{"id":583,"date":"2021-02-04T16:24:32","date_gmt":"2021-02-04T16:24:32","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=583"},"modified":"2022-03-16T22:29:50","modified_gmt":"2022-03-16T22:29:50","slug":"volume-and-the-slicing-method","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/volume-and-the-slicing-method\/","title":{"raw":"Slicing Method","rendered":"Slicing Method"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Determine the volume of a solid by integrating a cross-section (the slicing method)<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Volume and the Slicing Method<\/h2>\r\n<p id=\"fs-id1167794058130\">Just as area is the numerical measure of a two-dimensional region, volume is the numerical measure of a three-dimensional solid. Most of us have computed volumes of solids by using basic geometric formulas. The volume of a rectangular solid, for example, can be computed by multiplying length, width, and height: [latex]V=lwh.[\/latex] The formulas for the volume of a sphere [latex](V=\\frac{4}{3}\\pi {r}^{3}),[\/latex] a cone [latex](V=\\frac{1}{3}\\pi {r}^{2}h),[\/latex] and a pyramid [latex](V=\\frac{1}{3}Ah)[\/latex] have also been introduced. Although some of these formulas were derived using geometry alone, all these formulas can be obtained by using integration.<\/p>\r\n<p id=\"fs-id1167794118853\">We can also calculate the volume of a cylinder. Although most of us think of a cylinder as having a circular base, such as a soup can or a metal rod, in mathematics the word <em>cylinder<\/em> has a more general meaning. To discuss cylinders in this more general context, we first need to define some vocabulary.<\/p>\r\n<p id=\"fs-id1167793931500\">We define the <strong>cross-section<\/strong> of a solid to be the intersection of a plane with the solid. A <em>cylinder<\/em> is defined as any solid that can be generated by translating a plane region along a line perpendicular to the region, called the <em>axis<\/em> of the cylinder. Thus, all cross-sections perpendicular to the axis of a cylinder are identical. The solid shown in Figure 1 is an example of a cylinder with a noncircular base. To calculate the volume of a cylinder, then, we simply multiply the area of the cross-section by the height of the cylinder: [latex]V=A\u00b7h.[\/latex] In the case of a right circular cylinder (soup can), this becomes [latex]V=\\pi {r}^{2}h.[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"799\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11212740\/CNX_Calc_Figure_06_02_001.jpg\" alt=\"This graphic has two figures. The first figure is half of a cylinder, on the flat portion. The cylinder has a line through the center labeled \u201cx\u201d. Vertically cutting through the cylinder, perpendicular to the line is a plane. The second figure is a two dimensional cross section of the cylinder intersecting with the plane. It is a semi-circle.\" width=\"799\" height=\"412\" \/> Figure 1. Each cross-section of a particular cylinder is identical to the others.[\/caption]\r\n<p id=\"fs-id1167793374741\">If a solid does not have a constant cross-section (and it is not one of the other basic solids), we may not have a formula for its volume. In this case, we can use a definite integral to calculate the volume of the solid. We do this by slicing the solid into pieces, estimating the volume of each slice, and then adding those estimated volumes together. The slices should all be parallel to one another, and when we put all the slices together, we should get the whole solid. Consider, for example, the solid <em>S<\/em> shown in Figure 2, extending along the [latex]x\\text{-axis}\\text{.}[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"402\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11212743\/CNX_Calc_Figure_06_02_002.jpg\" alt=\"This figure is a graph of a 3-dimensional solid. It has one edge along the x-axis. The x-axis is part of the 2-dimensional coordinate system with the y-axis labeled. The edge of the solid along the x-axis starts at a point labeled \u201ca\u201d and stops at a point labeled \u201cb\u201d.\" width=\"402\" height=\"341\" \/> Figure 2. A solid with a varying cross-section.[\/caption]\r\n<p id=\"fs-id1167794051655\">We want to divide [latex]S[\/latex] into slices perpendicular to the [latex]x\\text{-axis}\\text{.}[\/latex] As we see later in the chapter, there may be times when we want to slice the solid in some other direction\u2014say, with slices perpendicular to the [latex]y[\/latex]-axis. The decision of which way to slice the solid is very important. If we make the wrong choice, the computations can get quite messy. Later in the chapter, we examine some of these situations in detail and look at how to decide which way to slice the solid. For the purposes of this section, however, we use slices perpendicular to the [latex]x\\text{-axis}\\text{.}[\/latex]<\/p>\r\n<p id=\"fs-id1167794142153\">Because the cross-sectional area is not constant, we let [latex]A(x)[\/latex] represent the area of the cross-section at point [latex]x.[\/latex] Now let [latex]P=\\left\\{{x}_{0},{x}_{1}\\text{\u2026},{X}_{n}\\right\\}[\/latex] be a regular partition of [latex]\\left[a,b\\right],[\/latex] and for [latex]i=1,2\\text{,\u2026}n,[\/latex] let [latex]{S}_{i}[\/latex] represent the slice of [latex]S[\/latex] stretching from [latex]{x}_{i-1}\\text{ to }{x}_{i}.[\/latex] The following figure shows the sliced solid with [latex]n=3.[\/latex]<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"402\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11212746\/CNX_Calc_Figure_06_02_003.jpg\" alt=\"This figure is a graph of a 3-dimensional solid. It has one edge along the x-axis. The x-axis is part of the 2-dimensional coordinate system with the y-axis labeled. The edge of the solid along the x-axis starts at a point labeled \u201ca=xsub0\u201d. The solid is divided up into smaller solids with slices at xsub1, xsub2, and stops at a point labeled \u201cb=xsub3\u201d. These smaller solids are labeled Ssub1, Ssub2, and Ssub3. They are also shaded.\" width=\"402\" height=\"341\" \/> Figure 3. The solid [latex]S[\/latex] has been divided into three slices perpendicular to the [latex]x\\text{-axis}.[\/latex][\/caption]\r\n<p id=\"fs-id1167793928471\">Finally, for [latex]i=1,2\\text{,\u2026}n,[\/latex] let [latex]{x}_{i}^{*}[\/latex] be an arbitrary point in [latex]\\left[{x}_{i-1},{x}_{i}\\right].[\/latex] Then the volume of slice [latex]{S}_{i}[\/latex] can be estimated by [latex]V({S}_{i})\\approx A({x}_{i}^{*})\\text{\u0394}x.[\/latex] Adding these approximations together, we see the volume of the entire solid [latex]S[\/latex] can be approximated by<\/p>\r\n\r\n<div id=\"fs-id1167793562402\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V(S)\\approx \\underset{i=1}{\\overset{n}{\\text{\u2211}}}A({x}_{i}^{*})\\text{\u0394}x[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793288503\">By now, we can recognize this as a Riemann sum, and our next step is to take the limit as [latex]n\\to \\infty .[\/latex] Then we have<\/p>\r\n\r\n<div id=\"fs-id1167794334680\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V(S)=\\underset{n\\to \\infty }{\\text{lim}}\\underset{i=1}{\\overset{n}{\\text{\u2211}}}A({x}_{i}^{*})\\text{\u0394}x=\\underset{a}{\\overset{b}{\\displaystyle\\int }}A(x)dx[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793638650\">The technique we have just described is called the <strong>slicing method<\/strong>. To apply it, we use the following strategy.<\/p>\r\n\r\n<div id=\"fs-id1167794210849\" class=\"textbox examples\">\r\n<h3>Problem-Solving Strategy: Finding Volumes by the Slicing Method<\/h3>\r\n<ol id=\"fs-id1167793274159\">\r\n \t<li>Examine the solid and determine the shape of a cross-section of the solid. It is often helpful to draw a picture if one is not provided.<\/li>\r\n \t<li>Determine a formula for the area of the cross-section.<\/li>\r\n \t<li>Integrate the area formula over the appropriate interval to get the volume.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<p id=\"fs-id1167793263412\">Recall that in this section, we assume the slices are perpendicular to the [latex]x\\text{-axis}\\text{.}[\/latex] Therefore, the area formula is in terms of [latex]x[\/latex] and the limits of integration lie on the [latex]x\\text{-axis}\\text{.}[\/latex] However, the problem-solving strategy shown here is valid regardless of how we choose to slice the solid.<\/p>\r\n\r\n<div id=\"fs-id1167793579586\" class=\"textbook exercises\">\r\n<h3>Example: Deriving the Formula for the Volume of a Pyramid<\/h3>\r\nWe know from geometry that the formula for the volume of a pyramid is [latex]V=\\frac{1}{3}Ah.[\/latex] If the pyramid has a square base, this becomes [latex]V=\\frac{1}{3}{a}^{2}h,[\/latex] where [latex]a[\/latex] denotes the length of one side of the base. We are going to use the slicing method to derive this formula.\r\n<div id=\"fs-id1167793510162\" class=\"exercise\">[reveal-answer q=\"fs-id1167793928284\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793928284\"]\r\n<p id=\"fs-id1167793928284\">We want to apply the slicing method to a pyramid with a square base. To set up the integral, consider the pyramid shown in Figure 4, oriented along the [latex]x\\text{-axis}\\text{.}[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"891\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11212749\/CNX_Calc_Figure_06_02_004.jpg\" alt=\"This figure has two graphs. The first graph, labeled \u201ca\u201d, is a pyramid on its side. The x-axis goes through the middle of the pyramid. The point of the top of the pyramid is at the origin of the x y coordinate system. The base of the pyramid is shaded and labeled \u201ca\u201d. Inside of the pyramid is a shaded rectangle labeled \u201cs\u201d. The distance from the y-axis to the base of the pyramid is labeled \u201ch\u201d. the distance the rectangle inside of the pyramid to the y-axis is labeled \u201cx\u201d. The second figure is a cross section of the pyramid with the x and y axes labeled. The cross section is a triangle with one side labeled \u201ca\u201d, perpendicular to the x-axis. The distance a is from the y-axis is h. There is another perpendicular line to the x-axis inside of the triangle. It is labeled \u201cs\u201d. It is x units from the y-axis.\" width=\"891\" height=\"315\" \/> Figure 4. (a) A pyramid with a square base is oriented along the x-axis. (b) A two-dimensional view of the pyramid is seen from the side.[\/caption]\r\n<p id=\"fs-id1167794071301\">We first want to determine the shape of a cross-section of the pyramid. We are know the base is a square, so the cross-sections are squares as well (step 1). Now we want to determine a formula for the area of one of these cross-sectional squares. Looking at Figure 4(b), and using a proportion, since these are similar triangles, we have<\/p>\r\n\r\n<div id=\"fs-id1167794077150\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{s}{a}=\\frac{x}{h}[\/latex]\u00a0 \u00a0or\u00a0 \u00a0[latex]s=\\frac{ax}{h}[\/latex]<\/div>\r\n<p id=\"fs-id1167794142328\">Therefore, the area of one of the cross-sectional squares is<\/p>\r\n\r\n<div id=\"fs-id1167794040742\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A(x)={s}^{2}={(\\frac{ax}{h})}^{2}[\/latex]\u00a0 (step [latex]2[\/latex])<\/div>\r\n<p id=\"fs-id1167793871619\">Then we find the volume of the pyramid by integrating from [latex]0\\text{ to }h[\/latex] (step [latex]3)\\text{:}[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1167793444900\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill V&amp; =\\underset{0}{\\overset{h}{\\displaystyle\\int }}A(x)dx\\hfill \\\\ &amp; =\\underset{0}{\\overset{h}{\\displaystyle\\int }}{(\\frac{ax}{h})}^{2}dx=\\frac{{a}^{2}}{{h}^{2}}\\underset{0}{\\overset{h}{\\displaystyle\\int }}{x}^{2}dx\\hfill \\\\ &amp; ={\\left[\\frac{{a}^{2}}{{h}^{2}}(\\frac{1}{3}{x}^{3})\\right]|}_{0}^{h}=\\frac{1}{3}{a}^{2}h.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1167794063528\">This is the formula we were looking for.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167793398724\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nUse the slicing method to derive the formula [latex]V=\\frac{1}{3}\\pi {r}^{2}h[\/latex] for the volume of a circular cone.\r\n<div id=\"fs-id1167793299677\" class=\"exercise\">\r\n<div id=\"fs-id1167794094176\" class=\"commentary\">\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1167794098203\">Use similar triangles, as in the last example.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<h2>Solids of Revolution and the Slicing Method<\/h2>\r\n<p id=\"fs-id1167793263466\">If a region in a plane is revolved around a line in that plane, the resulting solid is called a<strong> solid of revolution<\/strong>, as shown in the following figure.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"529\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11212754\/CNX_Calc_Figure_06_02_005.jpg\" alt=\"This figure has three graphs. The first graph, labeled \u201ca\u201d is a region in the x y plane. The region is created by a curve above the x-axis and the x-axis. The second graph, labeled \u201cb\u201d is the same region as in \u201ca\u201d, but it shows the region beginning to rotate around the x-axis. The third graph, labeled \u201cc\u201d is the solid formed by rotating the region from \u201ca\u201d completely around the x-axis, forming a solid.\" width=\"529\" height=\"1098\" \/> Figure 5. (a) This is the region that is revolved around the x-axis. (b) As the region begins to revolve around the axis, it sweeps out a solid of revolution. (c) This is the solid that results when the revolution is complete.[\/caption]\r\n<p id=\"fs-id1167793847672\">Solids of revolution are common in mechanical applications, such as machine parts produced by a lathe. We spend the rest of this section looking at solids of this type. The next example uses the slicing method to calculate the volume of a solid of revolution.<\/p>\r\n\r\n<div id=\"fs-id1167794036722\" class=\"textbox tryit\">\r\n<h3>Interactive<\/h3>\r\n<p id=\"fs-id1167793442787\"><a href=\"https:\/\/www.wolframalpha.com\/calculators\/integral-calculator\/\" target=\"_blank\" rel=\"noopener\">Use an online integral calculator to learn more.<\/a><\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1167794046315\" class=\"textbook exercises\">\r\n<h3>Example: Using the Slicing Method to find the Volume of a Solid of Revolution<\/h3>\r\nUse the slicing method to find the volume of the solid of revolution bounded by the graphs of [latex]f(x)={x}^{2}-4x+5,x=1,\\text{ and }x=4,[\/latex] and rotated about the [latex]x\\text{-axis}\\text{.}[\/latex]\r\n<div id=\"fs-id1167794167925\" class=\"exercise\">[reveal-answer q=\"fs-id1167793998058\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793998058\"]\r\n<p id=\"fs-id1167793998058\">Using the problem-solving strategy, we first sketch the graph of the quadratic function over the interval [latex]\\left[1,4\\right][\/latex] as shown in the following figure.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"304\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11212758\/CNX_Calc_Figure_06_02_006.jpg\" alt=\"This figure is a graph of the parabola f(x)=x^2-4x+5. The parabola is the top of a shaded region above the x-axis. The region is bounded to the left by a line at x=1 and to the right by a line at x=4.\" width=\"304\" height=\"314\" \/> Figure 6. A region used to produce a solid of revolution.[\/caption]\r\n<p id=\"fs-id1167794169307\">Next, revolve the region around the [latex]x[\/latex]-axis, as 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\/11212801\/CNX_Calc_Figure_06_02_007.jpg\" alt=\"This figure has two graphs of the parabola f(x)=x^2-4x+5. The parabola is the top of a shaded region above the x-axis. The region is bounded to the left by a line at x=1 and to the right by a line at x=4. The first graph has a shaded solid below the parabola. This solid has been formed by rotating the parabola around the x-axis. The second graph is the same as the first, with the solid being rotated to show the solid.\" width=\"740\" height=\"541\" \/> Figure 7. Two views, (a) and (b), of the solid of revolution produced by revolving the region in (Figure) about the [latex]x\\text{-axis}\\text{.}[\/latex][\/caption]\r\n<p id=\"fs-id1167793567021\">Since the solid was formed by revolving the region around the [latex]x\\text{-axis,}[\/latex] the cross-sections are circles (step 1). The area of the cross-section, then, is the area of a circle, and the radius of the circle is given by [latex]f(x).[\/latex] Use the formula for the area of the circle:<\/p>\r\n\r\n<div id=\"fs-id1167793938405\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A(x)=\\pi {r}^{2}=\\pi {\\left[f(x)\\right]}^{2}=\\pi {({x}^{2}-4x+5)}^{2}[\/latex]\u00a0 (step 2)<\/div>\r\n<p id=\"fs-id1167794068988\">The volume, then, is (step 3)<\/p>\r\n\r\n<div id=\"fs-id1167793618979\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill V&amp; =\\underset{a}{\\overset{h}{\\displaystyle\\int }}A(x)dx\\hfill \\\\ &amp; ={\\displaystyle\\int }_{1}^{4}\\pi {({x}^{2}-4x+5)}^{2}dx=\\pi {\\displaystyle\\int }_{1}^{4}({x}^{4}-8{x}^{3}+26{x}^{2}-40x+25)dx\\hfill \\\\ &amp; ={\\pi (\\frac{{x}^{5}}{5}-2{x}^{4}+\\frac{26{x}^{3}}{3}-20{x}^{2}+25x)|}_{1}^{4}=\\frac{78}{5}\\pi .\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1167793524919\">The volume is [latex]\\frac{78\\pi}{5}.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167794066617\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1167793887857\">Use the method of slicing to find the volume of the solid of revolution formed by revolving the region between the graph of the function [latex]f(x)=\\frac{1}{x}[\/latex] and the [latex]x\\text{-axis}[\/latex] over the interval [latex]\\left[1,2\\right][\/latex] around the [latex]x\\text{-axis}\\text{.}[\/latex] See the following figure.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"523\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11212804\/CNX_Calc_Figure_06_02_008.jpg\" alt=\"This figure has two graphs. The first graph is the curve f(x)=1\/x. It is a decreasing curve, above the x-axis in the first quadrant. The graph has a shaded region under the curve between x=1 and x=2. The second graph is the curve f(x)=1\/x in the first quadrant. Also, underneath this graph, there is a solid between x=1 and x=2 that has been formed by rotating the region from the first graph around the x-axis.\" width=\"523\" height=\"273\" \/> Figure 8.[\/caption]\r\n\r\n[reveal-answer q=\"fs-id1167793261323\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793261323\"]\r\n<p id=\"fs-id1167793261323\">[latex]\\frac{\\pi }{2}[\/latex]<\/p>\r\n\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1167794064666\">Use the problem-solving strategy presented earlier and follow the last example to help with step 2.<\/p>\r\n[\/hidden-answer]\r\n\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=495&amp;end=605&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\/2.2DeterminingVolumesBySlicing495to605_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<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]69239[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Determine the volume of a solid by integrating a cross-section (the slicing method)<\/li>\n<\/ul>\n<\/div>\n<h2>Volume and the Slicing Method<\/h2>\n<p id=\"fs-id1167794058130\">Just as area is the numerical measure of a two-dimensional region, volume is the numerical measure of a three-dimensional solid. Most of us have computed volumes of solids by using basic geometric formulas. The volume of a rectangular solid, for example, can be computed by multiplying length, width, and height: [latex]V=lwh.[\/latex] The formulas for the volume of a sphere [latex](V=\\frac{4}{3}\\pi {r}^{3}),[\/latex] a cone [latex](V=\\frac{1}{3}\\pi {r}^{2}h),[\/latex] and a pyramid [latex](V=\\frac{1}{3}Ah)[\/latex] have also been introduced. Although some of these formulas were derived using geometry alone, all these formulas can be obtained by using integration.<\/p>\n<p id=\"fs-id1167794118853\">We can also calculate the volume of a cylinder. Although most of us think of a cylinder as having a circular base, such as a soup can or a metal rod, in mathematics the word <em>cylinder<\/em> has a more general meaning. To discuss cylinders in this more general context, we first need to define some vocabulary.<\/p>\n<p id=\"fs-id1167793931500\">We define the <strong>cross-section<\/strong> of a solid to be the intersection of a plane with the solid. A <em>cylinder<\/em> is defined as any solid that can be generated by translating a plane region along a line perpendicular to the region, called the <em>axis<\/em> of the cylinder. Thus, all cross-sections perpendicular to the axis of a cylinder are identical. The solid shown in Figure 1 is an example of a cylinder with a noncircular base. To calculate the volume of a cylinder, then, we simply multiply the area of the cross-section by the height of the cylinder: [latex]V=A\u00b7h.[\/latex] In the case of a right circular cylinder (soup can), this becomes [latex]V=\\pi {r}^{2}h.[\/latex]<\/p>\n<div style=\"width: 809px\" 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\/11212740\/CNX_Calc_Figure_06_02_001.jpg\" alt=\"This graphic has two figures. The first figure is half of a cylinder, on the flat portion. The cylinder has a line through the center labeled \u201cx\u201d. Vertically cutting through the cylinder, perpendicular to the line is a plane. The second figure is a two dimensional cross section of the cylinder intersecting with the plane. It is a semi-circle.\" width=\"799\" height=\"412\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. Each cross-section of a particular cylinder is identical to the others.<\/p>\n<\/div>\n<p id=\"fs-id1167793374741\">If a solid does not have a constant cross-section (and it is not one of the other basic solids), we may not have a formula for its volume. In this case, we can use a definite integral to calculate the volume of the solid. We do this by slicing the solid into pieces, estimating the volume of each slice, and then adding those estimated volumes together. The slices should all be parallel to one another, and when we put all the slices together, we should get the whole solid. Consider, for example, the solid <em>S<\/em> shown in Figure 2, extending along the [latex]x\\text{-axis}\\text{.}[\/latex]<\/p>\n<div style=\"width: 412px\" 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\/11212743\/CNX_Calc_Figure_06_02_002.jpg\" alt=\"This figure is a graph of a 3-dimensional solid. It has one edge along the x-axis. The x-axis is part of the 2-dimensional coordinate system with the y-axis labeled. The edge of the solid along the x-axis starts at a point labeled \u201ca\u201d and stops at a point labeled \u201cb\u201d.\" width=\"402\" height=\"341\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. A solid with a varying cross-section.<\/p>\n<\/div>\n<p id=\"fs-id1167794051655\">We want to divide [latex]S[\/latex] into slices perpendicular to the [latex]x\\text{-axis}\\text{.}[\/latex] As we see later in the chapter, there may be times when we want to slice the solid in some other direction\u2014say, with slices perpendicular to the [latex]y[\/latex]-axis. The decision of which way to slice the solid is very important. If we make the wrong choice, the computations can get quite messy. Later in the chapter, we examine some of these situations in detail and look at how to decide which way to slice the solid. For the purposes of this section, however, we use slices perpendicular to the [latex]x\\text{-axis}\\text{.}[\/latex]<\/p>\n<p id=\"fs-id1167794142153\">Because the cross-sectional area is not constant, we let [latex]A(x)[\/latex] represent the area of the cross-section at point [latex]x.[\/latex] Now let [latex]P=\\left\\{{x}_{0},{x}_{1}\\text{\u2026},{X}_{n}\\right\\}[\/latex] be a regular partition of [latex]\\left[a,b\\right],[\/latex] and for [latex]i=1,2\\text{,\u2026}n,[\/latex] let [latex]{S}_{i}[\/latex] represent the slice of [latex]S[\/latex] stretching from [latex]{x}_{i-1}\\text{ to }{x}_{i}.[\/latex] The following figure shows the sliced solid with [latex]n=3.[\/latex]<\/p>\n<div style=\"width: 412px\" 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\/11212746\/CNX_Calc_Figure_06_02_003.jpg\" alt=\"This figure is a graph of a 3-dimensional solid. It has one edge along the x-axis. The x-axis is part of the 2-dimensional coordinate system with the y-axis labeled. The edge of the solid along the x-axis starts at a point labeled \u201ca=xsub0\u201d. The solid is divided up into smaller solids with slices at xsub1, xsub2, and stops at a point labeled \u201cb=xsub3\u201d. These smaller solids are labeled Ssub1, Ssub2, and Ssub3. They are also shaded.\" width=\"402\" height=\"341\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. The solid [latex]S[\/latex] has been divided into three slices perpendicular to the [latex]x\\text{-axis}.[\/latex]<\/p>\n<\/div>\n<p id=\"fs-id1167793928471\">Finally, for [latex]i=1,2\\text{,\u2026}n,[\/latex] let [latex]{x}_{i}^{*}[\/latex] be an arbitrary point in [latex]\\left[{x}_{i-1},{x}_{i}\\right].[\/latex] Then the volume of slice [latex]{S}_{i}[\/latex] can be estimated by [latex]V({S}_{i})\\approx A({x}_{i}^{*})\\text{\u0394}x.[\/latex] Adding these approximations together, we see the volume of the entire solid [latex]S[\/latex] can be approximated by<\/p>\n<div id=\"fs-id1167793562402\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V(S)\\approx \\underset{i=1}{\\overset{n}{\\text{\u2211}}}A({x}_{i}^{*})\\text{\u0394}x[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793288503\">By now, we can recognize this as a Riemann sum, and our next step is to take the limit as [latex]n\\to \\infty .[\/latex] Then we have<\/p>\n<div id=\"fs-id1167794334680\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V(S)=\\underset{n\\to \\infty }{\\text{lim}}\\underset{i=1}{\\overset{n}{\\text{\u2211}}}A({x}_{i}^{*})\\text{\u0394}x=\\underset{a}{\\overset{b}{\\displaystyle\\int }}A(x)dx[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793638650\">The technique we have just described is called the <strong>slicing method<\/strong>. To apply it, we use the following strategy.<\/p>\n<div id=\"fs-id1167794210849\" class=\"textbox examples\">\n<h3>Problem-Solving Strategy: Finding Volumes by the Slicing Method<\/h3>\n<ol id=\"fs-id1167793274159\">\n<li>Examine the solid and determine the shape of a cross-section of the solid. It is often helpful to draw a picture if one is not provided.<\/li>\n<li>Determine a formula for the area of the cross-section.<\/li>\n<li>Integrate the area formula over the appropriate interval to get the volume.<\/li>\n<\/ol>\n<\/div>\n<p id=\"fs-id1167793263412\">Recall that in this section, we assume the slices are perpendicular to the [latex]x\\text{-axis}\\text{.}[\/latex] Therefore, the area formula is in terms of [latex]x[\/latex] and the limits of integration lie on the [latex]x\\text{-axis}\\text{.}[\/latex] However, the problem-solving strategy shown here is valid regardless of how we choose to slice the solid.<\/p>\n<div id=\"fs-id1167793579586\" class=\"textbook exercises\">\n<h3>Example: Deriving the Formula for the Volume of a Pyramid<\/h3>\n<p>We know from geometry that the formula for the volume of a pyramid is [latex]V=\\frac{1}{3}Ah.[\/latex] If the pyramid has a square base, this becomes [latex]V=\\frac{1}{3}{a}^{2}h,[\/latex] where [latex]a[\/latex] denotes the length of one side of the base. We are going to use the slicing method to derive this formula.<\/p>\n<div id=\"fs-id1167793510162\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793928284\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793928284\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793928284\">We want to apply the slicing method to a pyramid with a square base. To set up the integral, consider the pyramid shown in Figure 4, oriented along the [latex]x\\text{-axis}\\text{.}[\/latex]<\/p>\n<div style=\"width: 901px\" 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\/11212749\/CNX_Calc_Figure_06_02_004.jpg\" alt=\"This figure has two graphs. The first graph, labeled \u201ca\u201d, is a pyramid on its side. The x-axis goes through the middle of the pyramid. The point of the top of the pyramid is at the origin of the x y coordinate system. The base of the pyramid is shaded and labeled \u201ca\u201d. Inside of the pyramid is a shaded rectangle labeled \u201cs\u201d. The distance from the y-axis to the base of the pyramid is labeled \u201ch\u201d. the distance the rectangle inside of the pyramid to the y-axis is labeled \u201cx\u201d. The second figure is a cross section of the pyramid with the x and y axes labeled. The cross section is a triangle with one side labeled \u201ca\u201d, perpendicular to the x-axis. The distance a is from the y-axis is h. There is another perpendicular line to the x-axis inside of the triangle. It is labeled \u201cs\u201d. It is x units from the y-axis.\" width=\"891\" height=\"315\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4. (a) A pyramid with a square base is oriented along the x-axis. (b) A two-dimensional view of the pyramid is seen from the side.<\/p>\n<\/div>\n<p id=\"fs-id1167794071301\">We first want to determine the shape of a cross-section of the pyramid. We are know the base is a square, so the cross-sections are squares as well (step 1). Now we want to determine a formula for the area of one of these cross-sectional squares. Looking at Figure 4(b), and using a proportion, since these are similar triangles, we have<\/p>\n<div id=\"fs-id1167794077150\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{s}{a}=\\frac{x}{h}[\/latex]\u00a0 \u00a0or\u00a0 \u00a0[latex]s=\\frac{ax}{h}[\/latex]<\/div>\n<p id=\"fs-id1167794142328\">Therefore, the area of one of the cross-sectional squares is<\/p>\n<div id=\"fs-id1167794040742\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A(x)={s}^{2}={(\\frac{ax}{h})}^{2}[\/latex]\u00a0 (step [latex]2[\/latex])<\/div>\n<p id=\"fs-id1167793871619\">Then we find the volume of the pyramid by integrating from [latex]0\\text{ to }h[\/latex] (step [latex]3)\\text{:}[\/latex]<\/p>\n<div id=\"fs-id1167793444900\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill V& =\\underset{0}{\\overset{h}{\\displaystyle\\int }}A(x)dx\\hfill \\\\ & =\\underset{0}{\\overset{h}{\\displaystyle\\int }}{(\\frac{ax}{h})}^{2}dx=\\frac{{a}^{2}}{{h}^{2}}\\underset{0}{\\overset{h}{\\displaystyle\\int }}{x}^{2}dx\\hfill \\\\ & ={\\left[\\frac{{a}^{2}}{{h}^{2}}(\\frac{1}{3}{x}^{3})\\right]|}_{0}^{h}=\\frac{1}{3}{a}^{2}h.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167794063528\">This is the formula we were looking for.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167793398724\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Use the slicing method to derive the formula [latex]V=\\frac{1}{3}\\pi {r}^{2}h[\/latex] for the volume of a circular cone.<\/p>\n<div id=\"fs-id1167793299677\" class=\"exercise\">\n<div id=\"fs-id1167794094176\" class=\"commentary\">\n<h4>Hint<\/h4>\n<p id=\"fs-id1167794098203\">Use similar triangles, as in the last example.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Solids of Revolution and the Slicing Method<\/h2>\n<p id=\"fs-id1167793263466\">If a region in a plane is revolved around a line in that plane, the resulting solid is called a<strong> solid of revolution<\/strong>, as shown in the following figure.<\/p>\n<div style=\"width: 539px\" 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\/11212754\/CNX_Calc_Figure_06_02_005.jpg\" alt=\"This figure has three graphs. The first graph, labeled \u201ca\u201d is a region in the x y plane. The region is created by a curve above the x-axis and the x-axis. The second graph, labeled \u201cb\u201d is the same region as in \u201ca\u201d, but it shows the region beginning to rotate around the x-axis. The third graph, labeled \u201cc\u201d is the solid formed by rotating the region from \u201ca\u201d completely around the x-axis, forming a solid.\" width=\"529\" height=\"1098\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 5. (a) This is the region that is revolved around the x-axis. (b) As the region begins to revolve around the axis, it sweeps out a solid of revolution. (c) This is the solid that results when the revolution is complete.<\/p>\n<\/div>\n<p id=\"fs-id1167793847672\">Solids of revolution are common in mechanical applications, such as machine parts produced by a lathe. We spend the rest of this section looking at solids of this type. The next example uses the slicing method to calculate the volume of a solid of revolution.<\/p>\n<div id=\"fs-id1167794036722\" class=\"textbox tryit\">\n<h3>Interactive<\/h3>\n<p id=\"fs-id1167793442787\"><a href=\"https:\/\/www.wolframalpha.com\/calculators\/integral-calculator\/\" target=\"_blank\" rel=\"noopener\">Use an online integral calculator to learn more.<\/a><\/p>\n<\/div>\n<div id=\"fs-id1167794046315\" class=\"textbook exercises\">\n<h3>Example: Using the Slicing Method to find the Volume of a Solid of Revolution<\/h3>\n<p>Use the slicing method to find the volume of the solid of revolution bounded by the graphs of [latex]f(x)={x}^{2}-4x+5,x=1,\\text{ and }x=4,[\/latex] and rotated about the [latex]x\\text{-axis}\\text{.}[\/latex]<\/p>\n<div id=\"fs-id1167794167925\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793998058\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793998058\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793998058\">Using the problem-solving strategy, we first sketch the graph of the quadratic function over the interval [latex]\\left[1,4\\right][\/latex] as shown in the following figure.<\/p>\n<div style=\"width: 314px\" 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\/11212758\/CNX_Calc_Figure_06_02_006.jpg\" alt=\"This figure is a graph of the parabola f(x)=x^2-4x+5. The parabola is the top of a shaded region above the x-axis. The region is bounded to the left by a line at x=1 and to the right by a line at x=4.\" width=\"304\" height=\"314\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 6. A region used to produce a solid of revolution.<\/p>\n<\/div>\n<p id=\"fs-id1167794169307\">Next, revolve the region around the [latex]x[\/latex]-axis, as 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\/11212801\/CNX_Calc_Figure_06_02_007.jpg\" alt=\"This figure has two graphs of the parabola f(x)=x^2-4x+5. The parabola is the top of a shaded region above the x-axis. The region is bounded to the left by a line at x=1 and to the right by a line at x=4. The first graph has a shaded solid below the parabola. This solid has been formed by rotating the parabola around the x-axis. The second graph is the same as the first, with the solid being rotated to show the solid.\" width=\"740\" height=\"541\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 7. Two views, (a) and (b), of the solid of revolution produced by revolving the region in (Figure) about the [latex]x\\text{-axis}\\text{.}[\/latex]<\/p>\n<\/div>\n<p id=\"fs-id1167793567021\">Since the solid was formed by revolving the region around the [latex]x\\text{-axis,}[\/latex] the cross-sections are circles (step 1). The area of the cross-section, then, is the area of a circle, and the radius of the circle is given by [latex]f(x).[\/latex] Use the formula for the area of the circle:<\/p>\n<div id=\"fs-id1167793938405\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A(x)=\\pi {r}^{2}=\\pi {\\left[f(x)\\right]}^{2}=\\pi {({x}^{2}-4x+5)}^{2}[\/latex]\u00a0 (step 2)<\/div>\n<p id=\"fs-id1167794068988\">The volume, then, is (step 3)<\/p>\n<div id=\"fs-id1167793618979\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill V& =\\underset{a}{\\overset{h}{\\displaystyle\\int }}A(x)dx\\hfill \\\\ & ={\\displaystyle\\int }_{1}^{4}\\pi {({x}^{2}-4x+5)}^{2}dx=\\pi {\\displaystyle\\int }_{1}^{4}({x}^{4}-8{x}^{3}+26{x}^{2}-40x+25)dx\\hfill \\\\ & ={\\pi (\\frac{{x}^{5}}{5}-2{x}^{4}+\\frac{26{x}^{3}}{3}-20{x}^{2}+25x)|}_{1}^{4}=\\frac{78}{5}\\pi .\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167793524919\">The volume is [latex]\\frac{78\\pi}{5}.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167794066617\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1167793887857\">Use the method of slicing to find the volume of the solid of revolution formed by revolving the region between the graph of the function [latex]f(x)=\\frac{1}{x}[\/latex] and the [latex]x\\text{-axis}[\/latex] over the interval [latex]\\left[1,2\\right][\/latex] around the [latex]x\\text{-axis}\\text{.}[\/latex] See the following figure.<\/p>\n<div style=\"width: 533px\" 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\/11212804\/CNX_Calc_Figure_06_02_008.jpg\" alt=\"This figure has two graphs. The first graph is the curve f(x)=1\/x. It is a decreasing curve, above the x-axis in the first quadrant. The graph has a shaded region under the curve between x=1 and x=2. The second graph is the curve f(x)=1\/x in the first quadrant. Also, underneath this graph, there is a solid between x=1 and x=2 that has been formed by rotating the region from the first graph around the x-axis.\" width=\"523\" height=\"273\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 8.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793261323\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793261323\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793261323\">[latex]\\frac{\\pi }{2}[\/latex]<\/p>\n<h4>Hint<\/h4>\n<p id=\"fs-id1167794064666\">Use the problem-solving strategy presented earlier and follow the last example to help with step 2.<\/p>\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=495&amp;end=605&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\/2.2DeterminingVolumesBySlicing495to605_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<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm69239\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=69239&theme=oea&iframe_resize_id=ohm69239&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\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-583\">\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 1. <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\/details\/books\/calculus-volume-1\">https:\/\/openstax.org\/details\/books\/calculus-volume-1<\/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-1\/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":7,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"2.2 Determining Volumes by Slicing\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-583","chapter","type-chapter","status-publish","hentry"],"part":65,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/583","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":22,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/583\/revisions"}],"predecessor-version":[{"id":4894,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/583\/revisions\/4894"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/65"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/583\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=583"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=583"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=583"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=583"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}