{"id":1193,"date":"2025-10-15T15:16:43","date_gmt":"2025-10-15T15:16:43","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=1193"},"modified":"2025-10-29T15:43:42","modified_gmt":"2025-10-29T15:43:42","slug":"double-integrals-over-basic-regions-learn-it-6","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/double-integrals-over-basic-regions-learn-it-6\/","title":{"raw":"Double Integrals over Basic Regions: Learn It 6","rendered":"Double Integrals over Basic Regions: Learn It 6"},"content":{"raw":"

Changing the Order of Integration<\/h2>\r\nAs we have already seen when we evaluate an iterated integral, sometimes one order of integration leads to a computation that is significantly simpler than the other order of integration. Sometimes the order of integration does not matter, but it is important to learn to recognize when a change in order will simplify our work.\r\n\r\n
Reverse the order of integration in the iterated integral [latex]\\displaystyle\\int_{x=0}^{x=\\sqrt2}\\displaystyle\\int_{y=0}^{y=2-x^2}xe^{x^2}dydx[\/latex]. Then evaluate the new iterated integral.[reveal-answer q=\"557138862\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"557138862\"]\r\n\r\nThe region as presented is of Type I. To reverse the order of integration, we must first express the region as Type II. Refer to\u00a0Figure 10.\r\n\r\n[caption id=\"attachment_1346\" align=\"aligncenter\" width=\"560\"]\"This Figure 10.\u00a0Converting a region from Type I to Type II.[\/caption]\r\n

We can see from the limits of integration that the region is bounded above by [latex]y=2-x^{2}[\/latex] and below by [latex]y=0[\/latex], where [latex]x[\/latex] is in the interval [latex][0,{\\sqrt{2}}][\/latex]. By reversing the order, we have the region bounded on the left by [latex]x=0[\/latex] and on the right by [latex]{x} = {{\\sqrt{2-y}}}[\/latex] where [latex]y[\/latex] is in the interval [latex][0,2][\/latex]. We solved [latex]y=2-x^2[\/latex] in terms of [latex]x[\/latex] to obtain [latex]{x} = {{\\sqrt{2-y}}}[\/latex].<\/p>\r\n

Hence<\/p>\r\n[latex]\\begin{align}\r\n\r\n\\displaystyle\\int_0^{\\sqrt2}\\displaystyle\\int_0^{2-x^2}xe^{x^2}dydx&=\\displaystyle\\int_0^2\\displaystyle\\int_0^{\\sqrt{2-y}}xe^{x^2}dxdy \\,\\,\\,\\,\\,\\text{ Reverse the order of integration then use substitution} \\\\\r\n\r\n&=\\displaystyle\\int_0^2\\left[\\frac12ex^2\\bigg|_0^{\\sqrt{2-y}}\\right]dy=\\displaystyle\\int_0^2\\frac12(e^{2-y}-1)dy=-\\frac12(e^{2-y}+y)\\bigg|_0^2 \\\\\r\n\r\n&=\\frac12(e^2-3).\r\n\r\n\\end{align}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

\r\n

Consider the iterated integral [latex]\\underset{R}{\\displaystyle\\iint}{f(x,y)}{dx} \\ {dy}[\/latex]\u00a0where [latex]{z} = {f(x,y)} = {x-2y}[\/latex]<\/span>\u00a0over a triangular region [latex]R[\/latex]<\/span>\u00a0that has sides on [latex]x=0,\\text{ }y=0[\/latex]\u00a0<\/span>and the line [latex]x+y=1[\/latex].\u00a0<\/span>Sketch the region, and then evaluate the iterated integral by<\/span><\/p>\r\n

a. integrating first with respect to [latex]y[\/latex], and then,<\/p>\r\n

b. integrating first with respect to [latex]x[\/latex].<\/p>\r\n[reveal-answer q=\"913592004\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"913592004\"]\r\n\r\nA sketch of the region appears in\u00a0Figure 11.\r\n\r\n[caption id=\"attachment_1347\" align=\"aligncenter\" width=\"262\"]\"The Figure 11.\u00a0A triangular region\u00a0[latex]R[\/latex]\u00a0<\/span>for integrating in two ways.[\/caption]\r\n

We can complete this integration in two different ways.<\/p>\r\n

a. One way to look at it is by first integrating [latex]y[\/latex] from [latex]y=0[\/latex] to [latex]y=1-x[\/latex] vertically and then integrating [latex]x[\/latex] from [latex]x=0[\/latex]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0to\u00a0[latex]x=1[\/latex]:<\/p>\r\n[latex]\\hspace{5cm}\\begin{align}\r\n\r\n\\underset{R}{\\displaystyle\\iint}f(x,y)dxdy&=\\displaystyle\\int_{x=0}^{x=1}\\displaystyle\\int_{y=0}^{y=1-x}(x-2y)dydx=\\displaystyle\\int_{x=0}^{x=1}[xy-2y^2]_{y=0}^{y=1-x}dx \\\\\r\n\r\n&=\\displaystyle\\int_{x=0}^{x=1}\\left[x(1-x)-(1-x)^2\\right]dx=\\displaystyle\\int_{x=0}^{x=1}[-1+3x-2x^2]dx \\\\\r\n\r\n&=\\left[-x+\\frac32x^2-\\frac23x^3\\right]_{x=0}^{x=1}=-\\frac16\r\n\r\n\\end{align}[\/latex]\r\n

b. The other way to do this problem is by first integrating [latex]x[\/latex] from [latex]x=0[\/latex] to [latex]x=1-y[\/latex] horizontally and then integrating [latex]y[\/latex] from [latex]y=0[\/latex] to\u00a0[latex]y=1[\/latex]:<\/p>\r\n[latex]\\hspace{5cm}\\begin{align}\r\n\r\n\\underset{R}{\\displaystyle\\iint}f(x,y)dxdy&=\\displaystyle\\int_{y=0}^{y=1}\\displaystyle\\int_{x=0}^{x=1-y}(x-2y)dxdy=\\displaystyle\\int_{y=0}^{y=1}[\\frac12x^2-2xy]_{x=0}^{x=1-y}dy \\\\\r\n\r\n&=\\displaystyle\\int_{y=0}^{y=1}\\left[\\frac12(1-y)^2-2y(1-y)\\right]dy=\\displaystyle\\int_{y=0}^{y=1}[\\frac12-3y+\\frac52y^2]dy \\\\\r\n\r\n&=\\left[\\frac12y-\\frac32y^2+\\frac56y^3\\right]_{y=0}^{y=1}=-\\frac16\r\n\r\n\\end{align}[\/latex]\r\n\r\n \r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"

Changing the Order of Integration<\/h2>\n

As we have already seen when we evaluate an iterated integral, sometimes one order of integration leads to a computation that is significantly simpler than the other order of integration. Sometimes the order of integration does not matter, but it is important to learn to recognize when a change in order will simplify our work.<\/p>\n

Reverse the order of integration in the iterated integral [latex]\\displaystyle\\int_{x=0}^{x=\\sqrt2}\\displaystyle\\int_{y=0}^{y=2-x^2}xe^{x^2}dydx[\/latex]. Then evaluate the new iterated integral.<\/p>\n