{"id":3977,"date":"2022-04-12T17:17:33","date_gmt":"2022-04-12T17:17:33","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=3977"},"modified":"2022-11-01T04:18:10","modified_gmt":"2022-11-01T04:18:10","slug":"calculating-volumes-areas-and-average-values","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/calculating-volumes-areas-and-average-values\/","title":{"raw":"Calculating Volumes, Areas, and Average Values","rendered":"Calculating Volumes, Areas, and Average Values"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul class=\"os-abstract\">\r\n \t<li><span class=\"os-abstract-content\">Use double integrals to calculate the volume of a region between two surfaces or the area of a plane region.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1167793948129\">We can use double integrals over general regions to compute volumes, areas, and average values. The methods are the same as those in\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/double-integrals-over-rectangular-regions\/\">Double Integrals over Rectangular Regions<\/a>, but without the restriction to a rectangular region, we can now solve a wider variety of problems.<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding the volume of a tetrahedron<\/h3>\r\nFind the volume of the solid bounded by the planes\u00a0[latex]x = 0[\/latex],\u00a0[latex]y = 0, \\ z = 0[\/latex], and\u00a0[latex]2x+3y+z=6[\/latex].\r\n\r\n[reveal-answer q=\"541521446\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"541521446\"]\r\n\r\nThe solid is a tetrahedron with the base on the [latex]xy[\/latex]-plane and a height [latex]z=6-2x-3y[\/latex] The base is the region [latex]D[\/latex] bounded by the lines, [latex]x=0[\/latex],\u00a0[latex]y=0[\/latex], and\u00a0[latex]2x+3y=6[\/latex] where [latex]z=0[\/latex] (Figure 1). Note that we can consider the region [latex]D[\/latex] as Type I or as Type II, and we can integrate in both ways.\r\n\r\n[caption id=\"attachment_1349\" align=\"aligncenter\" width=\"845\"]<img class=\"size-full wp-image-1349\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25060633\/5-2-12.jpeg\" alt=\"This figure shows a tetrahedron bounded by x = 0, y = 0, z = 0, and 2x + 3y = 6 (or z = 6 minus 2x minus 3y).\" width=\"845\" height=\"461\" \/> Figure 1.\u00a0A tetrahedron consisting of the three coordinate planes and the plane\u00a0[latex]\\small{z=6-2x-3y}[\/latex], with the base bound by\u00a0[latex]\\small{x=0}[\/latex],\u00a0[latex]\\small{y=0}[\/latex], and\u00a0[latex]\\small{2x+3y=6}[\/latex].[\/caption]\r\n<p id=\"fs-id1167794177172\">First, consider [latex]D[\/latex] as a Type I region, and hence [latex]{D} = {\\left \\{{(x,y)}{\\mid}{0} \\ {\\leq} \\ {x} \\ {\\leq} \\ {3,0} \\ {\\leq} \\ {y} \\ {\\leq} \\ {2-{\\frac{2}{3}}x} \\right \\}}[\/latex].<\/p>\r\n<p id=\"fs-id1167793589669\">Therefore, the volume is\r\n[latex]\\hspace{3cm}\\begin{align}\r\nV&amp;=\\displaystyle\\int_{x=0}^{x=3}\\displaystyle\\int_{y=0}^{y=2-(2x\/3)}(6-2x-3y)dydx=\\displaystyle\\int_{x=0}^{x=3}\\left[\\left(6y-2xy-\\frac32y^2\\right)\\Bigg|_{y=0}^{y=2-(2x\/3)}\\right]dx \\\\\r\n&amp;=\\displaystyle\\int_{x=0}^{x=3}\\left[\\frac23(x-3)^2\\right]dx=6\r\n\\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1167793423571\">Now consider [latex]D[\/latex] as a Type II region, so [latex]{D} = {\\left \\{{(x,y)}{\\mid}{0} \\ {\\leq} \\ {y} \\ {\\leq} \\ {2,0} \\ {\\leq} \\ {x} \\ {\\leq} \\ {3-{\\frac{3}{2}}y} \\right \\}}[\/latex]. In this calculation, the volume is\r\n[latex]\\hspace{3cm}\\begin{align}\r\nV&amp;=\\displaystyle\\int_{y=0}^{y=2}\\displaystyle\\int_{x=0}^{x=3-(3y\/2)}(6-2x-3y)dxdy=\\displaystyle\\int_{y=0}^{y=2}\\left[(6x-x^2-3xy)\\bigg|_{x=0}^{x=3-(3y\/2)}\\right]dy \\\\\r\n&amp;=\\displaystyle\\int_{y=0}^{y=2}\\left[\\frac94(y-2)^2\\right]dy=6\r\n\\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1167793570155\">Therefore, the volume is [latex]6[\/latex] cubic units.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nFind the volume of the solid bounded above by [latex]f(x, y)=10-2x+y[\/latex] over the region enclosed by the curves [latex]y=0[\/latex] and [latex]{y} = {e^x}[\/latex], where [latex]x[\/latex] is in the interval [latex][0,1][\/latex].\r\n\r\n[reveal-answer q=\"658122536\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"658122536\"]\r\n\r\n[latex]\\frac{e^2}4+10e-\\frac{49}4[\/latex] cubic units.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nFinding the area of a rectangular region is easy, but finding the area of a nonrectangular region is not so easy. As we have seen, we can use double integrals to find a rectangular area. As a matter of fact, this comes in very handy for finding the area of a general nonrectangular region, as stated in the next definition.\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">definition<\/h3>\r\n\r\n<hr \/>\r\n\r\nThe area of a plane-bounded region [latex]D[\/latex] is defined as the double integral\u00a0[latex]\\underset{D}{\\displaystyle\\iint}{1dA}[\/latex].\r\n\r\n<\/div>\r\nWe have already seen how to find areas in terms of single integration. Here we are seeing another way of finding areas by using double integrals, which can be very useful, as we will see in the later sections of this chapter.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding the area of a region<\/h3>\r\nFind the area of the region bounded below by the curve [latex]y=x^{2}[\/latex] and above by the line [latex]y=2x[\/latex] in the first quadrant (Figure 2).\r\n\r\n[caption id=\"attachment_1351\" align=\"aligncenter\" width=\"231\"]<img class=\"size-full wp-image-1351\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25060818\/5-2-13.jpeg\" alt=\"The line y = 2 x (also marked x = y\/2) is shown, as is y = x squared (also marked x = the square root of y). There are vertical and horizontal shadings giving for small stretch of this region, denoting that it can be treated as a Type I or Type II area.\" width=\"231\" height=\"315\" \/> Figure 2. The region bounded by\u00a0[latex]y=x^{2}[\/latex] and\u00a0[latex]y=2x[\/latex].[\/caption][reveal-answer q=\"501887643\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"501887643\"]\r\n<p id=\"fs-id1167793500470\">We just have to integrate the constant function [latex]f(x, y)=1[\/latex] over the region. Thus, the area [latex]A[\/latex] of the bounded region is\u00a0[latex]\\displaystyle\\int_{x=0}^{x=2}\\displaystyle\\int_{y=x^2}^{y=2x}dydx[\/latex] or\u00a0[latex]\\displaystyle\\int_{y=0}^{x=4}\\displaystyle\\int_{x=y\/2}^{x=\\sqrt{y}}dxdy[\/latex]:<\/p>\r\n<p style=\"text-align: center;\">[latex]A=\\underset{D}{\\displaystyle\\iint}1dxdy=\\displaystyle\\int_{x=0}^{x=2}\\displaystyle\\int_{y=x^2}^{y=2x}1dydx=\\displaystyle\\int_{x=0}^{x=2}\\left[y\\bigg|_{y=x^2}^{y=2x}\\right]dx=\\displaystyle\\int_{x=0}^{x=2}(2x-x^2)dx=x^2-\\frac{x^3}3\\bigg|_0^2=\\frac43.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nFind the area of a region bounded above by the curve [latex]y=x^{3}[\/latex] and below by [latex]y=0[\/latex] over the interval [latex][0,3][\/latex].\r\n\r\n[reveal-answer q=\"221145769\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"221145769\"]\r\n\r\n[latex]\\frac{81}4[\/latex] square units.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8197100&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=xCdMCWy1uac&amp;video_target=tpm-plugin-dgddicdg-xCdMCWy1uac\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP5.13_transcript.html\">transcript for \u201cCP 5.13\u201d here (opens in new window).<\/a><\/center>We can also use a double integral to find the average value of a function over a general region.\u00a0 First, recall how we find the average value of a function using single-variable calculus.\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Average Value of a Function (Single-variable version)<\/h3>\r\nIf [latex] f(x) [\/latex] is continuous on [latex] [a,b] [\/latex], then the average value of [latex] f(x) [\/latex] on [latex] [a,b] [\/latex] is\r\n\r\n[latex] f_{ave}=\\frac{1}{b-a} \\int_a^b f(x) dx [\/latex]\r\n\r\n<\/div>\r\nThe following definition is a direct extension of the formula above.\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">definition<\/h3>\r\n\r\n<hr \/>\r\n\r\nIf [latex]f(x, y)[\/latex] is integrable over a plane-bounded region [latex]D[\/latex] with positive area [latex]A(D)[\/latex] then the average value of the function is\r\n<p style=\"text-align: center;\">[latex]\\large{{f_{ave}} = {\\frac{1}{A(D)}}\\underset{D}{\\displaystyle\\iint}{f(x,y)}{dA}}[\/latex]<\/p>\r\nNote that the area is\u00a0[latex]{A(D)}=\\underset{D}{\\displaystyle\\iint}{1dA}[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding an average value<\/h3>\r\nFind the average value of the function [latex]f(x, y)=7xy^{2}[\/latex] on the region bounded by the line [latex]x=y[\/latex] and the curve [latex]x=\\sqrt{y}[\/latex] (Figure 3).\r\n\r\n[caption id=\"attachment_1353\" align=\"aligncenter\" width=\"380\"]<img class=\"size-full wp-image-1353\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25061017\/5-2-14.jpeg\" alt=\"The lines x = y and x = the square root of y bound a shaded region. There are horizontal dashed lines marked throughout the region.\" width=\"380\" height=\"385\" \/> Figure 3. The region bounded by\u00a0[latex]\\small{x=y}[\/latex] and [latex]\\small{x=\\sqrt{y}}[\/latex][\/caption][reveal-answer q=\"471992476\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"471992476\"]\r\n<p id=\"fs-id1167794207001\">First find the area [latex]A(D)[\/latex] where the region [latex]D[\/latex] is given by the figure. We have<\/p>\r\n[latex]A(d)=\\underset{D}{\\displaystyle\\iint}1dA=\\displaystyle\\int_{y=0}^{y=1}\\displaystyle\\int_{x=y}^{x=\\sqrt{y}}1dxdy \\displaystyle\\int_{y=0}^{y=1}\\left[x\\bigg|_{x=y}^{x=\\sqrt{y}}\\right]dy=\\frac23y^{3\/2}-\\frac{y^2}2\\Bigg|_0^1=\\frac16.[\/latex]\r\n<p id=\"fs-id1167793887440\">Then the average value of the given function over this region is<\/p>\r\n[latex]\\begin{align}\r\n\r\nf_{ave} &amp;= \\frac1{A(D)}\\underset{D}{\\displaystyle\\iint}f(x,y)dA = \\frac1{A(D)}\\displaystyle\\int_{y=0}^{y=1}\\displaystyle\\int_{x=y}^{x=\\sqrt{y}}yxy^2dxdy=\\frac1{1\/6}\\displaystyle\\int_{y=0}^{y=1}\\left[\\frac72x^2y^s\\Bigg|_{x=y}^{x=\\sqrt{y}}\\right]dy \\\\\r\n\r\n&amp;=6\\displaystyle\\int_{y=0}^{y=1}\\left[\\frac72y^2(y-y^2)\\right]dy=6\\displaystyle\\int_{y=0}^{y=1}\\left[\\frac72(y^3-y^4)\\right]dy=\\frac{42}2(\\frac{y^4}4-\\frac{y^5}5)\\bigg|_0^1=\\frac{42}{40}=\\frac{21}{20}.\r\n\r\n\\end{align}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nFind the average value of the function [latex]f(x, y)=xy[\/latex] over the triangle with vertices\u00a0[latex](0, 0)[\/latex],\u00a0[latex](1, 0)[\/latex], and\u00a0[latex](1, 3)[\/latex].\r\n\r\n[reveal-answer q=\"348629910\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"348629910\"]\r\n\r\n[latex]\\frac34[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul class=\"os-abstract\">\n<li><span class=\"os-abstract-content\">Use double integrals to calculate the volume of a region between two surfaces or the area of a plane region.<\/span><\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1167793948129\">We can use double integrals over general regions to compute volumes, areas, and average values. The methods are the same as those in\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/double-integrals-over-rectangular-regions\/\">Double Integrals over Rectangular Regions<\/a>, but without the restriction to a rectangular region, we can now solve a wider variety of problems.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: finding the volume of a tetrahedron<\/h3>\n<p>Find the volume of the solid bounded by the planes\u00a0[latex]x = 0[\/latex],\u00a0[latex]y = 0, \\ z = 0[\/latex], and\u00a0[latex]2x+3y+z=6[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q541521446\">Show Solution<\/span><\/p>\n<div id=\"q541521446\" class=\"hidden-answer\" style=\"display: none\">\n<p>The solid is a tetrahedron with the base on the [latex]xy[\/latex]-plane and a height [latex]z=6-2x-3y[\/latex] The base is the region [latex]D[\/latex] bounded by the lines, [latex]x=0[\/latex],\u00a0[latex]y=0[\/latex], and\u00a0[latex]2x+3y=6[\/latex] where [latex]z=0[\/latex] (Figure 1). Note that we can consider the region [latex]D[\/latex] as Type I or as Type II, and we can integrate in both ways.<\/p>\n<div id=\"attachment_1349\" style=\"width: 855px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1349\" class=\"size-full wp-image-1349\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25060633\/5-2-12.jpeg\" alt=\"This figure shows a tetrahedron bounded by x = 0, y = 0, z = 0, and 2x + 3y = 6 (or z = 6 minus 2x minus 3y).\" width=\"845\" height=\"461\" \/><\/p>\n<p id=\"caption-attachment-1349\" class=\"wp-caption-text\">Figure 1.\u00a0A tetrahedron consisting of the three coordinate planes and the plane\u00a0[latex]\\small{z=6-2x-3y}[\/latex], with the base bound by\u00a0[latex]\\small{x=0}[\/latex],\u00a0[latex]\\small{y=0}[\/latex], and\u00a0[latex]\\small{2x+3y=6}[\/latex].<\/p>\n<\/div>\n<p id=\"fs-id1167794177172\">First, consider [latex]D[\/latex] as a Type I region, and hence [latex]{D} = {\\left \\{{(x,y)}{\\mid}{0} \\ {\\leq} \\ {x} \\ {\\leq} \\ {3,0} \\ {\\leq} \\ {y} \\ {\\leq} \\ {2-{\\frac{2}{3}}x} \\right \\}}[\/latex].<\/p>\n<p id=\"fs-id1167793589669\">Therefore, the volume is<br \/>\n[latex]\\hspace{3cm}\\begin{align}  V&=\\displaystyle\\int_{x=0}^{x=3}\\displaystyle\\int_{y=0}^{y=2-(2x\/3)}(6-2x-3y)dydx=\\displaystyle\\int_{x=0}^{x=3}\\left[\\left(6y-2xy-\\frac32y^2\\right)\\Bigg|_{y=0}^{y=2-(2x\/3)}\\right]dx \\\\  &=\\displaystyle\\int_{x=0}^{x=3}\\left[\\frac23(x-3)^2\\right]dx=6  \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1167793423571\">Now consider [latex]D[\/latex] as a Type II region, so [latex]{D} = {\\left \\{{(x,y)}{\\mid}{0} \\ {\\leq} \\ {y} \\ {\\leq} \\ {2,0} \\ {\\leq} \\ {x} \\ {\\leq} \\ {3-{\\frac{3}{2}}y} \\right \\}}[\/latex]. In this calculation, the volume is<br \/>\n[latex]\\hspace{3cm}\\begin{align}  V&=\\displaystyle\\int_{y=0}^{y=2}\\displaystyle\\int_{x=0}^{x=3-(3y\/2)}(6-2x-3y)dxdy=\\displaystyle\\int_{y=0}^{y=2}\\left[(6x-x^2-3xy)\\bigg|_{x=0}^{x=3-(3y\/2)}\\right]dy \\\\  &=\\displaystyle\\int_{y=0}^{y=2}\\left[\\frac94(y-2)^2\\right]dy=6  \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1167793570155\">Therefore, the volume is [latex]6[\/latex] cubic units.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Find the volume of the solid bounded above by [latex]f(x, y)=10-2x+y[\/latex] over the region enclosed by the curves [latex]y=0[\/latex] and [latex]{y} = {e^x}[\/latex], where [latex]x[\/latex] is in the interval [latex][0,1][\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q658122536\">Show Solution<\/span><\/p>\n<div id=\"q658122536\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{e^2}4+10e-\\frac{49}4[\/latex] cubic units.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Finding the area of a rectangular region is easy, but finding the area of a nonrectangular region is not so easy. As we have seen, we can use double integrals to find a rectangular area. As a matter of fact, this comes in very handy for finding the area of a general nonrectangular region, as stated in the next definition.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">definition<\/h3>\n<hr \/>\n<p>The area of a plane-bounded region [latex]D[\/latex] is defined as the double integral\u00a0[latex]\\underset{D}{\\displaystyle\\iint}{1dA}[\/latex].<\/p>\n<\/div>\n<p>We have already seen how to find areas in terms of single integration. Here we are seeing another way of finding areas by using double integrals, which can be very useful, as we will see in the later sections of this chapter.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: finding the area of a region<\/h3>\n<p>Find the area of the region bounded below by the curve [latex]y=x^{2}[\/latex] and above by the line [latex]y=2x[\/latex] in the first quadrant (Figure 2).<\/p>\n<div id=\"attachment_1351\" style=\"width: 241px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1351\" class=\"size-full wp-image-1351\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25060818\/5-2-13.jpeg\" alt=\"The line y = 2 x (also marked x = y\/2) is shown, as is y = x squared (also marked x = the square root of y). There are vertical and horizontal shadings giving for small stretch of this region, denoting that it can be treated as a Type I or Type II area.\" width=\"231\" height=\"315\" \/><\/p>\n<p id=\"caption-attachment-1351\" class=\"wp-caption-text\">Figure 2. The region bounded by\u00a0[latex]y=x^{2}[\/latex] and\u00a0[latex]y=2x[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q501887643\">Show Solution<\/span><\/p>\n<div id=\"q501887643\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793500470\">We just have to integrate the constant function [latex]f(x, y)=1[\/latex] over the region. Thus, the area [latex]A[\/latex] of the bounded region is\u00a0[latex]\\displaystyle\\int_{x=0}^{x=2}\\displaystyle\\int_{y=x^2}^{y=2x}dydx[\/latex] or\u00a0[latex]\\displaystyle\\int_{y=0}^{x=4}\\displaystyle\\int_{x=y\/2}^{x=\\sqrt{y}}dxdy[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]A=\\underset{D}{\\displaystyle\\iint}1dxdy=\\displaystyle\\int_{x=0}^{x=2}\\displaystyle\\int_{y=x^2}^{y=2x}1dydx=\\displaystyle\\int_{x=0}^{x=2}\\left[y\\bigg|_{y=x^2}^{y=2x}\\right]dx=\\displaystyle\\int_{x=0}^{x=2}(2x-x^2)dx=x^2-\\frac{x^3}3\\bigg|_0^2=\\frac43.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Find the area of a region bounded above by the curve [latex]y=x^{3}[\/latex] and below by [latex]y=0[\/latex] over the interval [latex][0,3][\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q221145769\">Show Solution<\/span><\/p>\n<div id=\"q221145769\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{81}4[\/latex] square units.<\/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\" src=\"\/\/plugin.3playmedia.com\/show?mf=8197100&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=xCdMCWy1uac&amp;video_target=tpm-plugin-dgddicdg-xCdMCWy1uac\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP5.13_transcript.html\">transcript for \u201cCP 5.13\u201d here (opens in new window).<\/a><\/div>\n<p>We can also use a double integral to find the average value of a function over a general region.\u00a0 First, recall how we find the average value of a function using single-variable calculus.<\/p>\n<div class=\"textbox examples\">\n<h3>Recall: Average Value of a Function (Single-variable version)<\/h3>\n<p>If [latex]f(x)[\/latex] is continuous on [latex][a,b][\/latex], then the average value of [latex]f(x)[\/latex] on [latex][a,b][\/latex] is<\/p>\n<p>[latex]f_{ave}=\\frac{1}{b-a} \\int_a^b f(x) dx[\/latex]<\/p>\n<\/div>\n<p>The following definition is a direct extension of the formula above.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">definition<\/h3>\n<hr \/>\n<p>If [latex]f(x, y)[\/latex] is integrable over a plane-bounded region [latex]D[\/latex] with positive area [latex]A(D)[\/latex] then the average value of the function is<\/p>\n<p style=\"text-align: center;\">[latex]\\large{{f_{ave}} = {\\frac{1}{A(D)}}\\underset{D}{\\displaystyle\\iint}{f(x,y)}{dA}}[\/latex]<\/p>\n<p>Note that the area is\u00a0[latex]{A(D)}=\\underset{D}{\\displaystyle\\iint}{1dA}[\/latex].<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: finding an average value<\/h3>\n<p>Find the average value of the function [latex]f(x, y)=7xy^{2}[\/latex] on the region bounded by the line [latex]x=y[\/latex] and the curve [latex]x=\\sqrt{y}[\/latex] (Figure 3).<\/p>\n<div id=\"attachment_1353\" style=\"width: 390px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1353\" class=\"size-full wp-image-1353\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25061017\/5-2-14.jpeg\" alt=\"The lines x = y and x = the square root of y bound a shaded region. There are horizontal dashed lines marked throughout the region.\" width=\"380\" height=\"385\" \/><\/p>\n<p id=\"caption-attachment-1353\" class=\"wp-caption-text\">Figure 3. The region bounded by\u00a0[latex]\\small{x=y}[\/latex] and [latex]\\small{x=\\sqrt{y}}[\/latex]<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q471992476\">Show Solution<\/span><\/p>\n<div id=\"q471992476\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794207001\">First find the area [latex]A(D)[\/latex] where the region [latex]D[\/latex] is given by the figure. We have<\/p>\n<p>[latex]A(d)=\\underset{D}{\\displaystyle\\iint}1dA=\\displaystyle\\int_{y=0}^{y=1}\\displaystyle\\int_{x=y}^{x=\\sqrt{y}}1dxdy \\displaystyle\\int_{y=0}^{y=1}\\left[x\\bigg|_{x=y}^{x=\\sqrt{y}}\\right]dy=\\frac23y^{3\/2}-\\frac{y^2}2\\Bigg|_0^1=\\frac16.[\/latex]<\/p>\n<p id=\"fs-id1167793887440\">Then the average value of the given function over this region is<\/p>\n<p>[latex]\\begin{align}    f_{ave} &= \\frac1{A(D)}\\underset{D}{\\displaystyle\\iint}f(x,y)dA = \\frac1{A(D)}\\displaystyle\\int_{y=0}^{y=1}\\displaystyle\\int_{x=y}^{x=\\sqrt{y}}yxy^2dxdy=\\frac1{1\/6}\\displaystyle\\int_{y=0}^{y=1}\\left[\\frac72x^2y^s\\Bigg|_{x=y}^{x=\\sqrt{y}}\\right]dy \\\\    &=6\\displaystyle\\int_{y=0}^{y=1}\\left[\\frac72y^2(y-y^2)\\right]dy=6\\displaystyle\\int_{y=0}^{y=1}\\left[\\frac72(y^3-y^4)\\right]dy=\\frac{42}2(\\frac{y^4}4-\\frac{y^5}5)\\bigg|_0^1=\\frac{42}{40}=\\frac{21}{20}.    \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Find the average value of the function [latex]f(x, y)=xy[\/latex] over the triangle with vertices\u00a0[latex](0, 0)[\/latex],\u00a0[latex](1, 0)[\/latex], and\u00a0[latex](1, 3)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q348629910\">Show Solution<\/span><\/p>\n<div id=\"q348629910\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac34[\/latex]<\/p>\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-3977\">\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>CP 5.13. <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 3. <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-3\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-3\/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-3\/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 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