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Calculus 3 : Multiple Integration

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #51 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}((\frac{1}{2})^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}sin(z + 2))dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.41\text{ and }1.72\\&\text{and length }1.09\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.891,0.454)\text{ and }\overrightarrow{u_2}=(0.992,-0.125)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.78

-0.16

1.55

-2.33

Correct answer:

0.78

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}((\frac{1}{2})^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}sin(z + 2))dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}((\frac{1}{2})^{(\frac{(3\cdot r^{2})}{2})}\cdot rsin(z + 2))drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.454}{-0.891})=0.85\pi;\theta_2=arctan(\frac{-0.125}{0.992})=1.96\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[sin(az)]=-\frac{cos(az)}{a}\\&\int_{0}^{1.09}\int_{0.85\pi}^{1.96\pi}\int_{0.41}^{1.72}((\frac{1}{2})^{(\frac{(3\cdot r^{2})}{2})}\cdot rsin(z + 2))drd\theta dz=(-\frac{sin(z + 2)}{(3\cdot 2^{(\frac{(3\cdot r^{2})}{2})}ln(2))})d\theta dz|_{0.41}^{1.72}\\&\int_{0}^{1.09}\int_{0.85\pi}^{1.96\pi}(0.3816sin(z + 2))d\theta dz=(0.3816\theta sin(z + 2))dz|_{0.85\pi}^{1.96\pi}\\&\int_{0}^{1.09}(1.331sin(z + 2))dz=(-1.331cos(z + 2))|_{0}^{1.09}=0.78\end{align*}$

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Example Question #52 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(12cos(z + 1)cos(x^{2} + y^{2}))}{5})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.21\text{ and }1.59\\&\text{and length }1.48\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.454,0.891)\text{ and }\overrightarrow{u_2}=(0.729,-0.685)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

1.01

-1.01

-0.13

0.51

Correct answer:

0.51

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(12cos(z + 1)cos(x^{2} + y^{2}))}{5})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(12\cdot rcos(r^{2})cos(z + 1))}{5})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.891}{-0.454})=0.65\pi;\theta_2=arctan(\frac{-0.685}{0.729})=1.76\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[cos(az)]=\frac{sin(az)}{a}\\&\int_{0}^{1.48}\int_{0.65\pi}^{1.76\pi}\int_{0.21}^{1.59}(-\frac{(12\cdot rcos(r^{2})cos(z + 1))}{5})drd\theta dz=(-\frac{(6cos(z + 1)sin(r^{2}))}{5})d\theta dz|_{0.21}^{1.59}\\&\int_{0}^{1.48}\int_{0.65\pi}^{1.76\pi}(-0.638cos(z + 1))d\theta dz=(-0.638\theta cos(z + 1))dz|_{0.65\pi}^{1.76\pi}\\&\int_{0}^{1.48}(-2.225cos(z + 1))dz=(-2.225sin(z + 1))|_{0}^{1.48}=0.51\end{align*}$

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Example Question #53 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(2cos(z + 2)cos(x^{2} + y^{2}))}{15})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.19\text{ and }1.91\\&\text{and length }1.02\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.876,0.482)\text{ and }\overrightarrow{u_2}=(0.031,-1.000)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.23

-0.12

-0.06

0.01

Correct answer:

-0.06

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(2cos(z + 2)cos(x^{2} + y^{2}))}{15})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(2\cdot rcos(r^{2})cos(z + 2))}{15})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.482}{-0.876})=0.84\pi;\theta_2=arctan(\frac{-1.000}{0.031})=1.51\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[cos(az)]=\frac{sin(az)}{a}\\&\int_{0}^{1.02}\int_{0.84\pi}^{1.51\pi}\int_{0.19}^{1.91}(-\frac{(2\cdot rcos(r^{2})cos(z + 2))}{15})drd\theta dz=(-\frac{(cos(z + 2)sin(r^{2}))}{15})d\theta dz|_{0.19}^{1.91}\\&\int_{0}^{1.02}\int_{0.84\pi}^{1.51\pi}(0.03475cos(z + 2))d\theta dz=(0.03475\theta cos(z + 2))dz|_{0.84\pi}^{1.51\pi}\\&\int_{0}^{1.02}(0.07314cos(z + 2))dz=(0.07314sin(z + 2))|_{0}^{1.02}=-0.06\end{align*}$

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Example Question #511 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(9cos(2x^{2} + 2y^{2})cos(z + 1))}{4})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.35\\&\text{and length }1.2\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.771,0.637)\text{ and }\overrightarrow{u_2}=(0.992,-0.125)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.02

0.01

-0.03

0.13

Correct answer:

-0.03

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(9cos(2x^{2} + 2y^{2})cos(z + 1))}{4})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(9\cdot rcos(z + 1)cos(2\cdot r^{2}))}{4})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.637}{-0.771})=0.78\pi;\theta_2=arctan(\frac{-0.125}{0.992})=1.96\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[cos(az)]=\frac{sin(az)}{a}\\&\int_{0}^{1.2}\int_{0.78\pi}^{1.96\pi}\int_{0}^{1.35}(-\frac{(9\cdot rcos(z + 1)cos(2\cdot r^{2}))}{4})drd\theta dz=(-\frac{(9cos(z + 1)sin(2\cdot r^{2}))}{16})d\theta dz|_{0}^{1.35}\\&\int_{0}^{1.2}\int_{0.78\pi}^{1.96\pi}(0.2714cos(z + 1))d\theta dz=(0.2714\theta cos(z + 1))dz|_{0.78\pi}^{1.96\pi}\\&\int_{0}^{1.2}(1.006cos(z + 1))dz=(1.006sin(z + 1))|_{0}^{1.2}=-0.03\end{align*}$

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Example Question #512 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(2cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2})sin(z + 1))dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.24\text{ and }1.39\\&\text{and length }1.1\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.905,0.426)\text{ and }\overrightarrow{u_2}=(0.809,-0.588)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-25.96

-2.88

4.33

8.65

Correct answer:

8.65

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(2cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2})sin(z + 1))dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(2\cdot rsin(z + 1)cos(\frac{r^{2}}{2}))drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.426}{0.905})=0.14\pi;\theta_2=arctan(\frac{-0.588}{0.809})=1.8\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[sin(az)]=-\frac{cos(az)}{a}\\&\int_{0}^{1.1}\int_{0.14\pi}^{1.8\pi}\int_{0.24}^{1.39}(2\cdot rsin(z + 1)cos(\frac{r^{2}}{2}))drd\theta dz=(2sin(z + 1)sin(\frac{r^{2}}{2}))d\theta dz|_{0.24}^{1.39}\\&\int_{0}^{1.1}\int_{0.14\pi}^{1.8\pi}(1.588sin(z + 1))d\theta dz=(1.588\theta sin(z + 1))dz|_{0.14\pi}^{1.8\pi}\\&\int_{0}^{1.1}(8.28sin(z + 1))dz=(-8.28cos(z + 1))|_{0}^{1.1}=8.65\end{align*}$

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Example Question #513 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-cos(z + 2)e^{(-\frac{ x^{2}}{2}-\frac{ y^{2}}{2})})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&0.69\\&\text{and length }1.92\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.729,0.685)\text{ and }\overrightarrow{u_2}=(-0.707,-0.707)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-2.17

1.08

0.54

-0.22

Correct answer:

1.08

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-cos(z + 2)e^{(-\frac{ x^{2}}{2}-\frac{ y^{2}}{2})})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-rcos(z + 2)e^{(-\frac{r^{2}}{2})})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.685}{0.729})=0.24\pi;\theta_2=arctan(\frac{-0.707}{-0.707})=1.25\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[cos(az)]=\frac{sin(az)}{a}\\&\int_{0}^{1.92}\int_{0.24\pi}^{1.25\pi}\int_{0}^{0.69}(-rcos(z + 2)e^{(-\frac{r^{2}}{2})})drd\theta dz=(cos(z + 2)e^{(-\frac{r^{2}}{2})})d\theta dz|_{0}^{0.69}\\&\int_{0}^{1.92}\int_{0.24\pi}^{1.25\pi}(-0.2118cos(z + 2))d\theta dz=(-0.2118\theta cos(z + 2))dz|_{0.24\pi}^{1.25\pi}\\&\int_{0}^{1.92}(-0.6722cos(z + 2))dz=(-0.6722sin(z + 2))|_{0}^{1.92}=1.08\end{align*}$

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Example Question #514 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(5e^{(-2z)}e^{(x^{2} + y^{2})})}{41})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.13\text{ and }1.91\\&\text{and length }1.67\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.397,0.918)\text{ and }\overrightarrow{u_2}=(0.249,-0.969)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

16.72

-4.18

0.7

-0.7

Correct answer:

-4.18

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(5e^{(-2z)}e^{(x^{2} + y^{2})})}{41})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(5\cdot re^{(r^{2})}e^{(-2z)})}{41})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.918}{0.397})=0.37\pi;\theta_2=arctan(\frac{-0.969}{0.249})=1.58\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.67}\int_{0.37\pi}^{1.58\pi}\int_{0.13}^{1.91}(-\frac{(5\cdot re^{(r^{2})}e^{(-2z)})}{41})drd\theta dz=(-\frac{(5e^{(r^{2} - 2z)})}{82})d\theta dz|_{0.13}^{1.91}\\&\int_{0}^{1.67}\int_{0.37\pi}^{1.58\pi}(-2.28e^{(-2z)})d\theta dz=(-2.28\theta e^{(-2z)})dz|_{0.37\pi}^{1.58\pi}\\&\int_{0}^{1.67}(-8.665e^{(-2z)})dz=(4.333e^{(-2z)})|_{0}^{1.67}=-4.18\end{align*}$

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Example Question #515 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(8e^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})}\cdot (z + 1)^{2})}{3})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.21\\&\text{and length }0.84\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.279,0.960)\text{ and }\overrightarrow{u_2}=(-0.637,-0.771)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

10.88

32.63

-43.5

-5.44

Correct answer:

10.88

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(8e^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})}\cdot (z + 1)^{2})}{3})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(8\cdot re^{(\frac{r^{2}}{2})}\cdot (z + 1)^{2})}{3})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.960}{-0.279})=0.59\pi;\theta_2=arctan(\frac{-0.771}{-0.637})=1.28\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0}^{0.84}\int_{0.59\pi}^{1.28\pi}\int_{0}^{1.21}(\frac{(8\cdot re^{(\frac{r^{2}}{2})}\cdot (z + 1)^{2})}{3})drd\theta dz=(\frac{(8e^{(\frac{r^{2}}{2})}\cdot (z + 1)^{2})}{3})d\theta dz|_{0}^{1.21}\\&\int_{0}^{0.84}\int_{0.59\pi}^{1.28\pi}(2.878\cdot (z + 1)^{2})d\theta dz=(2.878\theta \cdot (z + 1)^{2})dz|_{0.59\pi}^{1.28\pi}\\&\int_{0}^{0.84}(6.239\cdot (z + 1)^{2})dz=(2.08\cdot (z + 1)^{3})|_{0}^{0.84}=10.88\end{align*}$

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Example Question #516 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-3sin(z + 1)sin(x^{2} + y^{2}))dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.19\text{ and }0.84\\&\text{and length }0.94\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.685,0.729)\text{ and }\overrightarrow{u_2}=(0.031,-1.000)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-1.26

-0.32

0.21

2.53

Correct answer:

-1.26

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-3sin(z + 1)sin(x^{2} + y^{2}))dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-3\cdot rsin(r^{2})sin(z + 1))drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.729}{0.685})=0.26\pi;\theta_2=arctan(\frac{-1.000}{0.031})=1.51\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[sin(az)]=-\frac{cos(az)}{a}\\&\int_{0}^{0.94}\int_{0.26\pi}^{1.51\pi}\int_{0.19}^{0.84}(-3\cdot rsin(r^{2})sin(z + 1))drd\theta dz=(\frac{(3cos(r^{2})sin(z + 1))}{2})d\theta dz|_{0.19}^{0.84}\\&\int_{0}^{0.94}\int_{0.26\pi}^{1.51\pi}(-0.3572sin(z + 1))d\theta dz=(-0.3572\theta sin(z + 1))dz|_{0.26\pi}^{1.51\pi}\\&\int_{0}^{0.94}(-1.403sin(z + 1))dz=(1.403cos(z + 1))|_{0}^{0.94}=-1.26\end{align*}$

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Example Question #517 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(17cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{(5\cdot (z + 1))})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.26\text{ and }1.21\\&\text{and length }1.99\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.031,1.000)\text{ and }\overrightarrow{u_2}=(0.368,-0.930)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-1.29

-38.83

7.77

46.6

Correct answer:

7.77

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(17cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{(5\cdot (z + 1))})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(17\cdot rcos(\frac{(2\cdot r^{2})}{3}))}{(5\cdot (z + 1))})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{1.000}{0.031})=0.49\pi;\theta_2=arctan(\frac{-0.930}{0.368})=1.62\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0}^{1.99}\int_{0.49\pi}^{1.62\pi}\int_{0.26}^{1.21}(\frac{(17\cdot rcos(\frac{(2\cdot r^{2})}{3}))}{(5\cdot (z + 1))})drd\theta dz=(\frac{(51sin(\frac{(2\cdot r^{2})}{3}))}{(20\cdot (z + 1))})d\theta dz|_{0.26}^{1.21}\\&\int_{0}^{1.99}\int_{0.49\pi}^{1.62\pi}(\frac{1.997}{(z + 1)})d\theta dz=(\frac{(1.997\theta )}{(z + 1)})dz|_{0.49\pi}^{1.62\pi}\\&\int_{0}^{1.99}(\frac{7.09}{(z + 1)})dz=(7.09ln(z + 1))|_{0}^{1.99}=7.77\end{align*}$

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If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

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A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

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Charles Cohn Varsity Tutors LLC
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St. Louis, MO 63105

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* URL of copyrighted work (where your original material is located)

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I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

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Calculus 3 : Multiple Integration

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