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Page 1

11-13.

Evaluate the double integral by first identifying it as the volume of a solid.

  1. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\iint_R 3dA,~~R={(x,y)~|~-2 \le x \le 2,~1 \le y \le 6}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_{-2}^2 3dx = 3(2-(-2)) = 12

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_1^6 12dy = 12(6-1) = 60

  2. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\iint_R (5-x)dA,~~R={(x,y)~|~0 \le x \le 5,~0 \le y \le 3}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_0^5 (5-x)dx
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= 5(5-0) - {1\over2}[x^2|_{x=0}^5
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= 25 - {1\over2}(25) = {25\over2}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_0^3 {25\over2}dy
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {25\over2}(3) = {75\over2}

  3. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\iint_R (4-2y)dA,~~R=[0,1] \times [0,1]

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_0^1 (4-2y)dy
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= 4(1-0) - 2{1\over2}(1-0)
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= 3

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_0^1 3dx = 3(1-0) = 3

  4. The integral \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\iint_R \sqrt{9-y^2}dA, where \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}R=[0,4] \times [0,2], represents the volume of a solid. Sketch the solid.

    Sketch of solid

Page 2

3-14.

Calculate the iterated integral.

  1. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_1^3\int_0^1 (1+4xy)dxdy

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_0^1 (1+4xy)dx = 1(1-0) + 4y{1\over2}(1-0)
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= 1 + 2y

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_1^3 (1+2y)dy = 1(3-1) + 2{1\over2}(3^2-1)
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= 2 + 8 = 10

  2. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_0^1\int_1^2 (4x^3-9x^2y^2)dxdy

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_1^2 (4x^3-9x^2y^2)dx = [x^4|{x=1}^2 - 3y^2[x^3|^2
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= (16-1) - 3y^2(8-1) = 15 - 21y^2

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_0^1 (15 - 21y^2)dy = 15(1-0) - 7[y^3|_{y=0}^1
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= 8

  3. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_0^2\int_0^{\pi/2} x\sin y\:dydx

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_0^{\pi/2} x\sin y\:dy = -x[\cos y|_{y=0}^{\pi/2}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= -x(0-1) = x

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_0^2 xdx = {1\over2}[x^2|_{x=0}^2
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= 2

  4. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_{\pi/6}^{\pi/2}\int_{-1}^5 \cos y\:dxdy

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_{-1}^5 \cos y\:dx = \cos y\:(5-(-1))
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= 6\cos y

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_{\pi/6}^{\pi/2} 6\cos y\:dy = 6[\sin y|_{y=\pi/6}^{\pi/2}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= 6(1-{1\over2}) = 3

  5. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_0^2\int_0^1 (2x+y)^8 dxdy

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_0^1 (2x+y)^8 dx = {9\over2}[(2x+y)^9|_{x=0}^1
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {9\over2}((2+y)^9 - y^9)

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}{9\over2}\int_0^2 (2+y)^9 dy - {9\over2}\int_0^2 y^9 dy
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {9\over20}[(2+y)^{10})|{y=0}^2 - {9\over20}[y^{10}|^2
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {9\over20}(4^{10} - 2^{10}) - {9\over20}(2^{10})
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {9\over20}(4^{10} - 2^{11})

  6. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_0^1\int_1^2 {xe^x \over y}dydx

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_1^2 {xe^x \over y}dy = xe^x[\ln|y|\:|_{y=1}^2
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= (\ln2 - e)xe^x

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int u\:dv = uv - \int v\:du
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~ u = x
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~ dv = e^x dx
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~ du = dx
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~ v = e^x

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}(\ln2-e)\int_0^1 xe^x dx
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= (\ln2-e)([xe^x|_{x=0}^1 - \int_0^1 e^xdx)
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= (\ln2-e)(e - (e - 1))
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= \ln2 - e

15-22.

Calculate the double integral.

  1. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\iint_R \cos(x+2y)dA,~~R={(x,y)~|~0 \le x \le \pi,~0 \le y \le \pi/2}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_0^\pi \cos(x+2y)dx = [\sin(x+2y)|_{x=0}^\pi
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= \sin(\pi+2y) - \sin(2y)

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_0^{\pi/2} \sin(\pi+2y)dy + \int_0^{\pi/2} \sin(2y)dy
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= -[\cos(\pi+2y)|{x=0}^{\pi/2} - [\cos(2y)|^{\pi/2}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= -\cos2\pi + \cos\pi - \cos\pi + \cos0
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= \cos0 - \cos2\pi = 0

  2. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\iint_R {x \over x^2+y^2}dA,~~R=[1,2] \times [0,1]

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~ u = x^2 + y^2
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~ du = 2x\:dx

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_1^2 {x \over x^2+y^2}dx
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= \int_{x=1}^2 {1\over 2u}du
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {1\over2}[\ln|u|\:|{x=1}^2
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {1\over2}[\ln(x^2 + y^2)|
    ^2
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {1\over2}\ln(4 + y^2) - {1\over2}\ln(1 + y^2)

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}{1\over2}\int_0^1 \ln(4+y^2)dy - {1\over2}\int_0^1 \ln(1+y^2)dy

Page 3

  1. Find the volume of the solid enclosed by the surface \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}z=1+e^x\sin y and the planes \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}x=\pm 1,y=0,y=\pi, and \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}z=0.

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_{-1}^1 (1 + e^x\sin y)dx
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= 1(1-(-1)) + e\sin y - {1\over e}\sin y
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= 2 + e\sin y - {1\over e}\sin y

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_0^\pi 2dy + \int_0^\pi (e - {1\over e})\sin y\:dy
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= 2\pi + (e-{1\over e})(\sin\pi - \sin0)
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= 2\pi

Page 4

1-6.

Evaluate the iterated integral.

  1. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_0^4\int_0^{\sqrt y} xy^2dxdy

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_0^{\sqrt y} xy^2dx
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {1\over2}y^2[x^2|_{x=0}^{\sqrt y}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= y^3

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_0^4 y^3dy
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {1\over4}[y^4|_{y=0}^4
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= 64

  2. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_0^1\int_{2x}^2 (x-y)dydx

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_{2x}^2 (x-y)dy
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= x[y|{y=2x}^2 - {1\over2}[y^2|^2
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= 2x - 2x^2 - 2 + 2x^2
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= 2x - 2

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_0^1 (2x-2)dx
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= [x^2|{x=0}^1 - 2[x|^1
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= 1 - 2(1)
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= -1

  3. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_0^1\int_{x^2}^x (1+2y)dydx

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_{x^2}^x (1+2y)dy
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= x - x^2 + [y^2|_{y=x^2}^x
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= x - x^2 + x^2 - x^4
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= x - x^4

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_0^1 (x-x^4)dx
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {1\over2}[x^2|{x=0}^1 - {1\over5}[x^5|^1
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {1\over2} - {1\over5} = {7\over10}

  4. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_0^2\int_y^{2y} xy\:dxdy

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_y^{2y} xy\:dx
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {1\over2}y[x^2|_{x=y}^{2y}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= 2y^3 - {1\over2}y^3 = {3\over2}y^3

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}{3\over2}\int_0^2 y^2dy
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {1\over2}[y^3|_{y=0}^2
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= 4

7-18.

Evaluate the double integral.

  1. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\iint_D x\cos y\:dA,~~D is bounded by \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}y=0,y=x^2,x=1

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_0^{x^2} x\cos y\:dy
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= x[\sin y|_{y=0}^{x^2}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= x\sin{x^2} - 0

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}x^2 = 0 \Rightarrow x = 0

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_0^1 x\sin{x^2}dx
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= -{1\over2}[\cos{x^2}|_0^1
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= -{1\over2}(\cos 1 - \cos 0)
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= -{1\over2}\cos 1 + {1\over2}

  2. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\iint_D (x+y)dA,~~D is bounded by \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}y=\sqrt x and \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}y=x^2

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_{x^2}^{\sqrt x} (x+y)dy
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= x(\sqrt x - x^2) + {1\over2}(x - x^4)
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= x^{3/2} - x^3 + {1\over2}x - {1\over2}x^4

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\sqrt x = x^2 \Rightarrow x \in {0,1}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_0^1 x^{3/2} - \int_0^1 x^3 + {1\over2}\int_0^1 x - {1\over2}\int_0^1 x^4
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {2\over5} - {1\over4} + {1\over4} - {1\over10}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {3\over10}

  3. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\iint_D y^3dA,~~D is the triangular region with verticies \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}(0,2),(1,1),(3,2)

    Plot of <span class="katex-math">\newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}D</span>

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}y = -x + 2 \Rightarrow x = 2 - y
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}y = {1\over2}x + {1\over2} \Rightarrow x = 2y - 1

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}D:~{(x,y)~|~1 \le y \le 2~\land~2 - y \le x \le 2y - 1}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_{2-y}^{2y-1} y^3dx
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= y^3(2y - 1 - 2 + y)
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= 3y^4 - 3y^3

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}3\int_1^2 y^4dy - 3\int_1^2 y^3dy
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {3\over5}(32-1) - {3\over4}(16-1)
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {93\over5} - {45\over4}

  4. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\iint_D xy^2dA,~~D is enclosed by \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}x=0 and \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}x=\sqrt{1-y^2}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}y^2\int_0^{\sqrt{1-y^2}} x\:dx
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {1\over2}y^2(1-y^2)
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {1\over2}y^2 - {1\over2}y^4

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}0=\sqrt{1-y^2} \Rightarrow y^2 = 1 \Rightarrow y \in {-1,1}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}{1\over2}\int_{-1}^1 y^2dy - {1\over2}\int_{-1}^1 y^4dy
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {1\over6}[y^3|{y=-1}^1 - {1\over10}[y^5|^1
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {1\over6}(2) - {1\over10}(2)
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {2\over15}

  5. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\iint_D (2x-y)dA,~~D is bounded by the circle with center the origin and radius 2

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}x^2 + y^2 = 4
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}D:~{(x,y)~|~-\sqrt{4-y^2} \le x \le \sqrt{4-y^2}~\land~-2 \le y \le 2}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}} (2x-y)dx

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= [x^2|_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}} - y[x|_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= -2y\sqrt{4-y^2}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~ u = 4 - y^2
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~ du = -2y\:dy

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\int_{-2}^2 (-2y)\sqrt{4-y^2}dy
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {2\over3}[(4-y^2)^{3/2}|_{y=-2}^2
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {2\over3}(0-0)
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= 0

  6. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\iint_D 2xy\:dA,~~D is the triangular region with verticies \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}(0,0),(1,2), and \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}(0,3)

    Plot of region

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}D:~{(x,y)~|~0 \le x \le 2~\land~{1\over2}x \le y \le 3-x}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}2x\int_{x/2}^{3-x} y\:dy
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= x((3-x)^2 - {1\over4}x^2)
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= x(9 - 6x + x^2 - {1\over4}x^2)
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= 9x - 6x^2 + {3\over4}x^3

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}9\int_0^2 x\:dx - 6\int_0^2 x^2dx + {3\over4}\int_0^2 x^3
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= 9({1\over2})(2^2-0) - 6({1\over3})(2^3-0) + {3\over4}({1\over4})(2^4-0)
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= 18 - 16 + 3 = 5

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