well. This is summarized in the following rule.<\/p>\n<div id=\"fs-id1167793571437\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">The Disk Method for Solids of Revolution around the [latex]y[\/latex]-axis<\/h3>\n<hr \/>\n<p id=\"fs-id1167794006397\">Let [latex]g(y)[\/latex] be continuous and nonnegative. 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] 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<p>&nbsp;<\/p>\n<\/div>\n<p id=\"fs-id1167793377534\">The next example shows how this rule works in practice.<\/p>\n<div id=\"fs-id1167793271607\" class=\"textbook exercises\">\n<h3>example: Using the Disk Method to Find the Volume of a Solid of Revolution 2<\/h3>\n<p>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 id=\"fs-id1167793271609\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793368748\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793368748\" class=\"hidden-answer\" style=\"display: none\"><\/div>\n<\/div>\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<div style=\"width: 750px\" 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\" \/><\/p>\n<p 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]<\/p>\n<\/div>\n<p id=\"fs-id1167794142277\">The region to be revolved and the full solid of revolution are depicted in the following figure.<\/p>\n<div style=\"width: 750px\" 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\" \/><\/p>\n<p 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]<\/p>\n<\/div>\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<\/div>\n<\/div>\n<div id=\"fs-id1167793455063\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Use the disk method to find the volume of the solid of revolution generated by rotating the region between the graph of [latex]g(y)=y[\/latex] and the [latex]y\\text{-axis}[\/latex] over the interval [latex]\\left[1,4\\right][\/latex] around the [latex]y\\text{-axis}\\text{.}[\/latex]<\/p>\n<div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q476774\">Hint<\/span><\/p>\n<div id=\"q476774\" class=\"hidden-answer\" style=\"display: none\">Use the procedure from the last example.<\/div>\n<\/div>\n<\/div>\n<div><\/div>\n<div id=\"fs-id1167793455066\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793960078\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793960078\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793960078\">[latex]21\\pi[\/latex] units<sup>3<\/sup><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm20075\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=20075&theme=oea&iframe_resize_id=ohm20075&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div id=\"fs-id1167794172336\" class=\"bc-section section\">\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. Sometimes, this is just a result of the way the region of revolution is shaped with respect to the axis of revolution. In other cases, cavities arise when the region of revolution is defined as the region between the graphs of two functions. A third way this can happen is when an axis of revolution other than the [latex]x\\text{-axis}[\/latex] or [latex]y\\text{-axis}[\/latex] is selected.<\/p>\n<p id=\"fs-id1167793831962\">When the solid of revolution has a cavity in the middle, the slices used to approximate the volume are not disks, but washers (disks with holes in the center). 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. The graph of the function and a representative washer are shown in Figure 13(a) and (b). The region of revolution and the resulting solid are shown in Figure 13(c) and (d).<\/p>\n<div style=\"width: 679px\" 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\" \/><\/p>\n<p 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.<\/p>\n<\/div>\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>&nbsp;<\/p>\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>&nbsp;<\/p>\n<p id=\"fs-id1167793465550\">Generalizing this process gives the <strong>washer method<\/strong>.<\/p>\n<div id=\"fs-id1167793397841\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">The Washer Method<\/h3>\n<hr \/>\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] 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] 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<p>&nbsp;<\/p>\n<\/div>\n<div id=\"fs-id1167794209450\" class=\"textbook exercises\">\n<h3>Example: Using the Washer Method<\/h3>\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 id=\"fs-id1167794209452\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794211097\">Show Solution<\/span><\/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<div style=\"width: 700px\" 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\" \/><\/p>\n<p 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.<\/p>\n<\/div>\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<\/div>\n<\/div>\n<div id=\"fs-id1167794324562\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the volume of a solid of revolution formed by revolving the region bounded by the graphs of [latex]f(x)=\\sqrt{x}[\/latex] and [latex]g(x)=\\frac{1}{x}[\/latex] over the interval [latex]\\left[1,3\\right][\/latex] around the [latex]x\\text{-axis}\\text{.}[\/latex]<\/p>\n<div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q888359\">Hint<\/span><\/p>\n<div id=\"q888359\" class=\"hidden-answer\" style=\"display: none\">Graph the functions to determine which graph forms the upper bound and which graph forms the lower bound, then use the procedure from the previous example.<\/div>\n<\/div>\n<\/div>\n<div><\/div>\n<div id=\"fs-id1167794324566\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794030044\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167794030044\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794030044\">[latex]\\frac{10\\pi }{3}[\/latex] units<sup>3<\/sup><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/7dG21yfKXHg?controls=0&amp;start=1315&amp;end=1476&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266833\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266833\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.2DeterminingVolumesBySlicing1315to1476_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;2.2 Determining Volumes by Slicing&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<p id=\"fs-id1167793940198\">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<div id=\"fs-id1167793565052\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">The Washer Method for Solids of Revolution around the [latex]y[\/latex]-axis<\/h3>\n<hr \/>\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] 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] 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<\/div>\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<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<div id=\"fs-id1167793912719\" class=\"textbook exercises\">\n<h3>Example: The Washer Method with a Different Axis of Revolution<\/h3>\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 id=\"fs-id1167793912721\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794218804\">Show Solution<\/span><\/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<div style=\"width: 750px\" 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\" \/><\/p>\n<p 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.<\/p>\n<\/div>\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. 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<\/div>\n<\/div>\n<div id=\"fs-id1167793571148\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\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+2[\/latex] and below by the [latex]x\\text{-axis}[\/latex] over the interval [latex]\\left[0,3\\right][\/latex] around the line [latex]y=-1.[\/latex]<\/p>\n<div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q541009\">Hint<\/span><\/p>\n<div id=\"q541009\" class=\"hidden-answer\" style=\"display: none\">Use the procedure from the last example.<\/div>\n<\/div>\n<\/div>\n<div><\/div>\n<div id=\"fs-id1167793571152\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793638825\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793638825\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793638825\">[latex]60\\pi[\/latex] units<sup>3<\/sup><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1168\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>2.2 Determining Volumes by Slicing. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 2. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":8,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 2\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"2.2 Determining Volumes by Slicing\",\"author\":\"Ryan 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