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Calculus 3 : Triple Integrals

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Example Questions

Example Question #561 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(2sin(3z)cos(x^{2} + y^{2}))}{13})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&0.67\\&\text{and length }0.69\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.941,0.339)\text{ and }\overrightarrow{u_2}=(0.031,-1.000)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.00641

0.06

-0.00641

-0.03

Correct answer:

-0.03

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(2sin(3z)cos(x^{2} + y^{2}))}{13})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(2\cdot rcos(r^{2})sin(3z))}{13})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.339}{-0.941})=0.89\pi;\theta_2=arctan(\frac{-1.000}{0.031})=1.51\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[sin(az)]=-\frac{cos(az)}{a}\\&\int_{0}^{0.69}\int_{0.89\pi}^{1.51\pi}\int_{0}^{0.67}(-\frac{(2\cdot rcos(r^{2})sin(3z))}{13})drd\theta dz=(-\frac{(sin(r^{2})sin(3z))}{13})d\theta dz|_{0}^{0.67}\\&\int_{0}^{0.69}\int_{0.89\pi}^{1.51\pi}(-0.03338sin(3z))d\theta dz=(-0.03338\theta sin(3z))dz|_{0.89\pi}^{1.51\pi}\\&\int_{0}^{0.69}(-0.06502sin(3z))dz=(0.02167cos(3z))|_{0}^{0.69}=-0.03\end{align*}$

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Example Question #562 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(3\cdot (\frac{3}{2})^{(x^{2} + y^{2})})}{49})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.49\text{ and }1.16\\&\text{and length }1.86\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.031,1.000)\text{ and }\overrightarrow{u_2}=(0.771,-0.637)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.12

-0.35

1.4

-0.17

Correct answer:

-0.35

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(3\cdot (\frac{3}{2})^{(x^{2} + y^{2})})}{49})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(3\cdot (\frac{3}{2})^{(r^{2})}\cdot r)}{49})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{1.000}{-0.031})=0.51\pi;\theta_2=arctan(\frac{-0.637}{0.771})=1.78\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0}^{1.86}\int_{0.51\pi}^{1.78\pi}\int_{0.49}^{1.16}(-\frac{(3\cdot (\frac{3}{2})^{(r^{2})}\cdot r)}{49})drd\theta dz=(-\frac{(3\cdot (\frac{3}{2})^{(r^{2})})}{(98ln(\frac{3}{2}))})d\theta dz|_{0.49}^{1.16}\\&\int_{0}^{1.86}\int_{0.51\pi}^{1.78\pi}(-0.04706)d\theta dz=(-0.04706\theta)dz|_{0.51\pi}^{1.78\pi}\\&\int_{0}^{1.86}(-0.1878)dz=(-0.1878z)|_{0}^{1.86}=-0.35\end{align*}$

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Example Question #563 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-7\cdot 3^ze^{(- 2x^{2} - 2y^{2})})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.1\\&\text{and length }1.89\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.339,0.941)\text{ and }\overrightarrow{u_2}=(-0.876,-0.482)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

122.44

12.24

-48.98

-24.49

Correct answer:

-24.49

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-7\cdot 3^ze^{(- 2x^{2} - 2y^{2})})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-7\cdot 3^z\cdot re^{(-2\cdot r^{2})})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.941}{0.339})=0.39\pi;\theta_2=arctan(\frac{-0.482}{-0.876})=1.16\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{1.89}\int_{0.39\pi}^{1.16\pi}\int_{0}^{1.1}(-7\cdot 3^z\cdot re^{(-2\cdot r^{2})})drd\theta dz=(\frac{(7\cdot 3^ze^{(-2\cdot r^{2})})}{4})d\theta dz|_{0}^{1.1}\\&\int_{0}^{1.89}\int_{0.39\pi}^{1.16\pi}(-1.594\cdot 3^z)d\theta dz=(-1.594\cdot 3^z\theta)dz|_{0.39\pi}^{1.16\pi}\\&\int_{0}^{1.89}(-3.857\cdot 3^z)dz=(-3.511\cdot 3^z)|_{0}^{1.89}=-24.49\end{align*}$

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Example Question #571 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(2e^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}e^{(-2z)})}{43})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.88\\&\text{and length }1.99\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.536,0.844)\text{ and }\overrightarrow{u_2}=(0.750,-0.661)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

31.21

-1.73

1.73

-5.2

Correct answer:

-5.2

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(2e^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}e^{(-2z)})}{43})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(2\cdot re^{(-2z)}e^{(\frac{(3\cdot r^{2})}{2})})}{43})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.844}{-0.536})=0.68\pi;\theta_2=arctan(\frac{-0.661}{0.750})=1.77\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.99}\int_{0.68\pi}^{1.77\pi}\int_{0}^{1.88}(-\frac{(2\cdot re^{(-2z)}e^{(\frac{(3\cdot r^{2})}{2})})}{43})drd\theta dz=(-\frac{(2e^{(\frac{(3\cdot r^{2})}{2}- 2z)})}{129})d\theta dz|_{0}^{1.88}\\&\int_{0}^{1.99}\int_{0.68\pi}^{1.77\pi}(-3.095e^{(-2z)})d\theta dz=(-3.095\theta e^{(-2z)})dz|_{0.68\pi}^{1.77\pi}\\&\int_{0}^{1.99}(-10.6e^{(-2z)})dz=(5.3e^{(-2z)})|_{0}^{1.99}=-5.2\end{align*}$

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Example Question #111 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-2sin(3z)e^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.45\text{ and }1.07\\&\text{and length }1.94\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.998,0.063)\text{ and }\overrightarrow{u_2}=(0.218,-0.976)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.03

0.38

-0.06

0.01

Correct answer:

-0.06

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.063}{0.998})=0.02\pi;\theta_2=arctan(\frac{-0.976}{0.218})=1.57\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[sin(az)]=-\frac{cos(az)}{a}\\&\int_{0}^{1.94}\int_{0.02\pi}^{1.57\pi}\int_{0.45}^{1.07}(-2\cdot rsin(3z)e^{(-\frac{(3\cdot r^{2})}{2})})drd\theta dz=(\frac{(2sin(3z)e^{(-\frac{(3\cdot r^{2})}{2})})}{3})d\theta dz|_{0.45}^{1.07}\\&\int_{0}^{1.94}\int_{0.02\pi}^{1.57\pi}(-0.3723sin(3z))d\theta dz=(-0.3723\theta sin(3z))dz|_{0.02\pi}^{1.57\pi}\\&\int_{0}^{1.94}(-1.813sin(3z))dz=(0.6044cos(3z))|_{0}^{1.94}=-0.06\end{align*}$

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Example Question #112 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(2sin(x^{2} + y^{2}))}{(3\cdot (z + 1)^{3})})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.04\\&\text{and length }0.9\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.729,0.685)\text{ and }\overrightarrow{u_2}=(-0.218,-0.976)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.4

0.07

-0.13

-0.27

Correct answer:

-0.13

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.685}{-0.729})=0.76\pi;\theta_2=arctan(\frac{-0.976}{-0.218})=1.43\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0}^{0.9}\int_{0.76\pi}^{1.43\pi}\int_{0}^{1.04}(-\frac{(2\cdot rsin(r^{2}))}{(3\cdot (z + 1)^{3})})drd\theta dz=(\frac{cos(r^{2})}{(3\cdot (3z + 3z^{2} + z^{3} + 1))})d\theta dz|_{0}^{1.04}\\&\int_{0}^{0.9}\int_{0.76\pi}^{1.43\pi}(-\frac{0.1767}{(z + 1)^{3}})d\theta dz=(-\frac{(0.1767\theta )}{(z + 1)^{3}})dz|_{0.76\pi}^{1.43\pi}\\&\int_{0}^{0.9}(-\frac{0.3719}{(z + 1)^{3}})dz=(\frac{0.186}{(z + 1)^{2}})|_{0}^{0.9}=-0.13\end{align*}$

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Example Question #113 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(15cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})sin(z + 2))}{2})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.28\text{ and }0.79\\&\text{and length }0.5\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.861,0.509)\text{ and }\overrightarrow{u_2}=(0.562,-0.827)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.6

-1.79

0.36

7.16

Correct answer:

-1.79

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(15cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})sin(z + 2))}{2})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(15\cdot rsin(z + 2)cos(\frac{(3\cdot r^{2})}{2}))}{2})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.509}{-0.861})=0.83\pi;\theta_2=arctan(\frac{-0.827}{0.562})=1.69\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[sin(az)]=-\frac{cos(az)}{a}\\&\int_{0}^{0.5}\int_{0.83\pi}^{1.69\pi}\int_{0.28}^{0.79}(-\frac{(15\cdot rsin(z + 2)cos(\frac{(3\cdot r^{2})}{2}))}{2})drd\theta dz=(-\frac{(5sin(z + 2)sin(\frac{(3\cdot r^{2})}{2}))}{2})d\theta dz|_{0.28}^{0.79}\\&\int_{0}^{0.5}\int_{0.83\pi}^{1.69\pi}(-1.72sin(z + 2))d\theta dz=(-1.72\theta sin(z + 2))dz|_{0.83\pi}^{1.69\pi}\\&\int_{0}^{0.5}(-4.647sin(z + 2))dz=(4.647cos(z + 2))|_{0}^{0.5}=-1.79\end{align*}$

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Example Question #114 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-31cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})sin(z + 1))dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.21\text{ and }0.8\\&\text{and length }0.73\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.809,0.588)\text{ and }\overrightarrow{u_2}=(-0.876,0.482)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

37.68

6.28

-12.56

-2.09

Correct answer:

-12.56

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-31cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})sin(z + 1))dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-31\cdot rsin(z + 1)cos(\frac{(2\cdot r^{2})}{3}))drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.588}{0.809})=0.2\pi;\theta_2=arctan(\frac{0.482}{-0.876})=0.84\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[sin(az)]=-\frac{cos(az)}{a}\\&\int_{0}^{0.73}\int_{0.2\pi}^{0.84\pi}\int_{0.21}^{0.8}(-31\cdot rsin(z + 1)cos(\frac{(2\cdot r^{2})}{3}))drd\theta dz=(-\frac{(93sin(z + 1)sin(\frac{(2\cdot r^{2})}{3}))}{4})d\theta dz|_{0.21}^{0.8}\\&\int_{0}^{0.73}\int_{0.2\pi}^{0.84\pi}(-8.938sin(z + 1))d\theta dz=(-8.938\theta sin(z + 1))dz|_{0.2\pi}^{0.84\pi}\\&\int_{0}^{0.73}(-17.97sin(z + 1))dz=(17.97cos(z + 1))|_{0}^{0.73}=-12.56\end{align*}$

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Example Question #115 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{7}{(\frac{3}{2})^{(x^{2} + y^{2})}})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.28\\&\text{and length }1.74\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.368,0.930)\text{ and }\overrightarrow{u_2}=(-0.339,-0.941)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-2.94

17.64

3.53

-52.91

Correct answer:

17.64

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{7}{(\frac{3}{2})^{(x^{2} + y^{2})}})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(7\cdot r)}{(\frac{3}{2})^{(r^{2})}})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.930}{-0.368})=0.62\pi;\theta_2=arctan(\frac{-0.941}{-0.339})=1.39\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0}^{1.74}\int_{0.62\pi}^{1.39\pi}\int_{0}^{1.28}(\frac{(7\cdot r)}{(\frac{3}{2})^{(r^{2})}})drd\theta dz=(-\frac{(7\cdot (\frac{2}{3})^{(r^{2})})}{(2ln(\frac{3}{2}))})d\theta dz|_{0}^{1.28}\\&\int_{0}^{1.74}\int_{0.62\pi}^{1.39\pi}(4.19)d\theta dz=(4.19\theta)dz|_{0.62\pi}^{1.39\pi}\\&\int_{0}^{1.74}(10.14)dz=(10.14z)|_{0}^{1.74}=17.64\end{align*}$

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Example Question #116 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(e^{(- x^{2} - y^{2})}\cdot (z + 1)^{2})}{5})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.3\text{ and }1.44\\&\text{and length }1.87\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.031,1.000)\text{ and }\overrightarrow{u_2}=(0.941,-0.339)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-2.62

15.7

0.52

-7.85

Correct answer:

-2.62

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{1.000}{0.031})=0.49\pi;\theta_2=arctan(\frac{-0.339}{0.941})=1.89\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0}^{1.87}\int_{0.49\pi}^{1.89\pi}\int_{0.3}^{1.44}(-\frac{(re^{(-r^{2})}\cdot (z + 1)^{2})}{5})drd\theta dz=(\frac{(e^{(-r^{2})}\cdot (z + 1)^{2})}{10})d\theta dz|_{0.3}^{1.44}\\&\int_{0}^{1.87}\int_{0.49\pi}^{1.89\pi}(-0.07882\cdot (z + 1)^{2})d\theta dz=(-0.07882\theta \cdot (z + 1)^{2})dz|_{0.49\pi}^{1.89\pi}\\&\int_{0}^{1.87}(-0.3467\cdot (z + 1)^{2})dz=(-0.1156\cdot (z + 1)^{3})|_{0}^{1.87}=-2.62\end{align*}$

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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.

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Calculus 3 : Triple Integrals

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