Skip to main content

Tutorial 10: Multiple Integrals

  1. Evaluate the following:

    a. ∫12∫23(2x−y)dydx\displaystyle\int_{1}^{2} \int_{2}^{3}(2 x-y) d y d x

    Solution
    ∫12∫23(2x−y)dydx=∫12[(2x)y−(y22)]23dx=∫12[((2x)(3)−322)−((2x)(2)−222)]dx=∫12[(6x−4.5)−(4x−2)]dx=∫122x−2.5dx=[x2−2.5x]12=[(4−5)−(1−2.5)]=4−5−1+2.5=0.5\begin{aligned} \int_1^2 \int_2^3(2 x-y) d y d x &=\int_1^2\left[(2 x) y-\left(\frac{y^2}{2}\right)\right]_2^3 d x \\ &=\int_1^2\left[\left((2 x)(3)-\frac{3^2}{2}\right)-\left((2 x)(2)-\frac{2^2}{2}\right)\right] d x \\ &=\int_1^2[(6 x-4.5)-(4 x-2)] d x=\int_1^2 2 x-2.5 d x \\ &=\left[x^2-2.5 x\right]_1^2=[(4-5)-(1-2.5)] \\ &=4-5-1+2.5=0.5 \end{aligned}

    b. ∫15∫−12(x−5y)dxdy\displaystyle\int_{1}^{5} \int_{-1}^{2}(x-5 y) d x d y

    Solution
    ∫15∫−12(x−5y)dxdy=∫15[x22−(5y)x]−12dy=∫15[(2−10y)−(0.5+5y)]dy=∫151.5−15ydy=[1.5y−15y22]15=[(7.5−187.5)−(1.5−7.5)]=−174\begin{aligned} \int_1^5 \int_{-1}^2(x-5 y) d x d y &=\int_1^5\left[\frac{x^2}{2}-(5 y) x\right]_{-1}^2 d y \\ &=\int_1^5[(2-10 y)-(0.5+5 y)] d y=\int_1^5 1.5-15 y d y \\ &=\left[1.5 y-\frac{15 y^2}{2}\right]_1^5=[(7.5-187.5)-(1.5-7.5)]=-174 \end{aligned}

    c. ∫13∫25(2x−3y)dxdy\displaystyle\int_{1}^{3} \int_{2}^{5}(2 x-3 y) d x d y

    Solution
    ∫13∫25(2x−3y)dxdy=∫13[2x22−(3y)x]25dy=∫13[(52−3(5)y)−(22−3(2)y)]dy=∫13[21−9y]dy=[21y−9y22]13=[(21(3)−9(3)22)−(21(1)−9(1)22)]=(63−40.5)−(21−4.5)=6\begin{aligned} \int_1^3 \int_2^5(2 x-3 y) d x d y&=\int_1^3\left[\frac{2 x^2}{2}-(3 y) x\right]_2^5 d y \\ &=\int_1^3\left[\left(5^2-3(5) y\right)-\left(2^2-3(2) y\right)\right] d y \\ &=\int_1^3[21-9 y] d y \\ &=\left[21 y-\frac{9 y^2}{2}\right]_1^3 \\ &=\left[\left(21(3)-\frac{9(3)^2}{2}\right)-\left(21(1)-\frac{9(1)^2}{2}\right)\right] \\ &=(63-40.5)-(21-4.5)=6 \end{aligned}

  2. Evaluate each of the following integrals over the given region DD :

    a. ∬D(1−12x2−12y2)dA\displaystyle\iint_{D}\left(1-\frac{1}{2} x^{2}-\frac{1}{2} y^{2}\right) d A \quad where D\mathrm{D} is the region given by 0≤x≤1,0≤y≤10 \leq \mathrm{x} \leq 1,0 \leq \mathrm{y} \leq 1

    Solution
    Figure for Solution 2a
    • Because the region R is a square, it is both vertically and horizontally simple, and you can use either order of integration.
    • Choose dydy dxdx by placing a vertical representative rectangle in the region, as shown in the figure at the right.
    ∬R(1−12x2−12y2)dA=∫01∫01(1−12x2−12y2)dydx=∫01[(1−12x2)y−y36]01dx=∫01(56−12x2)dx=[56x−x36]01=23\begin{aligned} \iint_R\left(1-\frac{1}{2} x^2-\frac{1}{2} y^2\right) d A &=\int_0^1 \int_0^1\left(1-\frac{1}{2} x^2-\frac{1}{2} y^2\right) d y d x \\ &=\int_0^1\left[\left(1-\frac{1}{2} x^2\right) y-\frac{y^3}{6}\right]_0^1 d x \\ &=\int_0^1\left(\frac{5}{6}-\frac{1}{2} x^2\right) d x \\ &=\left[\frac{5}{6} x-\frac{x^3}{6}\right]_0^1=\frac{2}{3} \end{aligned}

    b. ∬D(4xy−y3)dA\displaystyle\iint_{D}\left(4 x y-y^{3}\right) d A \quad where D\mathrm{D} is the region given by y=x0.5\mathrm{y}=\mathrm{x}^{0.5} and y=x3\mathrm{y}=\mathrm{x}^{3}

    Solution
    ∬D(4xy−y3)dA0≤x≤1 and x3≤y≤x0.5=∫01∫x3x0.54xy−y3dydx=∫01[4xy22−y44]x3x0.5dx=∫01[(4x(x0.5)22−(x0.5)44)−(4x(x3)22−(x3)44)]dx=∫01[(4x22−x24)−(4x72−x124)]dx=∫01[74x2−2x7+14x12]dx=[712x3−14x8+152x13]01=55156\begin{aligned} \iint_D\left(4 x y-y^3\right) d A \quad & 0 \leq x \leq 1 \text { and } x^3 \leq y \leq x^{0.5} \\ &=\int_0^1 \int_{x^3}^{x^{0.5}} 4 x y-y^3 d y d x \\ &=\int_0^1\left[\frac{4 x y^2}{2}-\frac{y^4}{4}\right]_{x^3}^{x^{0.5}} d x \\ &=\int_0^1\left[\left(\frac{4 x\left(x^{0.5}\right)^2}{2}-\frac{\left(x^{0.5}\right)^4}{4}\right)-\left(\frac{4 x\left(x^3\right)^2}{2}-\frac{\left(x^3\right)^4}{4}\right)\right] d x &=\int_0^1\left[\left(\frac{4 x^2}{2}-\frac{x^2}{4}\right)-\left(\frac{4 x^7}{2}-\frac{x^{12}}{4}\right)\right] d x \\ &=\int_0^1\left[\frac{7}{4} x^2-2 x^7+\frac{1}{4} x^{12}\right] d x \\ &=\left[\frac{7}{12} x^3-\frac{1}{4} x^8+\frac{1}{52} x^{13}\right]_0^1 \\ &=\frac{55}{156} \end{aligned}

  3. Evaluate the following integral:

    ∬D42y2−12xdA where D={(x,y)∣0≤x≤4,(x−2)2≤y≤6}\iint_{D} 42 y^{2}-12 x d A \quad \text { where } D=\left\{(x, y) \mid 0 \leq x \leq 4,(x-2)^{2} \leq y \leq 6\right\}
    Solution
    Figure for Solution 3
    ∬D42y2−12xdA=∫04∫(x−2)2642y2−12xdydx=∫04[14y3−12xy](x−2)26dx=∫043024−72x−14(x−2)6+12x(x−2)2dx=∫043024−24x−48x2+12x3−14(x−2)6dx=[3024x−12x2−16x3+3x4−2(x−2)7]04=11136\begin{aligned} \iint_D 42 y^2-12 x d A &=\int_0^4 \int_{(x-2)^2}^6 42 y^2-12 x d y d x \\ &=\int_0^4\left[14 y^3-12 x y\right]_{(x-2)^2}^6 d x \\ &=\int_0^4 3024-72 x-14(x-2)^6+12 x(x-2)^2 d x \\ &=\int_0^4 3024-24 x-48 x^2+12 x^3-14(x-2)^6 d x \\ &=\left[3024 x-12 x^2-16 x^3+3 x^4-2(x-2)^7\right]_0^4 \\ &=11136 \end{aligned}

  4. Evaluate the following integral over the indicated rectangle (a) by integrating with respect to xx first and (b) with respect to yy first.

    ∬R12x−18ydAR=[−1,4]×[2,3]\iint_{R} 12 x-18 y d A \quad R=[-1,4] \times[2,3]
    Solution

    (a)

    ∬R12x−18ydA=∫23∫−1412x−18ydxdy=∫23[6x2−18xy]−14dy=∫2390−90ydy=[90y−45y2]23=−135\begin{aligned} \iint_R 12 x-18 y d A &=\int_2^3 \int_{-1}^4 12 x-18 y d x d y \\ &=\int_2^3\left[6 x^2-18 x y\right]_{-1}^4 d y=\int_2^3 90-90 y d y \\ &=\left[90 y-45 y^2\right]_2^3=-135 \end{aligned}

    (b)

    ∬R12x−18ydA=∫−14∫2312x−18ydydx=∫−14[12xy−9y2]23dx=∫−1412x−45dx=[6x2−45x]−14=−135\begin{aligned} \iint_R 12 x-18 y d A &=\int_{-1}^4 \int_2^3 12 x-18 y d y d x \\ &=\int_{-1}^4\left[12 x y-9 y^2\right]_2^3 d x=\int_{-1}^4 12 x-45 d x \\ &=\left[6 x^2-45 x\right]_{-1}^4=-135 \end{aligned}

  5. Compute the following double integral over the indicated rectangle

    a. ∬R6yx−2y3dAR=[1,4]×[0,3]\displaystyle \iint_{R} 6 y \sqrt{x}-2 y^{3} d A \quad R=[1,4] \times[0,3]

    Solution
    ∬R6yx−2y3dA=∫03∫146yx12−2y3dxdy=∫03(4yx32−2xy3)∣14dy=∫0328y−6y3dy=(14y2−32y4)∣03=92\begin{aligned} \iint_R 6 y \sqrt{x}-2 y^3 d A &=\int_0^3 \int_1^4 6 y x^{\frac{1}{2}}-2 y^3 d x d y \\ &=\left.\int_0^3\left(4 y x^{\frac{3}{2}}-2 x y^3\right)\right|_1 ^4 d y=\int_0^3 28 y-6 y^3 d y \\ &=\left.\left(14 y^2-\frac{3}{2} y^4\right)\right|_0 ^3=\frac{9}{2} \end{aligned}

    b. ∬Ryey2−4xdAR=[0,2]×[0,8]\displaystyle\iint_{R} y e^{y^{2}-4 x} d A \quad R=[0,2] \times[0, \sqrt{8}]

    Solution

    Integrating with respect to yy first,

    ∬Ryey2−4xdA=∫02∫08yey2−4xdydx=∫02[12ey2−4x]08dx=∫0212(e8−4x−e−4x)dx=(12(−14e8−4x+14e−4x))∣02=18(e8+e−8−2)=372.3698\begin{aligned} \iint_R y e^{y^{2}-4 x} d A &=\int_0^2 \int_0^{\sqrt{8}} y e^{y^{2}-4 x} d y d x \\ &=\int_0^2\left[\frac{1}{2} e^{y^{2}-4 x}\right]_0^{\sqrt{8}} d x =\int_0^2 \frac{1}{2}\left(e^{8-4 x}-e^{-4 x}\right) d x \\ &=\left.\left(\frac{1}{2}\left(-\frac{1}{4} e^{8-4 x}+\frac{1}{4} e^{-4 x}\right)\right)\right|_0 ^2 \\ &=\frac{1}{8}\left(e^8+e^{-8}-2\right)=372.3698 \end{aligned}

  6. Evaluate:

    a. ∫−31∫−12∫016xyzdydxdz\displaystyle\int_{-3}^{1} \int_{-1}^{2} \int_{0}^{1} 6 x y z d y d x d z

    Solution
    ∫−31∫−12∫016xyzdydxdz=∫−31∫−12[3xy2z]01dxdz=∫−31∫−123xzdxdz=∫−31[32x2z]−12dz=∫−31[6z−32z]dz=∫−31[92z]dz=[94z2]−31=94−94(9)=94−814=−724=−18\begin{aligned} \int_{-3}^1 \int_{-1}^2 \int_0^1 6 x y z d y d x d z &=\int_{-3}^1 \int_{-1}^2\left[3 x y^2 z\right]_0^1 d x d z \\ &=\int_{-3}^1 \int_{-1}^2 3 x z d x d z \\ &=\int_{-3}^1\left[\frac{3}{2} x^2 z\right]_{-1}^2 d z=\int_{-3}^1\left[6 z-\frac{3}{2} z\right] d z \\ &=\int_{-3}^1\left[\frac{9}{2} z\right] d z=\left[\frac{9}{4} z^2\right]_{-3}^1 \\ &=\frac{9}{4}-\frac{9}{4}(9)=\frac{9}{4}-\frac{81}{4} \\ &=\frac{-72}{4}=-18 \end{aligned}

    b. ∫04∫−2−1∫12xydxdydz\displaystyle \int_{0}^{4} \int_{-2}^{-1} \int_{1}^{2} x y d x d y d z

    Solution
    ∫04∫−2−1∫12xydxdydz=∫04∫−2−1[x22y]12dydz=∫04∫−2−11.5ydydz=∫04[1.5y22]−2−1dz=∫04−2.25dz=[−2.25z]04=−9\begin{aligned} \int_0^4 \int_{-2}^{-1} \int_1^2 x y d x d y d z &=\int_0^4 \int_{-2}^{-1}\left[\frac{x^2}{2} y\right]_1^2 d y d z=\int_0^4 \int_{-2}^{-1} 1.5 y d y d z \\ &=\int_0^4\left[\frac{1.5 y^2}{2}\right]_{-2}^{-1} d z=\int_0^4-2.25 d z \\ &=[-2.25 z]_0^4=-9 \end{aligned}

  7. Evaluate:

    a. ∭Bxyz2dV0≤x≤1−1≤y≤20≤z≤3\displaystyle\iiint_{B} x y z^{2} d V\qquad\begin{aligned} &0 \leq x \leq 1 \\ &-1 \leq y \leq 2 \\ &0 \leq z \leq 3\end{aligned}

    Solution
    ∭Bxyz2dV=∫03∫−12∫01xyz2dxdydz=∫03∫−12[x2yz22]x=0x=1dydz=∫03∫−12yz22dydz=∫03[y2z24]y=−1y=2dz=∫033z24dz=[z34]03=274\begin{aligned} \iiint_{B} x y z^{2} d V&=\int_0^3 \int_{-1}^2 \int_0^1 x y z^2 d x d y d z\\ &=\int_0^3 \int_{-1}^2\left[\frac{x^2 y z^2}{2}\right]_{x=0}^{x=1} d y d z=\int_0^3 \int_{-1}^2 \frac{y z^2}{2} d y d z\\ &=\int_0^3\left[\frac{y^2 z^2}{4}\right]_{y=-1}^{y=2} d z=\int_0^3 \frac{3 z^2}{4} d z\\ &=\left[\frac{z^3}{4}\right]_0^3=\frac{27}{4} \end{aligned}

    b. ∭B4x2y−z3dzdydx2≤x≤3−1≤y≤40≤z≤1\displaystyle\iiint_{B} 4x^2y-z^3 dz dy dx \qquad\begin{aligned} &2 \leq x \leq 3 \\ &-1 \leq y \leq 4 \\ &0 \leq z \leq 1\end{aligned}

    Solution
    ∭B4x2y−z3dzdydx=∫23∫−14[4x2yz−14z4]01dydx=∫23∫−14(4x2y−14)dydx=∫23[2x2y2−14y]−14dx=∫23(30x2−54)dx=[10x3−54x]23=7554\begin{aligned} \iiint_{B} 4x^2y-z^3 dz dy dx&=\int_2^3 \int_{-1}^4\left[4 x^2 y z-\frac{1}{4} z^4\right]_0^1 d y d x \\ &=\int_2^3 \int_{-1}^4 \left(4 x^2 y-\frac{1}{4}\right) d y d x \\ &=\int_2^3\left[2 x^2 y^2-\frac{1}{4} y\right]_{-1}^4 d x=\int_2^3 \left(30 x^2-\frac{5}{4}\right) d x \\ &=\left[10 x^3-\frac{5}{4} x\right]_2^3=\frac{755}{4} \end{aligned}

  8. Integrate the function (x,y,z)=xy(x, y, z)=x y over the volume enclose by the planes z=x+yz=x+y and z=0z=0, and between the surfaces y=x2y=x^{2} and x=y2x=y^{2}.

    Solution

    Since zz is expressed as a function of (x,y)(x, y), the integration is done in the zz direction first. Considering the xyxy-plane, the two curves meet at (0,0)(0, 0) and (1,1)(1, 1). The integral is:

    ∫01∫x2x∫0x+yf(x,y,z)dzdydx=∫01∫x2x∫0x+yxydzdydx=∫01∫x2xxy[z]z=0z=x+ydydx=∫01∫x2xxy(x+y)dydx∫01∫x2xx2y+y2xdydx=∫01[x2y22+y3x3]y=x2y=xdx=12∫01(x2x2−x2(x2)2)dx=12∫01(x3−x6)dx+13∫01(xx−(x2)52−x7)dx=12[x44−x77]01+13[2x727−x88]01=12(14−17)+13(27−18)=328\begin{aligned} \int_0^1 \int_{x^2}^{\sqrt{x}} \int_0^{x+y} f(x, y, z) d z d y d x &=\int_0^1 \int_{x^2}^{\sqrt{x}} \int_0^{x+y} x y d z d y d x \\ &=\int_0^1 \int_{x^2}^{\sqrt{x}} x y[z]_{z=0}^{z=x+y} d y d x \\ &=\int_0^1 \int_{x^2}^{\sqrt{x}} x y(x+y) d y d x \int_0^1 \int_{x^2}^{\sqrt{x}} x^2 y+y^2 x d y d x \\ &=\int_0^1\left[\frac{x^2 y^2}{2}+\frac{y^3 x}{3}\right]_{y=x^2}^{y=\sqrt{x}} d x \\ &=\frac{1}{2} \int_0^1\left(x^2 \sqrt{x}{ }^2-x^2\left(x^2\right)^2\right) d x \\ &=\frac{1}{2} \int_0^1\left(x^3-x^6\right) d x+\frac{1}{3} \int_0^1\left(\sqrt{x} x-\left(x^2\right)^{\frac{5}{2}}-x^7\right) d x \\ &=\frac{1}{2}\left[\frac{x^4}{4}-\frac{x^7}{7}\right]_0^1+\frac{1}{3}\left[\frac{2 x^{\frac{7}{2}}}{7}-\frac{x^8}{8}\right]_0^1 \\ &=\frac{1}{2}\left(\frac{1}{4}-\frac{1}{7}\right)+\frac{1}{3}\left(\frac{2}{7}-\frac{1}{8}\right)=\frac{3}{28} \end{aligned}

  9. Find the mass for the following square lamina, represented by the unit square (shown below) with variable density ρ(x,y)=(x+y+2)g/cm2\rho(x, y)=(x+y+2) \mathrm{g} / \mathrm{cm}^{2}.

    Figure for Question 9

    Figure 1. A region R representing a lamina

    Solution
    M=∬R(x+y+2)dA=∫01∫01(x+y+2)dxdy=∫01[12x2+x(y+2)]01dy=∫0152+ydy=[52y+y22]01=52+12=3g\begin{aligned} M=\iint_R(x+y+2) d A &=\int_0^1 \int_0^1(x+y+2) d x d y \\ &=\int_0^1\left[\frac{1}{2} x^2+x(y+2)\right]_0^1 d y \\ &=\int_0^1 \frac{5}{2}+y d y \\ &=\left[\frac{5}{2} y+\frac{y^2}{2}\right]_0^1 \\ &=\frac{5}{2}+\frac{1}{2} \\ &=3 g \end{aligned}

  10. Find the mass and center of mass for the following solid region QQ bounded by x+2y+3zx+2 y+3 z =6=6 and the coordinate planes (as shown below) and has density ρ(x,y,z)=x2yz\rho(x, y, z)=x^{2} y z.

    Figure for Question 10

    Figure 2: Finding the mass of a three-dimensional solid Q.

    Solution

    The region QQ is a tetrahedron meeting the axes at the points (6,0,0),(0,3,0)(6,0,0),(0,3,0) and (0,0,2)(0,0,2). To find the limits of integration, let z=0z=0 in the slated plane z=13(6−x−2y)z=\frac{1}{3}(6-x-2 y).

    Then for xx and yy, find the projection of QQ onto the xyx y-plane, which is bounded by the axes and the line x+2y=6x+2 y=6. Therefore, the mass (m)(m) is:

    m=∭Qρ(x,y,z)dV=∫x=0x=6∫y=0y=12(6−x)∫z=0z=13(6−x−2y)x2yzdzdydx=10835\begin{aligned} m=\iiint_Q \rho(x, y, z) d V &=\int_{x=0}^{x=6} \int_{y=0}^{y=\frac{1}{2}(6-x)} \int_{z=0}^{z=\frac{1}{3}(6-x-2 y)} x^2 y z d z d y d x \\ &=\frac{108}{35} \end{aligned}

    To find the center of mass of Q\mathrm{Q}, we need to find the moments about the xyx y-plane, the xzx z-plane and the yzy z-plane:

    Mxy=∭Qzρ(x,y,z)dV=∫x=0x=6∫y=0y=12(6−x)∫z=0z=13(6−x−2y)x2yz2dzdydx=5435Mxz=∭Qyρ(x,y,z)dV=∫x=0x=6∫y=0y=12(6−x)∫z=0z=13(6−x−2y)x2y2zdzdydx=8135Myz=∭Qxρ(x,y,z)dV=∫x=0x=6∫y=0y=12(6−x)∫z=0z=13(6−x−2y)x3yzdzdydx=24335\begin{aligned} M_{x y}&=\iiint_Q z \rho(x, y, z) d V&&=\int_{x=0}^{x=6} \int_{y=0}^{y=\frac{1}{2}(6-x)} \int_{z=0}^{z=\frac{1}{3}(6-x-2 y)} x^2 y z^2 d z d y d x\\ &&&=\frac{54}{35} \\\\ M_{x z}&=\iiint_Q y \rho(x, y, z) d V&&=\int_{x=0}^{x=6} \int_{y=0}^{y=\frac{1}{2}(6-x)} \int_{z=0}^{z=\frac{1}{3}(6-x-2 y)} x^2 y^2 z d z d y d x\\ &&&=\frac{81}{35}\\\\ M_{y z}&=\iiint_Q x \rho(x, y, z) d V&&=\int_{x=0}^{x=6} \int_{y=0}^{y=\frac{1}{2}(6-x)} \int_{z=0}^{z=\frac{1}{3}(6-x-2 y)} x^3 y z d z d y d x\\ &&&=\frac{243}{35} \end{aligned}

    Hence the center of mass is:

    xˉ=Myzmyˉ=Mxzmzˉ=Mxymxˉ=Myzm=243/35108/35=243108yˉ=Mxzm=81/35108/35=81108zˉ=Mxym=54/35108/35=54108=2.25=0.75=0.5\begin{aligned} \bar{x}&=\frac{M_{y z}}{m} & \bar{y}&=\frac{M_{x z}}{m} & \bar{z}&=\frac{M_{x y}}{m} \\ \bar{x}&=\frac{M_{y z}}{m}=\frac{243 / 35}{108 / 35}=\frac{243}{108} & \bar{y}&=\frac{M_{x z}}{m}=\frac{81 / 35}{108 / 35}=\frac{81}{108} & \bar{z}&=\frac{M_{x y}}{m}=\frac{54 / 35}{108 / 35}=\frac{54}{108} \\ &=2.25 & &=0.75 & &=0.5 \end{aligned}

    The center of mass: (2.25,0.75,0.5)\quad(2.25,0.75,0.5)