{"id":678,"date":"2025-10-09T15:29:34","date_gmt":"2025-10-09T15:29:34","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=678"},"modified":"2025-11-11T17:58:37","modified_gmt":"2025-11-11T17:58:37","slug":"double-integrals-over-basic-regions-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/double-integrals-over-basic-regions-fresh-take\/","title":{"raw":"Double Integrals over Basic Regions: Fresh Take","rendered":"Double Integrals over Basic Regions: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Calculate double integrals when your region is a rectangle<\/li>\r\n \t<li>Find double integrals over more complex regions<\/li>\r\n \t<li>Switch the order of integration to make calculations easier<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2 data-type=\"title\">Volumes and 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\">Single integrals find areas under curves. Double integrals extend this idea to find volumes under surfaces. Instead of slicing a region into thin strips, you're dividing a rectangular base into small boxes and stacking them up to the surface.<\/p>\r\n<p class=\"whitespace-normal break-words\">For a function [latex]z = f(x, y)[\/latex] over a rectangular region [latex]R = [a, b] \\times [c, d][\/latex], the double integral is:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\large{\\underset{R}{\\displaystyle\\iint}f(x,y)dA=\\displaystyle\\lim_{m,n\\to\\infty}\\displaystyle\\sum_{i=1}^{m}\\sum_{j=1}^{n}f(x_i^*,y_j^*)\\Delta A}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">This represents the volume of the solid sitting above region [latex]R[\/latex] and below the surface [latex]z = f(x, y)[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">As you make the grid finer (more and smaller rectangles), the approximation improves. The exact volume is the limit as [latex]m[\/latex] and [latex]n[\/latex] approach infinity\u2014that limit is the double integral.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Key Notation<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Since [latex]dA = dx,dy = dy,dx[\/latex], you can write the double integral as:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{R}{\\displaystyle\\iint}f(x,y)dx,dy[\/latex] or [latex]\\underset{R}{\\displaystyle\\iint}f(x,y)dy,dx[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>When It Works<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">The double integral exists if [latex]f[\/latex] is continuous over [latex]R[\/latex] (or at least not too discontinuous). If [latex]f(x, y) \\geq 0[\/latex], the result is volume. If [latex]f[\/latex] is sometimes negative, think of it as signed volume\u2014parts below the [latex]xy[\/latex]-plane count as negative.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p id=\"fs-id1167794324430\">Use the same function [latex]z = f(x,y) = 3x^2 - y[\/latex] over the rectangular region [latex]R = [0,2] \\times [0,2][\/latex].<\/p>\r\n<p id=\"fs-id1167793473779\">Divide [latex]R[\/latex] into the same four squares with [latex]m = n = 2[\/latex] and choose the sample points as the upper left corner point of each square [latex](0,1), (1,1), (0,2),[\/latex] and [latex](1,2)[\/latex] (Figure 5) to approximate the signed volume of the solid [latex]S[\/latex] that lies above [latex]R[\/latex] and \u201cunder\u201d the graph of [latex]f[\/latex].<\/p>\r\n[reveal-answer q=\"265173902\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"265173902\"]\r\n<p style=\"text-align: center;\">[latex]\\large{V=\\displaystyle\\sum_{i=1}^2\\sum_{j=1}^2f(x_{ij}^*,y_{ij}^*)\\Delta A=0}[\/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=8197095&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=ZKNjm2OdGnU&amp;video_target=tpm-plugin-g6ue79lv-ZKNjm2OdGnU\" 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\/CP5.1_transcript.html\">transcript for \u201cCP 5.1\u201d here (opens in new window).<\/a><\/center><\/section>\r\n<h2 data-type=\"title\">Properties 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 behave much like single integrals\u2014they follow familiar algebraic rules that make calculations easier. Understanding these properties helps you break complex integrals into simpler pieces.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Linearity (Properties 1 and 2)<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Double integrals respect addition and scalar multiplication:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{R}{\\displaystyle\\iint}[f(x,y)+g(x,y)]dA=\\underset{R}{\\displaystyle\\iint}f(x,y)dA+\\underset{R}{\\displaystyle\\iint}g(x,y)dA[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{R}{\\displaystyle\\iint}cf(x,y)dA = c\\underset{R}{\\displaystyle\\iint}f(x,y)dA[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">You can split integrals of sums into separate integrals, and pull constants outside the integral. This works exactly like it did in single-variable calculus.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Additivity (Property 3)<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">If you split region [latex]R[\/latex] into non-overlapping subregions [latex]S[\/latex] and [latex]T[\/latex], the integral over [latex]R[\/latex] equals the sum of integrals over the pieces:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{R}{\\displaystyle\\iint}f(x,y)dA = \\underset{S}{\\displaystyle\\iint}f(x,y)dA + \\underset{T}{\\displaystyle\\iint}f(x,y)dA[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">This is useful when your region has a natural division or when part is easier to integrate than the whole.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Monotonicity (Property 4)<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">If [latex]f(x,y) \\geq g(x,y)[\/latex] everywhere on [latex]R[\/latex], then their integrals follow the same inequality. Bigger functions produce bigger integrals.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Bounding the Integral (Property 5)<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">If you know [latex]m \\leq f(x,y) \\leq M[\/latex] on region [latex]R[\/latex], then:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]m \\times A(R) \\leq \\underset{R}{\\displaystyle\\iint}f(x,y)dA \\leq M \\times A(R)[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">This gives quick estimates: multiply the region's area by the minimum and maximum function values.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Separable Functions (Property 6)<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">If [latex]f(x,y) = g(x)h(y)[\/latex] (a product of separate [latex]x[\/latex] and [latex]y[\/latex] functions), the double integral factors:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{R}{\\displaystyle\\iint}g(x)h(y)dA = \\left(\\displaystyle\\int_a^bg(x)dx\\right)\\left(\\displaystyle\\int_c^dh(y)dy\\right)[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">This turns a double integral into two single integrals\u2014often much easier to compute.<\/p>\r\n\r\n<\/div>\r\n<h2>Iterated 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\">Computing double integrals directly from the limit definition would be painfully tedious. Fortunately, Fubini's theorem gives us a practical method: break a double integral into two single integrals, integrating one variable at a time.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Key Idea: Iterated Integrals<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">An iterated integral lets you evaluate a double integral by integrating twice\u2014first with respect to one variable (treating the other as constant), then with respect to the second variable:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\displaystyle\\int_a^b\\int_c^df(x,y)dy,dx = \\displaystyle\\int_a^b\\left[\\displaystyle\\int_c^df(x,y)dy\\right]dx[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">The inner integral [latex]\\int_c^d f(x,y)dy[\/latex] treats [latex]x[\/latex] as a constant and integrates with respect to [latex]y[\/latex]. The result is a function of [latex]x[\/latex] alone, which you then integrate.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Fubini's Theorem<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">For a continuous function [latex]f(x,y)[\/latex] over rectangle [latex]R = [a,b] \\times [c,d][\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{R}{\\displaystyle\\iint}f(x,y)dA = \\displaystyle\\int_a^b\\displaystyle\\int_c^df(x,y)dy,dx = \\displaystyle\\int_c^d\\displaystyle\\int_a^bf(x,y)dx,dy[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Both orders of integration give the same answer. Choose whichever order makes the computation easier.<\/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>Identify your region<\/strong>: Note the bounds [latex]a \\leq x \\leq b[\/latex] and [latex]c \\leq y \\leq d[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Choose your order<\/strong>: Decide whether to integrate with respect to [latex]x[\/latex] or [latex]y[\/latex] first<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Integrate the inner integral<\/strong>: Treat one variable as constant and integrate with respect to the other<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Integrate the outer integral<\/strong>: Integrate the resulting expression with respect to the remaining variable<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>When Order Matters<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Sometimes one order is much easier than the other. If the inner integral looks complicated (requiring integration by parts or substitution that depends on both variables), try switching the order. A function like [latex]x\\sin(xy)[\/latex] integrates more easily with respect to [latex]y[\/latex] first than [latex]x[\/latex] first.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Critical Reminder<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">The limits of integration come directly from the region boundaries. For [latex]R = [a,b] \\times [c,d][\/latex], [latex]x[\/latex] ranges from [latex]a[\/latex] to [latex]b[\/latex] and [latex]y[\/latex] ranges from [latex]c[\/latex] to [latex]d[\/latex], regardless of integration order.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Use the properties of the double integral and Fubini\u2019s theorem to evaluate the integral\r\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\int_0^1\\int_{-1}^3(3-x+4y) dy\\ dx}[\/latex].<\/p>\r\n<\/li>\r\n \t<li>Show that [latex]0 \\leq \\underset{R}{\\displaystyle\\iint}\\sin \\pi x\\cos\\pi y\\ dA \\leq \\frac{1}{32}[\/latex] where [latex]R = \\left (0,\\frac{1}{4} \\right ) \\left (\\frac{1}{4}, \\frac{1}{2} \\right )[\/latex].<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"870681\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"870681\"]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>26<\/li>\r\n \t<li>Answers may vary.<\/li>\r\n<\/ol>\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=8197096&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=wf1Iv3IhDGI&amp;video_target=tpm-plugin-9dqvee1r-wf1Iv3IhDGI\" 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\/CP5.2_transcript.html\">transcript for \u201cCP 5.2\u201d here (opens in new window).<\/a><\/center><\/section><section class=\"textbox example\" aria-label=\"Example\">Evaluate\u00a0[latex]\\displaystyle\\int_{y=-3}^{y=2}\\displaystyle\\int_{x=3}^{x=5}(2-3x^2+y^2)dx\\ dy[\/latex].[reveal-answer q=\"241684052\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"241684052\"]\r\n<p style=\"text-align: center;\">[latex]\\large{-\\frac{1,340}3}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Evaluate the integral\u00a0[latex]\\underset{R}{\\displaystyle\\iint}xe^{xy}dA[\/latex] where\u00a0[latex]R = {[0,1]}{\\times}{[0,\\ln5]}[\/latex].[reveal-answer q=\"254888390\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"254888390\"]\r\n<p style=\"text-align: center;\">[latex]\\large{\\frac{4-\\ln5}{\\ln5}}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<h2 data-type=\"title\">General Regions of Integration<\/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\">Most real-world regions aren't rectangles. To integrate over irregular shapes, we classify them as Type I or Type II regions based on how their boundaries are defined. This determines which order of integration works naturally.<\/p>\r\n<p class=\"whitespace-normal break-words\">For any bounded region [latex]D[\/latex], we can enclose it in a rectangle [latex]R[\/latex]. We extend our function [latex]f(x,y)[\/latex] to be zero outside [latex]D[\/latex], allowing us to use rectangular integration techniques while only \"counting\" the area inside [latex]D[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Type I Regions: Vertical Slicing<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">A Type I region sits between two vertical lines [latex]x = a[\/latex] and [latex]x = b[\/latex], with top boundary [latex]y = g_2(x)[\/latex] and bottom boundary [latex]y = g_1(x)[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]D = {(x,y) \\mid a \\leq x \\leq b, g_1(x) \\leq y \\leq g_2(x)}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Think vertically: for each fixed [latex]x[\/latex]-value, [latex]y[\/latex] ranges between two functions of [latex]x[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Type II Regions: Horizontal Slicing<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">A Type II region sits between two horizontal lines [latex]y = c[\/latex] and [latex]y = d[\/latex], with left boundary [latex]x = h_1(y)[\/latex] and right boundary [latex]x = h_2(y)[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]D = {(x,y) \\mid c \\leq y \\leq d, h_1(y) \\leq x \\leq h_2(y)}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Think horizontally: for each fixed [latex]y[\/latex]-value, [latex]x[\/latex] ranges between two functions of [latex]y[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>How to Identify Region Type<\/strong><\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Type I<\/strong>: Draw vertical lines through the region. If each vertical line enters and exits the region just once (crossing top and bottom boundaries), it's Type I.<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Type II<\/strong>: Draw horizontal lines through the region. If each horizontal line enters and exits just once (crossing left and right boundaries), it's Type II.<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">Many regions can be described either way. The region between [latex]y = \\sqrt{x}[\/latex] and [latex]y = x^3[\/latex] for [latex]0 \\leq x \\leq 1[\/latex] is Type I with [latex]x^3 \\leq y \\leq \\sqrt{x}[\/latex]. Solve for [latex]x[\/latex] to get Type II: [latex]y^2 \\leq x \\leq \\sqrt[3]{y}[\/latex] for [latex]0 \\leq y \\leq 1[\/latex]. Choose whichever makes the integral easier to compute.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Consider the region in the first quadrant between the functions [latex]y=2x[\/latex] and [latex]y=x^{2}[\/latex]. Describe the region first as Type I and then as Type II.[reveal-answer q=\"984622178\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"984622178\"]Type I and Type II are expressed as [latex]\\{(x,y)|0\\leq{x}\\leq2,x^2\\leq{y}\\leq2x\\}[\/latex] and [latex]\\{(x,y)|0\\leq{y}\\leq4,\\frac12y\\leq{x}\\leq\\sqrt{y}\\}[\/latex], respectively.[\/hidden-answer]<\/section>\r\n<h2 data-type=\"title\">Double Integrals over Nonrectangular Regions<\/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\">Once you've identified your region as Type I, Type II, or a combination, you can set up the double integral using Fubini's theorem. The key is translating the region description into correct limits of integration.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Fubini's Theorem for General Regions<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">For a <strong>Type I<\/strong> region where [latex]a \\leq x \\leq b[\/latex] and [latex]g_1(x) \\leq y \\leq g_2(x)[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{D}{\\displaystyle\\iint}f(x,y)dA = \\displaystyle\\int_a^b\\left[\\displaystyle\\int_{g_1(x)}^{g_2(x)}f(x,y)dy\\right]dx[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">For a <strong>Type II<\/strong> region where [latex]c \\leq y \\leq d[\/latex] and [latex]h_1(y) \\leq x \\leq h_2(y)[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{D}{\\displaystyle\\iint}f(x,y)dA = \\displaystyle\\int_c^d\\left[\\displaystyle\\int_{h_1(y)}^{h_2(y)}f(x,y)dx\\right]dy[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>How the Limits Work<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">The outer integral limits are constants (the fixed interval). The inner integral limits are functions that depend on the outer variable. For Type I, [latex]y[\/latex] varies between two curves that depend on [latex]x[\/latex]. For Type II, [latex]x[\/latex] varies between two curves that depend on [latex]y[\/latex].<\/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> and identify all boundary curves<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Decide on region type<\/strong>: Choose Type I or Type II based on which gives simpler limits<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Write the bounds<\/strong>: Outer limits are constants, inner limits are functions<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Integrate<\/strong>: Work from the inside out, treating the outer variable as constant in the inner integral<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\">Sometimes one order makes integration much harder. If the inner integral looks difficult or impossible with your first choice, try describing the region the other way and switching the order of integration.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Complex Regions: Break Them Apart<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">If your region can't be described as a single Type I or Type II region, split it into pieces:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{D}{\\displaystyle\\iint}f(x,y)dA = \\underset{D_1}{\\displaystyle\\iint}f(x,y)dA + \\underset{D_2}{\\displaystyle\\iint}f(x,y)dA[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Evaluate each piece separately using whichever type fits that piece best, then add the results.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Sketch the region [latex]D[\/latex] and evaluate the iterated integral [latex]\\underset{D}{\\displaystyle\\iint}{xy} \\ {dy} \\ {dx}[\/latex] where [latex]D[\/latex] is the region bounded by the curves [latex]{y} = {\\cos{x}}[\/latex] and [latex]{y} = {\\sin{x}}[\/latex] in the interval [latex][{-3}{{\\pi}\/{4}},{{\\pi}\/{4}}][\/latex].[reveal-answer q=\"884561035\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"884561035\"][latex]\\pi\/4[\/latex][\/hidden-answer]<\/section><section class=\"textbox example\" aria-label=\"Example\">Consider the region bounded by the curves [latex]y = \\ln{x}[\/latex] and [latex]y = e^x[\/latex] in the interval [latex][1,2][\/latex]. Decompose the region into smaller regions of Type II.[reveal-answer q=\"710946832\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"710946832\"][latex]\\{(x,y)|0\\leq{y}\\leq1,1\\leq{x}\\leq{e^y}\\}\\cup\\{(x,y)|1\\leq{y}\\leq{e},1\\leq{x}\\leq2\\}\\cup\\{(x,y)|e\\leq{y}\\leq{e}^2,\\ln{y}\\leq{x}\\leq2\\}[\/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=8197098&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=JCwVItEtOkg&amp;video_target=tpm-plugin-72owa3pa-JCwVItEtOkg\" 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\/CP5.9_transcript.html\">transcript for \u201cCP 5.9\u201d here (opens in new window).<\/a><\/center><\/section>\r\n<h2 data-type=\"title\">Changing the Order of Integration<\/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\">Sometimes the order of integration dramatically affects how difficult the problem is to solve. Learning to reverse the order\u2014switching from [latex]dy,dx[\/latex] to [latex]dx,dy[\/latex] or vice versa\u2014is a valuable skill that can turn an impossible integral into a manageable one.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Why Change the Order?<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">An integral might be difficult or impossible to evaluate in one order but straightforward in the other. If you encounter an inner integral you can't solve, or one that requires complex techniques, switching the order of integration often provides a simpler path.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>How to Reverse the Order<\/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>Identify the current region type<\/strong>: Read the limits carefully. Is it Type I ([latex]dy,dx[\/latex]) or Type II ([latex]dx,dy[\/latex])?<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Sketch the region<\/strong>: Draw the boundaries to visualize what you're working with<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Rewrite as the opposite type<\/strong>: Convert Type I to Type II or vice versa by expressing the same region with different bounds<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Solve any equations needed<\/strong>: If the region is bounded by [latex]y = 2 - x^2[\/latex], solve for [latex]x[\/latex] to get [latex]x = \\sqrt{2-y}[\/latex] when converting<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Set up the new integral<\/strong>: Write the limits in the new order<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>Reading the Original Limits<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">From [latex]\\displaystyle\\int_{x=a}^{x=b}\\displaystyle\\int_{y=g_1(x)}^{y=g_2(x)}f(x,y)dy,dx[\/latex]:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">The region spans [latex]a \\leq x \\leq b[\/latex] (constants tell you the outer variable's range)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">For each [latex]x[\/latex], [latex]y[\/latex] goes from [latex]g_1(x)[\/latex] to [latex]g_2(x)[\/latex] (functions tell you the inner variable's range)<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Setting Up the New Limits<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">After sketching, determine:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">What constant values bound the new outer variable?<\/li>\r\n \t<li class=\"whitespace-normal break-words\">What functions (of the new outer variable) bound the new inner variable?<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Pro Tip<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Always sketch the region when changing order. Visual confirmation prevents mistakes with limits and helps you see which boundaries to use.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Evaluate the iterated integral [latex]\\underset{R}{\\displaystyle\\iint}{(x^2+y^2)}{dA}[\/latex] over the region [latex]D[\/latex] in the first quadrant between the functions [latex]y=2x[\/latex] and [latex]y=x^{2}[\/latex]. Evaluate the iterated integral by integrating first with respect to [latex]y[\/latex] and then integrating first with resect to [latex]x[\/latex].[reveal-answer q=\"867530921\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"867530921\"][latex]\\large{\\frac{216}{35}}[\/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=8197099&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=zcvSZfiNvqI&amp;video_target=tpm-plugin-zim0jvdd-zcvSZfiNvqI\" 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\/CP5.11_transcript.html\">transcript for \u201cCP 5.11\u201d here (opens in new window).<\/a><\/center><\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Calculate double integrals when your region is a rectangle<\/li>\n<li>Find double integrals over more complex regions<\/li>\n<li>Switch the order of integration to make calculations easier<\/li>\n<\/ul>\n<\/section>\n<h2 data-type=\"title\">Volumes and Double Integrals<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Single integrals find areas under curves. Double integrals extend this idea to find volumes under surfaces. Instead of slicing a region into thin strips, you&#8217;re dividing a rectangular base into small boxes and stacking them up to the surface.<\/p>\n<p class=\"whitespace-normal break-words\">For a function [latex]z = f(x, y)[\/latex] over a rectangular region [latex]R = [a, b] \\times [c, d][\/latex], the double integral is:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\large{\\underset{R}{\\displaystyle\\iint}f(x,y)dA=\\displaystyle\\lim_{m,n\\to\\infty}\\displaystyle\\sum_{i=1}^{m}\\sum_{j=1}^{n}f(x_i^*,y_j^*)\\Delta A}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">This represents the volume of the solid sitting above region [latex]R[\/latex] and below the surface [latex]z = f(x, y)[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">As you make the grid finer (more and smaller rectangles), the approximation improves. The exact volume is the limit as [latex]m[\/latex] and [latex]n[\/latex] approach infinity\u2014that limit is the double integral.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Key Notation<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Since [latex]dA = dx,dy = dy,dx[\/latex], you can write the double integral as:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{R}{\\displaystyle\\iint}f(x,y)dx,dy[\/latex] or [latex]\\underset{R}{\\displaystyle\\iint}f(x,y)dy,dx[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>When It Works<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The double integral exists if [latex]f[\/latex] is continuous over [latex]R[\/latex] (or at least not too discontinuous). If [latex]f(x, y) \\geq 0[\/latex], the result is volume. If [latex]f[\/latex] is sometimes negative, think of it as signed volume\u2014parts below the [latex]xy[\/latex]-plane count as negative.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p id=\"fs-id1167794324430\">Use the same function [latex]z = f(x,y) = 3x^2 - y[\/latex] over the rectangular region [latex]R = [0,2] \\times [0,2][\/latex].<\/p>\n<p id=\"fs-id1167793473779\">Divide [latex]R[\/latex] into the same four squares with [latex]m = n = 2[\/latex] and choose the sample points as the upper left corner point of each square [latex](0,1), (1,1), (0,2),[\/latex] and [latex](1,2)[\/latex] (Figure 5) to approximate the signed volume of the solid [latex]S[\/latex] that lies above [latex]R[\/latex] and \u201cunder\u201d the graph of [latex]f[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q265173902\">Show Solution<\/button><\/p>\n<div id=\"q265173902\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\large{V=\\displaystyle\\sum_{i=1}^2\\sum_{j=1}^2f(x_{ij}^*,y_{ij}^*)\\Delta A=0}[\/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=8197095&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=ZKNjm2OdGnU&amp;video_target=tpm-plugin-g6ue79lv-ZKNjm2OdGnU\" 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\/CP5.1_transcript.html\">transcript for \u201cCP 5.1\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<h2 data-type=\"title\">Properties 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 behave much like single integrals\u2014they follow familiar algebraic rules that make calculations easier. Understanding these properties helps you break complex integrals into simpler pieces.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Linearity (Properties 1 and 2)<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Double integrals respect addition and scalar multiplication:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{R}{\\displaystyle\\iint}[f(x,y)+g(x,y)]dA=\\underset{R}{\\displaystyle\\iint}f(x,y)dA+\\underset{R}{\\displaystyle\\iint}g(x,y)dA[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{R}{\\displaystyle\\iint}cf(x,y)dA = c\\underset{R}{\\displaystyle\\iint}f(x,y)dA[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">You can split integrals of sums into separate integrals, and pull constants outside the integral. This works exactly like it did in single-variable calculus.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Additivity (Property 3)<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">If you split region [latex]R[\/latex] into non-overlapping subregions [latex]S[\/latex] and [latex]T[\/latex], the integral over [latex]R[\/latex] equals the sum of integrals over the pieces:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{R}{\\displaystyle\\iint}f(x,y)dA = \\underset{S}{\\displaystyle\\iint}f(x,y)dA + \\underset{T}{\\displaystyle\\iint}f(x,y)dA[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">This is useful when your region has a natural division or when part is easier to integrate than the whole.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Monotonicity (Property 4)<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">If [latex]f(x,y) \\geq g(x,y)[\/latex] everywhere on [latex]R[\/latex], then their integrals follow the same inequality. Bigger functions produce bigger integrals.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Bounding the Integral (Property 5)<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">If you know [latex]m \\leq f(x,y) \\leq M[\/latex] on region [latex]R[\/latex], then:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]m \\times A(R) \\leq \\underset{R}{\\displaystyle\\iint}f(x,y)dA \\leq M \\times A(R)[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">This gives quick estimates: multiply the region&#8217;s area by the minimum and maximum function values.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Separable Functions (Property 6)<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">If [latex]f(x,y) = g(x)h(y)[\/latex] (a product of separate [latex]x[\/latex] and [latex]y[\/latex] functions), the double integral factors:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{R}{\\displaystyle\\iint}g(x)h(y)dA = \\left(\\displaystyle\\int_a^bg(x)dx\\right)\\left(\\displaystyle\\int_c^dh(y)dy\\right)[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">This turns a double integral into two single integrals\u2014often much easier to compute.<\/p>\n<\/div>\n<h2>Iterated Integrals<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Computing double integrals directly from the limit definition would be painfully tedious. Fortunately, Fubini&#8217;s theorem gives us a practical method: break a double integral into two single integrals, integrating one variable at a time.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Key Idea: Iterated Integrals<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">An iterated integral lets you evaluate a double integral by integrating twice\u2014first with respect to one variable (treating the other as constant), then with respect to the second variable:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\displaystyle\\int_a^b\\int_c^df(x,y)dy,dx = \\displaystyle\\int_a^b\\left[\\displaystyle\\int_c^df(x,y)dy\\right]dx[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">The inner integral [latex]\\int_c^d f(x,y)dy[\/latex] treats [latex]x[\/latex] as a constant and integrates with respect to [latex]y[\/latex]. The result is a function of [latex]x[\/latex] alone, which you then integrate.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Fubini&#8217;s Theorem<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">For a continuous function [latex]f(x,y)[\/latex] over rectangle [latex]R = [a,b] \\times [c,d][\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{R}{\\displaystyle\\iint}f(x,y)dA = \\displaystyle\\int_a^b\\displaystyle\\int_c^df(x,y)dy,dx = \\displaystyle\\int_c^d\\displaystyle\\int_a^bf(x,y)dx,dy[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Both orders of integration give the same answer. Choose whichever order makes the computation easier.<\/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>Identify your region<\/strong>: Note the bounds [latex]a \\leq x \\leq b[\/latex] and [latex]c \\leq y \\leq d[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Choose your order<\/strong>: Decide whether to integrate with respect to [latex]x[\/latex] or [latex]y[\/latex] first<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Integrate the inner integral<\/strong>: Treat one variable as constant and integrate with respect to the other<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Integrate the outer integral<\/strong>: Integrate the resulting expression with respect to the remaining variable<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>When Order Matters<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Sometimes one order is much easier than the other. If the inner integral looks complicated (requiring integration by parts or substitution that depends on both variables), try switching the order. A function like [latex]x\\sin(xy)[\/latex] integrates more easily with respect to [latex]y[\/latex] first than [latex]x[\/latex] first.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Critical Reminder<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The limits of integration come directly from the region boundaries. For [latex]R = [a,b] \\times [c,d][\/latex], [latex]x[\/latex] ranges from [latex]a[\/latex] to [latex]b[\/latex] and [latex]y[\/latex] ranges from [latex]c[\/latex] to [latex]d[\/latex], regardless of integration order.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Use the properties of the double integral and Fubini\u2019s theorem to evaluate the integral\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\int_0^1\\int_{-1}^3(3-x+4y) dy\\ dx}[\/latex].<\/p>\n<\/li>\n<li>Show that [latex]0 \\leq \\underset{R}{\\displaystyle\\iint}\\sin \\pi x\\cos\\pi y\\ dA \\leq \\frac{1}{32}[\/latex] where [latex]R = \\left (0,\\frac{1}{4} \\right ) \\left (\\frac{1}{4}, \\frac{1}{2} \\right )[\/latex].<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q870681\">Show Solution<\/button><\/p>\n<div id=\"q870681\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>26<\/li>\n<li>Answers may vary.<\/li>\n<\/ol>\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=8197096&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=wf1Iv3IhDGI&amp;video_target=tpm-plugin-9dqvee1r-wf1Iv3IhDGI\" 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\/CP5.2_transcript.html\">transcript for \u201cCP 5.2\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Evaluate\u00a0[latex]\\displaystyle\\int_{y=-3}^{y=2}\\displaystyle\\int_{x=3}^{x=5}(2-3x^2+y^2)dx\\ dy[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q241684052\">Show Solution<\/button><\/p>\n<div id=\"q241684052\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\large{-\\frac{1,340}3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Evaluate the integral\u00a0[latex]\\underset{R}{\\displaystyle\\iint}xe^{xy}dA[\/latex] where\u00a0[latex]R = {[0,1]}{\\times}{[0,\\ln5]}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q254888390\">Show Solution<\/button><\/p>\n<div id=\"q254888390\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\large{\\frac{4-\\ln5}{\\ln5}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<h2 data-type=\"title\">General Regions of Integration<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Most real-world regions aren&#8217;t rectangles. To integrate over irregular shapes, we classify them as Type I or Type II regions based on how their boundaries are defined. This determines which order of integration works naturally.<\/p>\n<p class=\"whitespace-normal break-words\">For any bounded region [latex]D[\/latex], we can enclose it in a rectangle [latex]R[\/latex]. We extend our function [latex]f(x,y)[\/latex] to be zero outside [latex]D[\/latex], allowing us to use rectangular integration techniques while only &#8220;counting&#8221; the area inside [latex]D[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Type I Regions: Vertical Slicing<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">A Type I region sits between two vertical lines [latex]x = a[\/latex] and [latex]x = b[\/latex], with top boundary [latex]y = g_2(x)[\/latex] and bottom boundary [latex]y = g_1(x)[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]D = {(x,y) \\mid a \\leq x \\leq b, g_1(x) \\leq y \\leq g_2(x)}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Think vertically: for each fixed [latex]x[\/latex]-value, [latex]y[\/latex] ranges between two functions of [latex]x[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Type II Regions: Horizontal Slicing<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">A Type II region sits between two horizontal lines [latex]y = c[\/latex] and [latex]y = d[\/latex], with left boundary [latex]x = h_1(y)[\/latex] and right boundary [latex]x = h_2(y)[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]D = {(x,y) \\mid c \\leq y \\leq d, h_1(y) \\leq x \\leq h_2(y)}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Think horizontally: for each fixed [latex]y[\/latex]-value, [latex]x[\/latex] ranges between two functions of [latex]y[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>How to Identify Region Type<\/strong><\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Type I<\/strong>: Draw vertical lines through the region. If each vertical line enters and exits the region just once (crossing top and bottom boundaries), it&#8217;s Type I.<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Type II<\/strong>: Draw horizontal lines through the region. If each horizontal line enters and exits just once (crossing left and right boundaries), it&#8217;s Type II.<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">Many regions can be described either way. The region between [latex]y = \\sqrt{x}[\/latex] and [latex]y = x^3[\/latex] for [latex]0 \\leq x \\leq 1[\/latex] is Type I with [latex]x^3 \\leq y \\leq \\sqrt{x}[\/latex]. Solve for [latex]x[\/latex] to get Type II: [latex]y^2 \\leq x \\leq \\sqrt[3]{y}[\/latex] for [latex]0 \\leq y \\leq 1[\/latex]. Choose whichever makes the integral easier to compute.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Consider the region in the first quadrant between the functions [latex]y=2x[\/latex] and [latex]y=x^{2}[\/latex]. Describe the region first as Type I and then as Type II.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q984622178\">Show Solution<\/button><\/p>\n<div id=\"q984622178\" class=\"hidden-answer\" style=\"display: none\">Type I and Type II are expressed as [latex]\\{(x,y)|0\\leq{x}\\leq2,x^2\\leq{y}\\leq2x\\}[\/latex] and [latex]\\{(x,y)|0\\leq{y}\\leq4,\\frac12y\\leq{x}\\leq\\sqrt{y}\\}[\/latex], respectively.<\/div>\n<\/div>\n<\/section>\n<h2 data-type=\"title\">Double Integrals over Nonrectangular Regions<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Once you&#8217;ve identified your region as Type I, Type II, or a combination, you can set up the double integral using Fubini&#8217;s theorem. The key is translating the region description into correct limits of integration.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Fubini&#8217;s Theorem for General Regions<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">For a <strong>Type I<\/strong> region where [latex]a \\leq x \\leq b[\/latex] and [latex]g_1(x) \\leq y \\leq g_2(x)[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{D}{\\displaystyle\\iint}f(x,y)dA = \\displaystyle\\int_a^b\\left[\\displaystyle\\int_{g_1(x)}^{g_2(x)}f(x,y)dy\\right]dx[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">For a <strong>Type II<\/strong> region where [latex]c \\leq y \\leq d[\/latex] and [latex]h_1(y) \\leq x \\leq h_2(y)[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{D}{\\displaystyle\\iint}f(x,y)dA = \\displaystyle\\int_c^d\\left[\\displaystyle\\int_{h_1(y)}^{h_2(y)}f(x,y)dx\\right]dy[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>How the Limits Work<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The outer integral limits are constants (the fixed interval). The inner integral limits are functions that depend on the outer variable. For Type I, [latex]y[\/latex] varies between two curves that depend on [latex]x[\/latex]. For Type II, [latex]x[\/latex] varies between two curves that depend on [latex]y[\/latex].<\/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> and identify all boundary curves<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Decide on region type<\/strong>: Choose Type I or Type II based on which gives simpler limits<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Write the bounds<\/strong>: Outer limits are constants, inner limits are functions<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Integrate<\/strong>: Work from the inside out, treating the outer variable as constant in the inner integral<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\">Sometimes one order makes integration much harder. If the inner integral looks difficult or impossible with your first choice, try describing the region the other way and switching the order of integration.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Complex Regions: Break Them Apart<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">If your region can&#8217;t be described as a single Type I or Type II region, split it into pieces:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{D}{\\displaystyle\\iint}f(x,y)dA = \\underset{D_1}{\\displaystyle\\iint}f(x,y)dA + \\underset{D_2}{\\displaystyle\\iint}f(x,y)dA[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Evaluate each piece separately using whichever type fits that piece best, then add the results.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Sketch the region [latex]D[\/latex] and evaluate the iterated integral [latex]\\underset{D}{\\displaystyle\\iint}{xy} \\ {dy} \\ {dx}[\/latex] where [latex]D[\/latex] is the region bounded by the curves [latex]{y} = {\\cos{x}}[\/latex] and [latex]{y} = {\\sin{x}}[\/latex] in the interval [latex][{-3}{{\\pi}\/{4}},{{\\pi}\/{4}}][\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q884561035\">Show Solution<\/button><\/p>\n<div id=\"q884561035\" class=\"hidden-answer\" style=\"display: none\">[latex]\\pi\/4[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Consider the region bounded by the curves [latex]y = \\ln{x}[\/latex] and [latex]y = e^x[\/latex] in the interval [latex][1,2][\/latex]. Decompose the region into smaller regions of Type II.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q710946832\">Show Solution<\/button><\/p>\n<div id=\"q710946832\" class=\"hidden-answer\" style=\"display: none\">[latex]\\{(x,y)|0\\leq{y}\\leq1,1\\leq{x}\\leq{e^y}\\}\\cup\\{(x,y)|1\\leq{y}\\leq{e},1\\leq{x}\\leq2\\}\\cup\\{(x,y)|e\\leq{y}\\leq{e}^2,\\ln{y}\\leq{x}\\leq2\\}[\/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=8197098&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=JCwVItEtOkg&amp;video_target=tpm-plugin-72owa3pa-JCwVItEtOkg\" 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\/CP5.9_transcript.html\">transcript for \u201cCP 5.9\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<h2 data-type=\"title\">Changing the Order of Integration<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Sometimes the order of integration dramatically affects how difficult the problem is to solve. Learning to reverse the order\u2014switching from [latex]dy,dx[\/latex] to [latex]dx,dy[\/latex] or vice versa\u2014is a valuable skill that can turn an impossible integral into a manageable one.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Why Change the Order?<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">An integral might be difficult or impossible to evaluate in one order but straightforward in the other. If you encounter an inner integral you can&#8217;t solve, or one that requires complex techniques, switching the order of integration often provides a simpler path.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>How to Reverse the Order<\/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>Identify the current region type<\/strong>: Read the limits carefully. Is it Type I ([latex]dy,dx[\/latex]) or Type II ([latex]dx,dy[\/latex])?<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Sketch the region<\/strong>: Draw the boundaries to visualize what you&#8217;re working with<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Rewrite as the opposite type<\/strong>: Convert Type I to Type II or vice versa by expressing the same region with different bounds<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Solve any equations needed<\/strong>: If the region is bounded by [latex]y = 2 - x^2[\/latex], solve for [latex]x[\/latex] to get [latex]x = \\sqrt{2-y}[\/latex] when converting<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Set up the new integral<\/strong>: Write the limits in the new order<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>Reading the Original Limits<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">From [latex]\\displaystyle\\int_{x=a}^{x=b}\\displaystyle\\int_{y=g_1(x)}^{y=g_2(x)}f(x,y)dy,dx[\/latex]:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">The region spans [latex]a \\leq x \\leq b[\/latex] (constants tell you the outer variable&#8217;s range)<\/li>\n<li class=\"whitespace-normal break-words\">For each [latex]x[\/latex], [latex]y[\/latex] goes from [latex]g_1(x)[\/latex] to [latex]g_2(x)[\/latex] (functions tell you the inner variable&#8217;s range)<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Setting Up the New Limits<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">After sketching, determine:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">What constant values bound the new outer variable?<\/li>\n<li class=\"whitespace-normal break-words\">What functions (of the new outer variable) bound the new inner variable?<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Pro Tip<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Always sketch the region when changing order. Visual confirmation prevents mistakes with limits and helps you see which boundaries to use.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Evaluate the iterated integral [latex]\\underset{R}{\\displaystyle\\iint}{(x^2+y^2)}{dA}[\/latex] over the region [latex]D[\/latex] in the first quadrant between the functions [latex]y=2x[\/latex] and [latex]y=x^{2}[\/latex]. Evaluate the iterated integral by integrating first with respect to [latex]y[\/latex] and then integrating first with resect to [latex]x[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q867530921\">Show Solution<\/button><\/p>\n<div id=\"q867530921\" class=\"hidden-answer\" style=\"display: none\">[latex]\\large{\\frac{216}{35}}[\/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=8197099&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=zcvSZfiNvqI&amp;video_target=tpm-plugin-zim0jvdd-zcvSZfiNvqI\" 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\/CP5.11_transcript.html\">transcript for \u201cCP 5.11\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":13,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-678","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\/678","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":10,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/678\/revisions"}],"predecessor-version":[{"id":2168,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/678\/revisions\/2168"}],"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\/678\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=678"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=678"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=678"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=678"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}