double integrals over nonrectangular regions<\/h3>\r\n
Suppose [latex]g(x, y)[\/latex] is the extension to the rectangle [latex]R[\/latex] of the function [latex]f(x, y)[\/latex] defined on the regions [latex]D[\/latex] and [latex]R[\/latex] inside [latex]R[\/latex]. Then [latex]g(x, y)[\/latex] is integrable and we define the double integral of [latex]f(x, y)[\/latex] over [latex]D[\/latex] by:<\/p>\r\n
[latex]\\underset{D}{\\displaystyle\\iint} {f(x,y)dA} = \\underset{R}{\\displaystyle\\iint} {g(x,y)dA}[\/latex].<\/p>\r\n\r\n<\/section>\r\n
The right-hand side of this equation is what we have seen before, so this theorem is reasonable because [latex]R[\/latex] is a rectangle and [latex]\\underset{R}{\\displaystyle\\iint}g(x,y)dA[\/latex] has been discussed in the preceding section. Also, the equality works because the values of [latex]g(x, y)[\/latex] are [latex]0[\/latex] for any point [latex](x, y)[\/latex] that lies outside [latex]D[\/latex], and hence these points do not add anything to the integral. However, it is important that the rectangle [latex]R[\/latex] contains the region [latex]D[\/latex].<\/p>\r\n
If the region [latex]D[\/latex] is bounded by smooth curves on a plane and we are able to describe it as Type I or Type II or a mix of both, then we can use the following theorem and not have to find a rectangle [latex]R[\/latex] containing the region.<\/p>\r\n\r\n For a function [latex]f(x, y)[\/latex] that is continuous on a region [latex]D[\/latex] of Type I, we have:<\/p>\r\n [latex]\\underset{D}{\\displaystyle\\iint}{f(x,y)dA} = \\underset{D}{\\displaystyle\\iint}{f(x,y)dy \\ dx}=\\displaystyle\\int_a^b\\left[\\displaystyle\\int_{g_1(x)}^{g_2(x)}f(x,y)dy\\right]dx.[\/latex]<\/p>\r\nSimilarly, for a function [latex]f(x, y)[\/latex] that is continuous on a region [latex]D[\/latex] of Type II, we have:\r\n [latex]\\underset{D}{\\displaystyle\\iint}{f(x,y)dA} = \\underset{D}{\\displaystyle\\iint}{f(x,y)dx \\ dy}=\\displaystyle\\int_c^d\\left[\\displaystyle\\int_{h_1(y)}^{h_2(y)}f(x,y)dx\\right]dy.[\/latex]<\/p>\r\n\r\n<\/section>The integral in each of these expressions is an iterated integral, similar to those we have seen before. Notice that, in the inner integral in the first expression, we integrate [latex]f(x, y)[\/latex] with [latex]x[\/latex] being held constant and the limits of integration being [latex]g_1(x)[\/latex] and [latex]g_2(x)[\/latex]. In the inner integral in the second expression, we integrate [latex]f(x, y)[\/latex] with [latex]y[\/latex] being held constant and the limits of integration are [latex]h_1(x)[\/latex] and\u00a0[latex]h_2(x)[\/latex].\r\n\r\n [latex]\\large{\\underset{D}{\\displaystyle\\iint}{x^2}{e^{xy}}{dA} = \\displaystyle\\int_{x=0}^{x=2}\\displaystyle\\int_{y=1\/2x}^{y=1}x^2e^{xy}dydx.}[\/latex]<\/p>\r\n\r\n[caption id=\"attachment_1324\" align=\"aligncenter\" width=\"251\"] [latex]\\large{\\underset{D}{\\displaystyle\\iint}{x^2}{e^{xy}}{dA} = \\displaystyle\\int_{y=0}^{y=1}\\displaystyle\\int_{x=0}^{x=2u}{x^2}{e^{xy}}{dx} \\ {dy}}[\/latex]<\/p>\r\nHowever, if we integrate first with respect to [latex]x[\/latex], this integral is lengthy to compute because we have to use integration by parts twice.\r\n\r\n As these examples show, choosing the right region type can significantly simplify your calculations. Another useful technique is breaking complex regions into simpler pieces.<\/p>\r\n\r\n<\/section> Property of Double Integrals<\/strong><\/p>\r\n Recall from Double Integrals over Rectangular Regions that all properties of double integrals are also valid for a function defined on a nonrectangular bounded region on a plane. In particular, property 3 states:<\/p>\r\n If [latex]R = S \\cup T[\/latex] and [latex]S \\cap T = \\emptyset[\/latex] except at their boundaries, then:<\/p>\r\n [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\r\n<\/section>Similarly, we have the following property of double integrals over a nonrectangular bounded region on a plane.\r\n\r\n [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\r\n<\/section>This theorem is particularly useful for nonrectangular regions because it allows us to split a region into a union of regions of Type I and Type II. Then we can compute the double integral on each piece in a convenient way, as in the next example.\r\n\r\n [latex]\\underset{D}{\\displaystyle\\iint}(2x+5y)dA[\/latex].<\/p>\r\n\r\n\r\n[caption id=\"attachment_1343\" align=\"aligncenter\" width=\"342\"] To evaluate a double integral over a general nonrectangular region, we need to first understand the region and be able to express it as Type I or Type II or a combination of both. Without understanding the regions, we will not be able to determine the limits of integration in double integrals. As a first step, let us look at the following theorem.<\/p>\n Suppose [latex]g(x, y)[\/latex] is the extension to the rectangle [latex]R[\/latex] of the function [latex]f(x, y)[\/latex] defined on the regions [latex]D[\/latex] and [latex]R[\/latex] inside [latex]R[\/latex]. Then [latex]g(x, y)[\/latex] is integrable and we define the double integral of [latex]f(x, y)[\/latex] over [latex]D[\/latex] by:<\/p>\n [latex]\\underset{D}{\\displaystyle\\iint} {f(x,y)dA} = \\underset{R}{\\displaystyle\\iint} {g(x,y)dA}[\/latex].<\/p>\n<\/section>\n The right-hand side of this equation is what we have seen before, so this theorem is reasonable because [latex]R[\/latex] is a rectangle and [latex]\\underset{R}{\\displaystyle\\iint}g(x,y)dA[\/latex] has been discussed in the preceding section. Also, the equality works because the values of [latex]g(x, y)[\/latex] are [latex]0[\/latex] for any point [latex](x, y)[\/latex] that lies outside [latex]D[\/latex], and hence these points do not add anything to the integral. However, it is important that the rectangle [latex]R[\/latex] contains the region [latex]D[\/latex].<\/p>\n If the region [latex]D[\/latex] is bounded by smooth curves on a plane and we are able to describe it as Type I or Type II or a mix of both, then we can use the following theorem and not have to find a rectangle [latex]R[\/latex] containing the region.<\/p>\n For a function [latex]f(x, y)[\/latex] that is continuous on a region [latex]D[\/latex] of Type I, we have:<\/p>\n [latex]\\underset{D}{\\displaystyle\\iint}{f(x,y)dA} = \\underset{D}{\\displaystyle\\iint}{f(x,y)dy \\ dx}=\\displaystyle\\int_a^b\\left[\\displaystyle\\int_{g_1(x)}^{g_2(x)}f(x,y)dy\\right]dx.[\/latex]<\/p>\n Similarly, for a function [latex]f(x, y)[\/latex] that is continuous on a region [latex]D[\/latex] of Type II, we have:<\/p>\n [latex]\\underset{D}{\\displaystyle\\iint}{f(x,y)dA} = \\underset{D}{\\displaystyle\\iint}{f(x,y)dx \\ dy}=\\displaystyle\\int_c^d\\left[\\displaystyle\\int_{h_1(y)}^{h_2(y)}f(x,y)dx\\right]dy.[\/latex]<\/p>\n<\/section>\n The integral in each of these expressions is an iterated integral, similar to those we have seen before. Notice that, in the inner integral in the first expression, we integrate [latex]f(x, y)[\/latex] with [latex]x[\/latex] being held constant and the limits of integration being [latex]g_1(x)[\/latex] and [latex]g_2(x)[\/latex]. In the inner integral in the second expression, we integrate [latex]f(x, y)[\/latex] with [latex]y[\/latex] being held constant and the limits of integration are [latex]h_1(x)[\/latex] and\u00a0[latex]h_2(x)[\/latex].<\/p>\nFubini's theorem (strong form)<\/h3>\r\n
![\"A](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23155437\/5-2-5.jpeg\")
Figure 5. We can express region [latex]\\small{D}[\/latex] as a Type I region and integrate from [latex]\\small{y=\\frac{1}{2}x}[\/latex] to\u00a0[latex]\\small{y=1}[\/latex], between the lines\u00a0[latex]\\small{x=0}[\/latex] and\u00a0[latex]\\small{x=2}[\/latex].[\/caption]\r\nTherefore, we have\r\n[latex]\\begin{align}\\displaystyle\\int_{x=0}^{x=2}\\displaystyle\\int_{y=\\frac12x}^{y=1}x^2e^{xy}dydx &=\\int_{x=0}^{x=2}[\\int_{y=\\frac12x}^{y=1}x^2e^{xy}dydx] &\\text{Iterated for a Type I region.} \\\\&=\\int_{x=0}^{x=2}[x^2\\frac{e^xy}x]\\Bigg|_{y=\\frac12x}^{y=1}dx &\\text{Integrate with respect to }y\\text{ using }u\\text{-substitution} \\\\& &\\text{ with }u=xy\\text{ where }x\\text{ is held constant} \\\\&=\\int_{x=0}^{x=2}[xe^x-xe^{\\frac{x^2}2}]dx &\\text{Integrate with respect to }x\\text{ using }u\\text{-substitution with}u=\\frac12x^2 \\\\&=[xe^x-e^x-e^{\\frac12x^2}]\\bigg|_{x=0}^{x=2}=2\\end{align}[\/latex][\/hidden-answer]<\/section>In the previous example, we could have looked at the region in another way, such as [latex]{D} = {\\left \\{{(x,y)}{\\mid}{0} \\ {\\leq} \\ {y} \\ {\\leq} \\ {1,0} \\ {\\leq} \\ {x} \\ {\\leq} \\ {2y} \\right \\}}[\/latex] (Figure 6).\r\n\r\n[caption id=\"attachment_1340\" align=\"aligncenter\" width=\"229\"]![\"A](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25055920\/5-2-6.jpeg\")
Figure 6.[\/caption]\r\n\r\nThis is a Type II region and the integral would then look like:\r\n![\"This](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25060113\/5-2-7.jpeg\")
Figure 7.\u00a0The region [latex]D[\/latex] in this example can be either (a) Type I or (b) Type II.[\/caption]Choosing this order of integration, we have[latex]\\begin{align}\\underset{D}{\\displaystyle\\iint}(3x^2+y^2)dA&=\\displaystyle\\int_{y=-2}^{y=3}\\displaystyle\\int_{x=y^2-3}^{x=y+3}(3x^2+y^2)dxdy &\\quad &\\text{Interated integral, Type II region}.\\\\&=\\displaystyle\\int_{y=-2}^{y=3}(x^2+xy^2\\Bigg|_{y^2-3}^{y+3}dy &\\quad &\\text{Integrate with respect to }x. \\\\&=\\displaystyle\\int_{y=-2}^{y=3}\\left((y+3)^2+(y+3)y^2-(y^2-3)^3-(y^2-3)y^2\\right)dy \\\\&=\\displaystyle\\int_{-2}^{3}(54+27y-12y^2+2y^3+8y^4-y^6)dy &\\quad &\\text{Integrate with respect to }y. \\\\&=\\left[54y+\\frac{27y^2}2-4y^3+\\frac{y^4}2+\\frac{8y^5}5-\\frac{y^7}7\\right]\\bigg|_{-2}^3 \\\\&=\\frac{2,375}7.\\end{align}[\/latex][\/hidden-answer]<\/section>decomposing regions into smaller regions<\/h3>\r\nSuppose the region [latex]D[\/latex] can be expressed as [latex]{D} = {D_1} \\ {\\cup} \\ {D_2}[\/latex] where [latex]D_1[\/latex] and [latex]D_2[\/latex] do not overlap except at their boundaries. Then:\r\n
![\"A](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25060228\/5-2-8.jpeg\")
Figure 8.\u00a0This region can be decomposed into a union of three regions of Type I or Type II.[\/caption]\r\n\r\n[reveal-answer q=\"589644316\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"589644316\"]\r\n\r\nThe region [latex]D[\/latex] is not easy to decompose into any one type; it is actually a combination of different types. So we can write it as a union of three regions [latex]D_1[\/latex], [latex]D_2[\/latex], and [latex]D_3[\/latex] where,\r\n\r\n[latex]{D_1} = {\\left \\{{(x,y)}{\\mid}{-2} \\ {\\leq} \\ {x} \\ {\\leq} \\ {0,0} \\ {\\leq} \\ {y} \\ {\\leq} \\ {(x+y)^2} \\right \\}}[\/latex],\r\n\r\n[latex]{D_2} = {\\left \\{{(x,y)}{\\mid}{0} \\ {\\leq} \\ {y} \\ {\\leq} \\ {4,0} \\ {\\leq} \\ {x} \\ {\\leq} \\ {(y-{\\frac{1}{16}y^3})} \\right \\}}[\/latex].\r\n\r\nThese regions are illustrated more clearly in\u00a0Figure 9.\r\n\r\n[caption id=\"attachment_1344\" align=\"aligncenter\" width=\"342\"]![\"The](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25060322\/5-2-9.jpeg\")
Figure 9.\u00a0Breaking the region into three subregions makes it easier to set up the integration.[\/caption]\r\n\r\nHere [latex]D_1[\/latex] is Type I\u00a0and [latex]D_2[\/latex] and [latex]D_3[\/latex] are both of Type II. Hence,\r\n\r\n[latex]\\begin{align}\r\n\r\n\\underset{D}{\\displaystyle\\iint}(2x+5y)dA&=\\underset{D_1}{\\displaystyle\\iint}(2x+5y)dA+\\underset{D_2}{\\displaystyle\\iint}(2x+5y)dA+\\underset{D_3}{\\displaystyle\\iint}(2x+5y)dA \\\\\r\n\r\n&=\\displaystyle\\int_{x=-2}^{x=0}\\displaystyle\\int_{y=0}^{y=(x+2)^2}(2x+5y)dydx+\\displaystyle\\int_{y=0}^{y=4}\\displaystyle\\int_{x=0}^{x=y-(1\/16)y^3}(2x+5y)dxdy+\\displaystyle\\int_{y=-4}^{y=0}\\displaystyle\\int_{x=-2}^{x=y-(1\/16)y^3}(2x+5y)dxdy \\\\\r\n\r\n&=\\displaystyle\\int_{x=-2}^{x=0}\\left[\\frac12(2+x)^2(20+24x+5x^2)\\right] \\\\\r\n\r\n&\\,\\,\\,+\\displaystyle\\int_{y=0}^{y=4}\\left[\\frac1{256}y^6-\\frac7{16}y^4+6y^2\\right] +\\displaystyle\\int_{y=-4}^{y=0}\\left[\\frac1{256}y^6-\\frac7{16}y^4+6y^2+10y-4\\right] \\\\\r\n\r\n&=\\frac{40}3+\\frac{1,664}{35}-\\frac{1,696}{35}=\\frac{1,304}{105}.\r\n\r\n\\end{align}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"Double Integrals over Nonrectangular Regions<\/h2>\n
double integrals over nonrectangular regions<\/h3>\n
Fubini’s theorem (strong form)<\/h3>\n