{"id":688,"date":"2025-10-09T15:30:38","date_gmt":"2025-10-09T15:30:38","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=688"},"modified":"2025-11-11T18:06:21","modified_gmt":"2025-11-11T18:06:21","slug":"applications-of-double-integrals-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/applications-of-double-integrals-fresh-take\/","title":{"raw":"Applications of Double Integrals: Fresh Take","rendered":"Applications of Double Integrals: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Use a double integral to calculate the area of a region, volume under a surface, or average value of a function over a plane region<\/li>\r\n \t<li>Use double integrals to calculate the volume of a region between two surfaces or the area of a plane region<\/li>\r\n \t<li>Solve problems involving improper double integrals<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Applications of Double Integrals<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Double integrals do more than just find volumes\u2014they're versatile tools for calculating areas, average values, and analyzing real-world data over regions. Understanding these applications helps you see why double integrals matter beyond abstract computation.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Finding Area<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">To find the area of a region [latex]R[\/latex], integrate the constant function [latex]f(x,y) = 1[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]A(R) = \\underset{R}{\\displaystyle\\iint}1\\,dA[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">This works because you're essentially \"summing up\" tiny area elements [latex]dA[\/latex] across the entire region. For a rectangle, this reduces to length times width.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Finding Volume<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">When [latex]f(x,y) \\geq 0[\/latex] over region [latex]R[\/latex], the double integral gives the volume of the solid under the surface [latex]z = f(x,y)[\/latex] and above [latex]R[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]V = \\underset{R}{\\displaystyle\\iint}f(x,y)\\,dA[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">The function value [latex]f(x,y)[\/latex] represents the height at each point, and [latex]dA[\/latex] is the base area element.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Average Value<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">The average value of a function over region [latex]R[\/latex] is the total integral divided by the region's area:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]f_{\\text{ave}} = \\dfrac{1}{\\text{Area }R}\\underset{R}{\\displaystyle\\iint}f(x,y)\\,dA[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">This extends the single-variable concept where you divided by interval length. Now you divide by area instead.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Find the volume of the solid bounded above by the graph of [latex]f(x,y) = xy\\sin(x^2y)[\/latex] and below by the [latex]xy[\/latex]-plane on the rectangular region [latex]R = [0,1] \\times [0, \\pi][\/latex].[reveal-answer q=\"225713860\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"225713860\"]\r\n<p style=\"text-align: center;\">[latex]\\LARGE{\\frac{\\pi}2}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8197097&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=OKpuV-ocq0A&amp;video_target=tpm-plugin-9m1n4yac-OKpuV-ocq0A\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP5.5_transcript.html\">transcript for \u201cCP 5.5\u201d here (opens in new window).<\/a><\/center><\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<p id=\"fs-id1167793221660\">A contour map is shown for a function [latex]f(x,y)[\/latex] on the rectangle [latex]{R} = {[-3,6]} {\\times} {[-1,4]}[\/latex].<\/p>\r\n\r\n\r\n[caption id=\"attachment_1315\" align=\"aligncenter\" width=\"379\"]<img class=\"size-full wp-image-1315\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23143822\/5-1-9.jpeg\" alt=\"A contour map is shown with the highest point being about 18 and centered near (4, negative 1). From this point, the values decrease to 16, 14, 12, 10, 8, and 6 roughly every 0.5 to 1 distance. The lowest point is negative four near (negative 3, 4). There is a local minimum of 2 near (negative 1, 0).\" width=\"379\" height=\"234\" \/> Figure 4.[\/caption]\r\n<ol id=\"fs-id1167793502441\" type=\"a\">\r\n \t<li>Use the midpoint rule with [latex]m = 3[\/latex] and [latex]n = 2[\/latex] to estimate the value of [latex]\\underset{R}{\\displaystyle\\iint}{f(x,y)dA}[\/latex].<\/li>\r\n \t<li>Estimate the average value of the function [latex]f(x,y)[\/latex].<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"387460142\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"387460142\"]\r\n\r\nAnswers to both parts a. and b. may vary.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<h2>Calculating Volumes, Areas, and Average Values<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Now that you can integrate over non-rectangular regions, you can tackle practical problems involving irregular shapes. The same setup works for different calculations\u2014just change what you're integrating.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Volume Under a Surface<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">For a solid bounded by coordinate planes and a tilted plane (like a tetrahedron), identify the region [latex]D[\/latex] where the base sits, then integrate the height function. A plane like [latex]2x + 3y + z = 6[\/latex] gives height [latex]z = 6 - 2x - 3y[\/latex]. The volume is then [latex]\\underset{D}{\\displaystyle\\iint}(6 - 2x - 3y)\\,dA[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Area of Curved Regions<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">For regions bounded by curves like [latex]y = x^2[\/latex] and [latex]y = 2x[\/latex], double integrals provide an alternative to single-variable methods. Set up [latex]\\underset{D}{\\displaystyle\\iint}1\\,dA[\/latex] using the curves as your limits. This approach shines when the region is easier to describe with two variables than with a single integral.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Computing Average Values<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Finding an average requires two steps: calculate the total integral of your function, then divide by the region's area. For [latex]f(x,y) = 7xy^2[\/latex] over a curved region, you need both [latex]\\underset{D}{\\displaystyle\\iint}7xy^2\\,dA[\/latex] and [latex]A(D) = \\underset{D}{\\displaystyle\\iint}1\\,dA[\/latex]. The average is their ratio.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Sketch the region<\/strong>: Identify all boundary curves and intersection points<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Choose region type<\/strong>: Decide whether Type I or Type II gives simpler limits<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Set up bounds<\/strong>: Determine constant limits (outer integral) and function limits (inner integral)<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Choose your integrand<\/strong>: Use [latex]f(x,y)[\/latex] for volume, 1 for area<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Integrate<\/strong>: Work inside out, treating the outer variable as constant in the inner integral<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>For average value<\/strong>: Compute both [latex]\\iint f,dA[\/latex] and [latex]\\iint 1,dA[\/latex], then divide<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>Strategic Choices<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">When a region can be described both ways, try the simpler one first. For a tetrahedron base that's a right triangle, you might find [latex]0 \\leq x \\leq 3[\/latex], [latex]0 \\leq y \\leq 2 - \\frac{2x}{3}[\/latex] easier than the reversed bounds. The algebra often guides your choice\u2014whichever avoids messy fractions or complicated limits usually works better.<\/p>\r\n<p class=\"whitespace-normal break-words\"><\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">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].[reveal-answer q=\"658122536\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"658122536\"][latex]\\frac{e^2}4+10e-\\frac{49}4[\/latex] cubic units.[\/hidden-answer]<\/section><section class=\"textbox example\" aria-label=\"Example\">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].[reveal-answer q=\"221145769\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"221145769\"][latex]\\frac{81}4[\/latex] square units.[\/hidden-answer]<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<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\" data-mce-fragment=\"1\"><\/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><\/section><section class=\"textbox example\" aria-label=\"Example\">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].[reveal-answer q=\"348629910\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"348629910\"][latex]\\frac34[\/latex][\/hidden-answer]<\/section>\r\n<h2>Improper Double Integrals<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Just as in single-variable calculus, double integrals can be \"improper\" when either the region is unbounded or the function has discontinuities. These integrals require limits to evaluate properly, but many converge to finite values.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Two Types of Improper Double Integrals<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">An improper double integral occurs when:<\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>The region [latex]D[\/latex] is unbounded<\/strong> (extends to infinity in some direction)<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>The function [latex]f[\/latex] is unbounded<\/strong> (has discontinuities or vertical asymptotes within [latex]D[\/latex])<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>Fubini's Theorem Still Applies<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">For nonnegative functions with finitely many discontinuities in the interior, you can still convert double integrals to iterated integrals. The key restriction: the function must be nonnegative over [latex]D[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Handling Unbounded Regions<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">When integrating over an unbounded region like the first quadrant [latex](x \\geq 0, y \\geq 0)[\/latex], use limits:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{R}{\\displaystyle\\iint}f(x,y)dA = \\displaystyle\\lim_{(b,d)\\to(\\infty,\\infty)}\\displaystyle\\int_a^b\\displaystyle\\int_c^d f(x,y)dy\\,dx[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Evaluate the integral with finite bounds first, then take the limit as those bounds approach infinity.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>When Functions Have Discontinuities<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">If [latex]f[\/latex] has a discontinuity at a point inside [latex]D[\/latex] (like [latex](0,0)[\/latex]), you may still be able to evaluate the integral by switching the order of integration to avoid the problematic point in a specific limit.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Probability Applications<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Joint density functions in probability often lead to improper integrals. For independent random variables [latex]X[\/latex] and [latex]Y[\/latex], the joint density is [latex]f(x,y) = f_1(x)f_2(y)[\/latex]. Probabilities are computed as:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]P((X,Y) \\in D) = \\underset{D}{\\displaystyle\\iint}f(x,y)dA[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Expected values require integrating over the entire sample space (often unbounded):<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]E(X) = \\underset{S}{\\displaystyle\\iint}xf(x,y)dA[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">These improper integrals often converge because exponential density functions decay rapidly.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Evaluate the improper integral [latex]\\underset{D}{\\displaystyle\\iint}{\\dfrac{y}{\\sqrt{1-x^2-y^2}}}{dA}[\/latex] where\u00a0[latex]{D} = {\\left \\{{(x,y)}{x} \\ {\\geq} \\ {0,y} \\ {\\geq} \\ {0,x^2+y^2} \\ {\\leq} \\ {1} \\right \\}}[\/latex].[reveal-answer q=\"634193357\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"634193357\"][latex]\\frac{\\pi}4[\/latex][\/hidden-answer]<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8197101&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=Dn3nDdB-HME&amp;video_target=tpm-plugin-0ht940zz-Dn3nDdB-HME\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP5.15_transcript.html\">transcript for \u201cCP 5.15\u201d here (opens in new window).<\/a><\/center><\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<p id=\"fs-id1167793499076\">The joint density function for two random variables [latex]X[\/latex] and [latex]Y[\/latex] is given by<\/p>\r\n<p style=\"text-align: left;\">[latex]\\hspace{6cm}f(x,y) = \\begin{align}\r\n&amp;\\frac1{16,250}(x^2+y^2) &amp;\\,&amp;\\text{ if } 0 \\leq x \\leq 15,0 \\leq y \\leq 10 \\\\\r\n&amp;0 &amp;\\,&amp;\\text{otherwise}\r\n\\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1167793462449\">Find the probability that [latex]X[\/latex] is at most 10 and [latex]Y[\/latex] is at least 5.<\/p>\r\n[reveal-answer q=\"889466210\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"889466210\"]\r\n\r\n[latex]\\frac{55}{72}\\approx0.7638[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Use a double integral to calculate the area of a region, volume under a surface, or average value of a function over a plane region<\/li>\n<li>Use double integrals to calculate the volume of a region between two surfaces or the area of a plane region<\/li>\n<li>Solve problems involving improper double integrals<\/li>\n<\/ul>\n<\/section>\n<h2>Applications of Double Integrals<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Double integrals do more than just find volumes\u2014they&#8217;re versatile tools for calculating areas, average values, and analyzing real-world data over regions. Understanding these applications helps you see why double integrals matter beyond abstract computation.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Finding Area<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">To find the area of a region [latex]R[\/latex], integrate the constant function [latex]f(x,y) = 1[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]A(R) = \\underset{R}{\\displaystyle\\iint}1\\,dA[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">This works because you&#8217;re essentially &#8220;summing up&#8221; tiny area elements [latex]dA[\/latex] across the entire region. For a rectangle, this reduces to length times width.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Finding Volume<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">When [latex]f(x,y) \\geq 0[\/latex] over region [latex]R[\/latex], the double integral gives the volume of the solid under the surface [latex]z = f(x,y)[\/latex] and above [latex]R[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]V = \\underset{R}{\\displaystyle\\iint}f(x,y)\\,dA[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">The function value [latex]f(x,y)[\/latex] represents the height at each point, and [latex]dA[\/latex] is the base area element.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Average Value<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The average value of a function over region [latex]R[\/latex] is the total integral divided by the region&#8217;s area:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]f_{\\text{ave}} = \\dfrac{1}{\\text{Area }R}\\underset{R}{\\displaystyle\\iint}f(x,y)\\,dA[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">This extends the single-variable concept where you divided by interval length. Now you divide by area instead.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Find the volume of the solid bounded above by the graph of [latex]f(x,y) = xy\\sin(x^2y)[\/latex] and below by the [latex]xy[\/latex]-plane on the rectangular region [latex]R = [0,1] \\times [0, \\pi][\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q225713860\">Show Solution<\/button><\/p>\n<div id=\"q225713860\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\LARGE{\\frac{\\pi}2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8197097&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=OKpuV-ocq0A&amp;video_target=tpm-plugin-9m1n4yac-OKpuV-ocq0A\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/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\/CP5.5_transcript.html\">transcript for \u201cCP 5.5\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p id=\"fs-id1167793221660\">A contour map is shown for a function [latex]f(x,y)[\/latex] on the rectangle [latex]{R} = {[-3,6]} {\\times} {[-1,4]}[\/latex].<\/p>\n<figure id=\"attachment_1315\" aria-describedby=\"caption-attachment-1315\" style=\"width: 379px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1315\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23143822\/5-1-9.jpeg\" alt=\"A contour map is shown with the highest point being about 18 and centered near (4, negative 1). From this point, the values decrease to 16, 14, 12, 10, 8, and 6 roughly every 0.5 to 1 distance. The lowest point is negative four near (negative 3, 4). There is a local minimum of 2 near (negative 1, 0).\" width=\"379\" height=\"234\" \/><figcaption id=\"caption-attachment-1315\" class=\"wp-caption-text\">Figure 4.<\/figcaption><\/figure>\n<ol id=\"fs-id1167793502441\" type=\"a\">\n<li>Use the midpoint rule with [latex]m = 3[\/latex] and [latex]n = 2[\/latex] to estimate the value of [latex]\\underset{R}{\\displaystyle\\iint}{f(x,y)dA}[\/latex].<\/li>\n<li>Estimate the average value of the function [latex]f(x,y)[\/latex].<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q387460142\">Show Solution<\/button><\/p>\n<div id=\"q387460142\" class=\"hidden-answer\" style=\"display: none\">\n<p>Answers to both parts a. and b. may vary.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<h2>Calculating Volumes, Areas, and Average Values<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Now that you can integrate over non-rectangular regions, you can tackle practical problems involving irregular shapes. The same setup works for different calculations\u2014just change what you&#8217;re integrating.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Volume Under a Surface<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">For a solid bounded by coordinate planes and a tilted plane (like a tetrahedron), identify the region [latex]D[\/latex] where the base sits, then integrate the height function. A plane like [latex]2x + 3y + z = 6[\/latex] gives height [latex]z = 6 - 2x - 3y[\/latex]. The volume is then [latex]\\underset{D}{\\displaystyle\\iint}(6 - 2x - 3y)\\,dA[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Area of Curved Regions<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">For regions bounded by curves like [latex]y = x^2[\/latex] and [latex]y = 2x[\/latex], double integrals provide an alternative to single-variable methods. Set up [latex]\\underset{D}{\\displaystyle\\iint}1\\,dA[\/latex] using the curves as your limits. This approach shines when the region is easier to describe with two variables than with a single integral.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Computing Average Values<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Finding an average requires two steps: calculate the total integral of your function, then divide by the region&#8217;s area. For [latex]f(x,y) = 7xy^2[\/latex] over a curved region, you need both [latex]\\underset{D}{\\displaystyle\\iint}7xy^2\\,dA[\/latex] and [latex]A(D) = \\underset{D}{\\displaystyle\\iint}1\\,dA[\/latex]. The average is their ratio.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Sketch the region<\/strong>: Identify all boundary curves and intersection points<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Choose region type<\/strong>: Decide whether Type I or Type II gives simpler limits<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Set up bounds<\/strong>: Determine constant limits (outer integral) and function limits (inner integral)<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Choose your integrand<\/strong>: Use [latex]f(x,y)[\/latex] for volume, 1 for area<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Integrate<\/strong>: Work inside out, treating the outer variable as constant in the inner integral<\/li>\n<li class=\"whitespace-normal break-words\"><strong>For average value<\/strong>: Compute both [latex]\\iint f,dA[\/latex] and [latex]\\iint 1,dA[\/latex], then divide<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>Strategic Choices<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">When a region can be described both ways, try the simpler one first. For a tetrahedron base that&#8217;s a right triangle, you might find [latex]0 \\leq x \\leq 3[\/latex], [latex]0 \\leq y \\leq 2 - \\frac{2x}{3}[\/latex] easier than the reversed bounds. The algebra often guides your choice\u2014whichever avoids messy fractions or complicated limits usually works better.<\/p>\n<p class=\"whitespace-normal break-words\">\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q658122536\">Show Solution<\/button><\/p>\n<div id=\"q658122536\" class=\"hidden-answer\" style=\"display: none\">[latex]\\frac{e^2}4+10e-\\frac{49}4[\/latex] cubic units.<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q221145769\">Show Solution<\/button><\/p>\n<div id=\"q221145769\" class=\"hidden-answer\" style=\"display: none\">[latex]\\frac{81}4[\/latex] square units.<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/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\" data-mce-fragment=\"1\"><\/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<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q348629910\">Show Solution<\/button><\/p>\n<div id=\"q348629910\" class=\"hidden-answer\" style=\"display: none\">[latex]\\frac34[\/latex]<\/div>\n<\/div>\n<\/section>\n<h2>Improper Double Integrals<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Just as in single-variable calculus, double integrals can be &#8220;improper&#8221; when either the region is unbounded or the function has discontinuities. These integrals require limits to evaluate properly, but many converge to finite values.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Two Types of Improper Double Integrals<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">An improper double integral occurs when:<\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>The region [latex]D[\/latex] is unbounded<\/strong> (extends to infinity in some direction)<\/li>\n<li class=\"whitespace-normal break-words\"><strong>The function [latex]f[\/latex] is unbounded<\/strong> (has discontinuities or vertical asymptotes within [latex]D[\/latex])<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>Fubini&#8217;s Theorem Still Applies<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">For nonnegative functions with finitely many discontinuities in the interior, you can still convert double integrals to iterated integrals. The key restriction: the function must be nonnegative over [latex]D[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Handling Unbounded Regions<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">When integrating over an unbounded region like the first quadrant [latex](x \\geq 0, y \\geq 0)[\/latex], use limits:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{R}{\\displaystyle\\iint}f(x,y)dA = \\displaystyle\\lim_{(b,d)\\to(\\infty,\\infty)}\\displaystyle\\int_a^b\\displaystyle\\int_c^d f(x,y)dy\\,dx[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Evaluate the integral with finite bounds first, then take the limit as those bounds approach infinity.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>When Functions Have Discontinuities<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">If [latex]f[\/latex] has a discontinuity at a point inside [latex]D[\/latex] (like [latex](0,0)[\/latex]), you may still be able to evaluate the integral by switching the order of integration to avoid the problematic point in a specific limit.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Probability Applications<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Joint density functions in probability often lead to improper integrals. For independent random variables [latex]X[\/latex] and [latex]Y[\/latex], the joint density is [latex]f(x,y) = f_1(x)f_2(y)[\/latex]. Probabilities are computed as:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]P((X,Y) \\in D) = \\underset{D}{\\displaystyle\\iint}f(x,y)dA[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Expected values require integrating over the entire sample space (often unbounded):<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]E(X) = \\underset{S}{\\displaystyle\\iint}xf(x,y)dA[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">These improper integrals often converge because exponential density functions decay rapidly.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Evaluate the improper integral [latex]\\underset{D}{\\displaystyle\\iint}{\\dfrac{y}{\\sqrt{1-x^2-y^2}}}{dA}[\/latex] where\u00a0[latex]{D} = {\\left \\{{(x,y)}{x} \\ {\\geq} \\ {0,y} \\ {\\geq} \\ {0,x^2+y^2} \\ {\\leq} \\ {1} \\right \\}}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q634193357\">Show Solution<\/button><\/p>\n<div id=\"q634193357\" class=\"hidden-answer\" style=\"display: none\">[latex]\\frac{\\pi}4[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8197101&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=Dn3nDdB-HME&amp;video_target=tpm-plugin-0ht940zz-Dn3nDdB-HME\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/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\/CP5.15_transcript.html\">transcript for \u201cCP 5.15\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p id=\"fs-id1167793499076\">The joint density function for two random variables [latex]X[\/latex] and [latex]Y[\/latex] is given by<\/p>\n<p style=\"text-align: left;\">[latex]\\hspace{6cm}f(x,y) = \\begin{align}  &\\frac1{16,250}(x^2+y^2) &\\,&\\text{ if } 0 \\leq x \\leq 15,0 \\leq y \\leq 10 \\\\  &0 &\\,&\\text{otherwise}  \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1167793462449\">Find the probability that [latex]X[\/latex] is at most 10 and [latex]Y[\/latex] is at least 5.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q889466210\">Show Solution<\/button><\/p>\n<div id=\"q889466210\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{55}{72}\\approx0.7638[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":18,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-688","chapter","type-chapter","status-publish","hentry"],"part":432,"module-header":"fresh_take","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/688","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":12,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/688\/revisions"}],"predecessor-version":[{"id":2177,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/688\/revisions\/2177"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/432"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/688\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=688"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=688"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=688"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=688"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}