{"id":1216,"date":"2025-10-15T15:27:34","date_gmt":"2025-10-15T15:27:34","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=1216"},"modified":"2025-10-29T15:56:34","modified_gmt":"2025-10-29T15:56:34","slug":"applications-of-double-integrals-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/applications-of-double-integrals-learn-it-2\/","title":{"raw":"Applications of Double Integrals: Learn It 2","rendered":"Applications of Double Integrals: Learn It 2"},"content":{"raw":"

Calculating Volumes, Areas, and Average Values<\/h2>\r\n

We can use double integrals over general regions to compute volumes, areas, and average values. The methods are the same as those we used for rectangular regions, but without that restriction, we can now solve a wider variety of problems.<\/p>\r\n\r\n

Find the volume of the solid bounded by the planes\u00a0[latex]x = 0[\/latex],\u00a0[latex]y = 0, \\ z = 0[\/latex], and\u00a0[latex]2x+3y+z=6[\/latex].[reveal-answer q=\"541521446\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"541521446\"]The solid is a tetrahedron with the base on the [latex]xy[\/latex]-plane and a height [latex]z=6-2x-3y[\/latex] The base is the region [latex]D[\/latex] bounded by the lines, [latex]x=0[\/latex],\u00a0[latex]y=0[\/latex], and\u00a0[latex]2x+3y=6[\/latex] where [latex]z=0[\/latex] (Figure 1). Note that we can consider the region [latex]D[\/latex] as Type I or as Type II, and we can integrate in both ways.[caption id=\"attachment_1349\" align=\"aligncenter\" width=\"845\"]\"This Figure 1.\u00a0A tetrahedron consisting of the three coordinate planes and the plane\u00a0[latex]\\small{z=6-2x-3y}[\/latex], with the base bound by\u00a0[latex]\\small{x=0}[\/latex],\u00a0[latex]\\small{y=0}[\/latex], and\u00a0[latex]\\small{2x+3y=6}[\/latex].[\/caption]\r\n

First, consider [latex]D[\/latex] as a Type I region, and hence [latex]{D} = {\\left \\{{(x,y)}{\\mid}{0} \\ {\\leq} \\ {x} \\ {\\leq} \\ {3,0} \\ {\\leq} \\ {y} \\ {\\leq} \\ {2-{\\frac{2}{3}}x} \\right \\}}[\/latex].<\/p>\r\n

Therefore, the volume is\r\n[latex]\\hspace{3cm}\\begin{align}\r\nV&=\\displaystyle\\int_{x=0}^{x=3}\\displaystyle\\int_{y=0}^{y=2-(2x\/3)}(6-2x-3y)dydx=\\displaystyle\\int_{x=0}^{x=3}\\left[\\left(6y-2xy-\\frac32y^2\\right)\\Bigg|_{y=0}^{y=2-(2x\/3)}\\right]dx \\\\\r\n&=\\displaystyle\\int_{x=0}^{x=3}\\left[\\frac23(x-3)^2\\right]dx=6\r\n\\end{align}[\/latex]<\/p>\r\n

Now consider [latex]D[\/latex] as a Type II region, so [latex]{D} = {\\left \\{{(x,y)}{\\mid}{0} \\ {\\leq} \\ {y} \\ {\\leq} \\ {2,0} \\ {\\leq} \\ {x} \\ {\\leq} \\ {3-{\\frac{3}{2}}y} \\right \\}}[\/latex]. In this calculation, the volume is\r\n[latex]\\hspace{3cm}\\begin{align}\r\nV&=\\displaystyle\\int_{y=0}^{y=2}\\displaystyle\\int_{x=0}^{x=3-(3y\/2)}(6-2x-3y)dxdy=\\displaystyle\\int_{y=0}^{y=2}\\left[(6x-x^2-3xy)\\bigg|_{x=0}^{x=3-(3y\/2)}\\right]dy \\\\\r\n&=\\displaystyle\\int_{y=0}^{y=2}\\left[\\frac94(y-2)^2\\right]dy=6\r\n\\end{align}[\/latex]<\/p>\r\n

Therefore, the volume is [latex]6[\/latex] cubic units.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n

Finding Areas Using Double Integrals<\/h3>\r\n

Finding the area of a rectangular region is easy, but finding the area of a nonrectangular region is not so easy. As we have seen, we can use double integrals to find a rectangular area. This approach is even more useful for finding the area of a general nonrectangular region, as stated in the next definition.<\/p>\r\n\r\n

\r\n

area of a nonrectangular region<\/h3>\r\nThe area of a plane-bounded region [latex]D[\/latex] is defined as the double integral:\r\n

[latex]\\underset{D}{\\displaystyle\\iint}{1dA}[\/latex].<\/p>\r\n\r\n<\/section>We have already seen how to find areas in terms of single integration. Here we are seeing another way of finding areas by using double integrals, which can be very useful.\r\n\r\n

Find the area of the region bounded below by the curve [latex]y=x^{2}[\/latex] and above by the line [latex]y=2x[\/latex] in the first quadrant (Figure 2).[caption id=\"attachment_1351\" align=\"aligncenter\" width=\"231\"]\"The Figure 2. The region bounded by\u00a0[latex]y=x^{2}[\/latex] and\u00a0[latex]y=2x[\/latex].[\/caption][reveal-answer q=\"501887643\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"501887643\"]\r\n

We just have to integrate the constant function [latex]f(x, y)=1[\/latex] over the region. Thus, the area [latex]A[\/latex] of the bounded region is\u00a0[latex]\\displaystyle\\int_{x=0}^{x=2}\\displaystyle\\int_{y=x^2}^{y=2x}dydx[\/latex] or\u00a0[latex]\\displaystyle\\int_{y=0}^{x=4}\\displaystyle\\int_{x=y\/2}^{x=\\sqrt{y}}dxdy[\/latex]:<\/p>\r\n

[latex]A=\\underset{D}{\\displaystyle\\iint}1dxdy=\\displaystyle\\int_{x=0}^{x=2}\\displaystyle\\int_{y=x^2}^{y=2x}1dydx=\\displaystyle\\int_{x=0}^{x=2}\\left[y\\bigg|_{y=x^2}^{y=2x}\\right]dx=\\displaystyle\\int_{x=0}^{x=2}(2x-x^2)dx=x^2-\\frac{x^3}3\\bigg|_0^2=\\frac43.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n

Average Value over General Regions<\/h3>\r\n

We can also use a double integral to find the average value of a function over a general region. This extends our single-variable approach to two dimensions.<\/p>\r\n\r\n

Average Value of a Function (Single-variable version)\r\n[latex]\\\\[\/latex]\r\n<\/strong>If [latex] f(x) [\/latex] is continuous on [latex] [a,b] [\/latex], then the average value of [latex] f(x) [\/latex] on [latex] [a,b] [\/latex] is:\r\n

[latex] f_{ave}=\\frac{1}{b-a} \\int_a^b f(x) dx [\/latex]<\/p>\r\n\r\n<\/section>The following definition is a direct extension of the formula above.\r\n\r\n

\r\n

average value of a function over a general region<\/h3>\r\nIf [latex]f(x, y)[\/latex] is integrable over a plane-bounded region [latex]D[\/latex] with positive area [latex]A(D)[\/latex] then the average value of the function is:\r\n

[latex]\\large{{f_{ave}} = {\\frac{1}{A(D)}}\\underset{D}{\\displaystyle\\iint}{f(x,y)}{dA}}[\/latex]<\/p>\r\nNote that the area is:\r\n

[latex]{A(D)}=\\underset{D}{\\displaystyle\\iint}{1dA}[\/latex].<\/p>\r\n\r\n<\/section>

Find the average value of the function [latex]f(x, y)=7xy^{2}[\/latex] on the region bounded by the line [latex]x=y[\/latex] and the curve [latex]x=\\sqrt{y}[\/latex] (Figure 3).[caption id=\"attachment_1353\" align=\"aligncenter\" width=\"380\"]\"The Figure 3. The region bounded by\u00a0[latex]\\small{x=y}[\/latex] and [latex]\\small{x=\\sqrt{y}}[\/latex][\/caption][reveal-answer q=\"471992476\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"471992476\"]\r\n

First find the area [latex]A(D)[\/latex] where the region [latex]D[\/latex] is given by the figure. We have<\/p>\r\n[latex]A(d)=\\underset{D}{\\displaystyle\\iint}1dA=\\displaystyle\\int_{y=0}^{y=1}\\displaystyle\\int_{x=y}^{x=\\sqrt{y}}1dxdy \\displaystyle\\int_{y=0}^{y=1}\\left[x\\bigg|_{x=y}^{x=\\sqrt{y}}\\right]dy=\\frac23y^{3\/2}-\\frac{y^2}2\\Bigg|_0^1=\\frac16.[\/latex]\r\n

Then the average value of the given function over this region is<\/p>\r\n[latex]\\begin{align}\r\n\r\nf_{ave} &= \\frac1{A(D)}\\underset{D}{\\displaystyle\\iint}f(x,y)dA = \\frac1{A(D)}\\displaystyle\\int_{y=0}^{y=1}\\displaystyle\\int_{x=y}^{x=\\sqrt{y}}yxy^2dxdy=\\frac1{1\/6}\\displaystyle\\int_{y=0}^{y=1}\\left[\\frac72x^2y^s\\Bigg|_{x=y}^{x=\\sqrt{y}}\\right]dy \\\\\r\n\r\n&=6\\displaystyle\\int_{y=0}^{y=1}\\left[\\frac72y^2(y-y^2)\\right]dy=6\\displaystyle\\int_{y=0}^{y=1}\\left[\\frac72(y^3-y^4)\\right]dy=\\frac{42}2(\\frac{y^4}4-\\frac{y^5}5)\\bigg|_0^1=\\frac{42}{40}=\\frac{21}{20}.\r\n\r\n\\end{align}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"

Calculating Volumes, Areas, and Average Values<\/h2>\n

We can use double integrals over general regions to compute volumes, areas, and average values. The methods are the same as those we used for rectangular regions, but without that restriction, we can now solve a wider variety of problems.<\/p>\n

Find the volume of the solid bounded by the planes\u00a0[latex]x = 0[\/latex],\u00a0[latex]y = 0, \\ z = 0[\/latex], and\u00a0[latex]2x+3y+z=6[\/latex].<\/p>\n