Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 3 : Multiple Integration

Study concepts, example questions & explanations for Calculus 3

Create An Account

All Calculus 3 Resources

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

Example Questions

Example Question #11 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{5}{(39\cdot (x^{2} + y^{2}))}+\frac{ (4sin(2x^{2} + 2y^{2})e^{(-2z)})}{3})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin with radii }\\&0.4\text{ and }1.4\\&\text{and length }0.62\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.685,0.729)\text{ and }\overrightarrow{u_2}=(0.218,-0.976)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-2.44

-0.2

0.3

1.22

Correct answer:

1.22

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{5}{(39\cdot (x^{2} + y^{2}))}+\frac{ (4sin(2x^{2} + 2y^{2})e^{(-2z)})}{3})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{5}{(39\cdot r)}+\frac{ (4\cdot re^{(-2z)}sin(2\cdot r^{2}))}{3})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.729}{0.685})=0.26\pi;\theta_2=arctan(\frac{-0.976}{0.218})=1.57\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{0.62}\int_{0.26\pi}^{1.57\pi}\int_{0.4}^{1.4}(\frac{5}{(39\cdot r)}+\frac{ (4\cdot re^{(-2z)}sin(2\cdot r^{2}))}{3})drd\theta dz=(\frac{(5ln(r))}{39}-\frac{ (2cos(r^{2})^{2}e^{(-2z)})}{3})d\theta dz|_{0.4}^{1.4}\\&\int_{0}^{0.62}\int_{0.26\pi}^{1.57\pi}(0.5538e^{(-2z)} + 0.1606)d\theta dz=(\theta\cdot (0.5538e^{(-2z)} + 0.1606))dz|_{0.26\pi}^{1.57\pi}\\&\int_{0}^{0.62}(2.279e^{(-2z)} + 0.661)dz=(0.661z - 1.139e^{(-2z)})|_{0}^{0.62}=1.22\end{align*}$

Report an Error

Example Question #11 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(4cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}) - 21cos(z + 2)e^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.81\\&\text{and length }1.37\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.685,0.729)\text{ and }\overrightarrow{u_2}=(-0.982,0.187)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-892.16

297.39

99.13

-74.35

Correct answer:

297.39

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(4cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}) - 21cos(z + 2)e^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(4\cdot rcos(\frac{(3\cdot r^{2})}{2}) - 21\cdot rcos(z + 2)e^{(\frac{(2\cdot r^{2})}{3})})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.729}{0.685})=0.26\pi;\theta_2=arctan(\frac{0.187}{-0.982})=0.94\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[cos(az)]=\frac{sin(az)}{a}\\&\int_{0}^{1.37}\int_{0.26\pi}^{0.94\pi}\int_{0}^{1.81}(4\cdot rcos(\frac{(3\cdot r^{2})}{2}) - 21\cdot rcos(z + 2)e^{(\frac{(2\cdot r^{2})}{3})})drd\theta dz=(\frac{(4sin(\frac{(3\cdot r^{2})}{2}))}{3}-\frac{ (63cos(z + 2)e^{(\frac{(2\cdot r^{2})}{3})})}{4})d\theta dz|_{0}^{1.81}\\&\int_{0}^{1.37}\int_{0.26\pi}^{0.94\pi}(- 124.1cos(z + 2) - 1.306)d\theta dz=(-\theta\cdot (124.1cos(z + 2) + 1.306))dz|_{0.26\pi}^{0.94\pi}\\&\int_{0}^{1.37}(- 265.2cos(z + 2) - 2.791)dz=(- 2.791z - 265.2sin(z + 2))|_{0}^{1.37}=297.39\end{align*}$

Report an Error

Example Question #461 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(4cos(x^{2} + y^{2}))}{9}+\frac{ (cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})e^{(-z)})}{4})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&0.86\\&\text{and length }1.1\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.941,0.339)\text{ and }\overrightarrow{u_2}=(0.588,-0.809)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.37

-5.59

5.59

1.12

Correct answer:

1.12

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(4cos(x^{2} + y^{2}))}{9}+\frac{ (cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})e^{(-z)})}{4})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(4\cdot rcos(r^{2}))}{9}+\frac{ (re^{(-z)}cos(\frac{(2\cdot r^{2})}{3}))}{4})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.339}{0.941})=0.11\pi;\theta_2=arctan(\frac{-0.809}{0.588})=1.7\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.1}\int_{0.11\pi}^{1.7\pi}\int_{0}^{0.86}(\frac{(4\cdot rcos(r^{2}))}{9}+\frac{ (re^{(-z)}cos(\frac{(2\cdot r^{2})}{3}))}{4})drd\theta dz=(\frac{(2sin(r^{2}))}{9}+\frac{ (3e^{(-z)}sin(\frac{(2\cdot r^{2})}{3}))}{16})d\theta dz|_{0}^{0.86}\\&\int_{0}^{1.1}\int_{0.11\pi}^{1.7\pi}(0.08875e^{(-z)} + 0.1498)d\theta dz=(\theta\cdot (0.08875e^{(-z)} + 0.1498))dz|_{0.11\pi}^{1.7\pi}\\&\int_{0}^{1.1}(0.4433e^{(-z)} + 0.7482)dz=(0.7482z - 0.4433e^{(-z)})|_{0}^{1.1}=1.12\end{align*}$

Report an Error

Example Question #11 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(3e^{(- 2x^{2} - 2y^{2})} -\frac{ cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2})}{(11\cdot 2^{(\frac{z}{2})})})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&0.84\\&\text{and length }1.77\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.809,0.588)\text{ and }\overrightarrow{u_2}=(-0.960,-0.279)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

2.69

-0.54

-5.38

8.07

Correct answer:

2.69

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(3e^{(- 2x^{2} - 2y^{2})} -\frac{ cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2})}{(11\cdot 2^{(\frac{z}{2})})})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(3\cdot re^{(-2\cdot r^{2})} -\frac{ (rcos(\frac{r^{2}}{2}))}{(11\cdot 2^{(\frac{z}{2})})})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.588}{0.809})=0.2\pi;\theta_2=arctan(\frac{-0.279}{-0.960})=1.09\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{1.77}\int_{0.2\pi}^{1.09\pi}\int_{0}^{0.84}(3\cdot re^{(-2\cdot r^{2})} -\frac{ (rcos(\frac{r^{2}}{2}))}{(11\cdot 2^{(\frac{z}{2})})})drd\theta dz=(-\frac{ (3e^{(-2\cdot r^{2})})}{4}-\frac{ sin(\frac{r^{2}}{2})}{(11\cdot 2^{(\frac{z}{2})})})d\theta dz|_{0}^{0.84}\\&\int_{0}^{1.77}\int_{0.2\pi}^{1.09\pi}(0.5671 -\frac{ 0.03141}{2^{(0.5z)}})d\theta dz=(-1.0\theta\cdot (\frac{0.03141}{2^{(0.5z)}}- 0.5671))dz|_{0.2\pi}^{1.09\pi}\\&\int_{0}^{1.77}(1.586 -\frac{ 0.08783}{2^{(0.5z)}})dz=(1.586z +\frac{ 0.2534}{2^{(0.5z)}})|_{0}^{1.77}=2.69\end{align*}$

Report an Error

Example Question #11 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(4e^{(x^{2} + y^{2})})}{25}+\frac{ (6cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})e^{(-2z)})}{23})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin with radii }\\&0.13\text{ and }0.99\\&\text{and length }1.7\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.536,0.844)\text{ and }\overrightarrow{u_2}=(-0.536,-0.844)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.83

-4.16

-0.42

0.14

Correct answer:

0.83

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(4e^{(x^{2} + y^{2})})}{25}+\frac{ (6cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})e^{(-2z)})}{23})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(4\cdot re^{(r^{2})})}{25}+\frac{ (6\cdot re^{(-2z)}cos(\frac{(3\cdot r^{2})}{2}))}{23})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.844}{0.536})=0.32\pi;\theta_2=arctan(\frac{-0.844}{-0.536})=1.32\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.7}\int_{0.32\pi}^{1.32\pi}\int_{0.13}^{0.99}(\frac{(4\cdot re^{(r^{2})})}{25}+\frac{ (6\cdot re^{(-2z)}cos(\frac{(3\cdot r^{2})}{2}))}{23})drd\theta dz=(\frac{(2e^{(r^{2})})}{25}+\frac{ (2e^{(-2z)}sin(\frac{(3\cdot r^{2})}{2}))}{23})d\theta dz|_{0.13}^{0.99}\\&\int_{0}^{1.7}\int_{0.32\pi}^{1.32\pi}(0.08431e^{(-2z)} + 0.1318)d\theta dz=(\theta\cdot (0.08431e^{(-2z)} + 0.1318))dz|_{0.32\pi}^{1.32\pi}\\&\int_{0}^{1.7}(0.2649e^{(-2z)} + 0.4141)dz=(0.4141z - 0.1324e^{(-2z)})|_{0}^{1.7}=0.83\end{align*}$

Report an Error

Example Question #12 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(19zsin(2x^{2} + 2y^{2}))}{2}-\frac{ sin(x^{2} + y^{2})}{47})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin with radii }\\&0.27\text{ and }1.58\\&\text{and length }1.44\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.982,0.187)\text{ and }\overrightarrow{u_2}=(-0.918,-0.397)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.97

1.45

-11.61

5.81

Correct answer:

5.81

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(19zsin(2x^{2} + 2y^{2}))}{2}-\frac{ sin(x^{2} + y^{2})}{47})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(19\cdot rzsin(2\cdot r^{2}))}{2}-\frac{ (rsin(r^{2}))}{47})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.187}{0.982})=0.06\pi;\theta_2=arctan(\frac{-0.397}{-0.918})=1.13\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0}^{1.44}\int_{0.06\pi}^{1.13\pi}\int_{0.27}^{1.58}(\frac{(19\cdot rzsin(2\cdot r^{2}))}{2}-\frac{ (rsin(r^{2}))}{47})drd\theta dz=(-\frac{(cos(r^{2})\cdot (893zcos(r^{2}) - 2))}{188})d\theta dz|_{0.27}^{1.58}\\&\int_{0}^{1.44}\int_{0.06\pi}^{1.13\pi}(1.693z - 0.01911)d\theta dz=(\theta\cdot (1.693z - 0.01911))dz|_{0.06\pi}^{1.13\pi}\\&\int_{0}^{1.44}(5.689z - 0.06424)dz=(0.993\cdot (1.693z - 0.01911)^{2})|_{0}^{1.44}=5.81\end{align*}$

Report an Error

Example Question #472 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (3e^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})})}{4}- 17cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2})e^{(-z)})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin with radii }\\&0.38\text{ and }1.63\\&\text{and length }0.76\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.588,0.809)\text{ and }\overrightarrow{u_2}=(0.661,-0.750)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

62.59

7.82

-31.3

-15.65

Correct answer:

-31.3

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{ (3e^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})})}{4}- 17cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2})e^{(-z)})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (3\cdot re^{(\frac{r^{2}}{2})})}{4}- 17\cdot re^{(-z)}cos(\frac{r^{2}}{2}))drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.809}{-0.588})=0.7\pi;\theta_2=arctan(\frac{-0.750}{0.661})=1.73\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{0.76}\int_{0.7\pi}^{1.73\pi}\int_{0.38}^{1.63}(-\frac{ (3\cdot re^{(\frac{r^{2}}{2})})}{4}- 17\cdot re^{(-z)}cos(\frac{r^{2}}{2}))drd\theta dz=(-\frac{ (3e^{(\frac{r^{2}}{2})})}{4}- 17e^{(-z)}sin(\frac{r^{2}}{2}))d\theta dz|_{0.38}^{1.63}\\&\int_{0}^{0.76}\int_{0.7\pi}^{1.73\pi}(- 15.28e^{(-z)} - 2.025)d\theta dz=(-\theta\cdot (15.28e^{(-z)} + 2.025))dz|_{0.7\pi}^{1.73\pi}\\&\int_{0}^{0.76}(- 49.43e^{(-z)} - 6.553)dz=(49.43e^{(-z)} - 6.553z)|_{0}^{0.76}=-31.3\end{align*}$

Report an Error

Example Question #18 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(39\cdot (\frac{2}{3})^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}e^{(2z)})}{2}-\frac{ (\frac{2}{(\frac{1}{2})^{(x^{2} + y^{2})}})}{11})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&0.73\\&\text{and length }1.01\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.861,0.509)\text{ and }\overrightarrow{u_2}=(0.397,-0.918)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

33.14

-198.82

-22.09

66.27

Correct answer:

66.27

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(39\cdot (\frac{2}{3})^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}e^{(2z)})}{2}-\frac{ (\frac{2}{(\frac{1}{2})^{(x^{2} + y^{2})}})}{11})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(39\cdot (\frac{2}{3})^{(\frac{(3\cdot r^{2})}{2})}\cdot re^{(2z)})}{2}-\frac{ (2\cdot r)}{(11\cdot (\frac{1}{2})^{(r^{2})})})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.509}{0.861})=0.17\pi;\theta_2=arctan(\frac{-0.918}{0.397})=1.63\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.01}\int_{0.17\pi}^{1.63\pi}\int_{0}^{0.73}(\frac{(39\cdot (\frac{2}{3})^{(\frac{(3\cdot r^{2})}{2})}\cdot re^{(2z)})}{2}-\frac{ (2\cdot r)}{(11\cdot (\frac{1}{2})^{(r^{2})})})drd\theta dz=(\frac{(13\cdot (\frac{2}{3})^{(\frac{(3\cdot r^{2})}{2})}e^{(2z)})}{(2ln(\frac{2}{3}))}-\frac{ 2^{(r^{2})}}{(11ln(2))})d\theta dz|_{0}^{0.73}\\&\int_{0}^{1.01}\int_{0.17\pi}^{1.63\pi}(4.438e^{(2z)} - 0.0586)d\theta dz=(1.0\theta\cdot (4.438e^{(2z)} - 0.0586))dz|_{0.17\pi}^{1.63\pi}\\&\int_{0}^{1.01}(20.36e^{(2z)} - 0.2688)dz=(10.18e^{(2z)} - 0.2688z)|_{0}^{1.01}=66.27\end{align*}$

Report an Error

Example Question #15 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(22e^{(-2z)})}{(2x^{2} + 2y^{2})}-\frac{ cos(x^{2} + y^{2})}{11})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin with radii }\\&0.24\text{ and }1.58\\&\text{and length }1.12\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.125,0.992)\text{ and }\overrightarrow{u_2}=(0.454,-0.891)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-6.9

8.63

-69.04

34.52

Correct answer:

34.52

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(22e^{(-2z)})}{(2x^{2} + 2y^{2})}-\frac{ cos(x^{2} + y^{2})}{11})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(11e^{(-2z)})}{r}-\frac{ (rcos(r^{2}))}{11})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.992}{0.125})=0.46\pi;\theta_2=arctan(\frac{-0.891}{0.454})=1.65\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.12}\int_{0.46\pi}^{1.65\pi}\int_{0.24}^{1.58}(\frac{(11e^{(-2z)})}{r}-\frac{ (rcos(r^{2}))}{11})drd\theta dz=(11e^{(-2z)}ln(r) -\frac{ sin(r^{2})}{22})d\theta dz|_{0.24}^{1.58}\\&\int_{0}^{1.12}\int_{0.46\pi}^{1.65\pi}(20.73e^{(-2z)} - 0.02472)d\theta dz=(\theta\cdot (20.73e^{(-2z)} - 0.02472))dz|_{0.46\pi}^{1.65\pi}\\&\int_{0}^{1.12}(77.5e^{(-2z)} - 0.09241)dz=(- 0.09241z - 38.75e^{(-2z)})|_{0}^{1.12}=34.52\end{align*}$

Report an Error

Example Question #473 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(e^{(z)}sin(x^{2} + y^{2}))}{10}-\frac{ (9cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{2})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.11\\&\text{and length }1.17\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.951,0.309)\text{ and }\overrightarrow{u_2}=(-0.707,0.707)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

1.44

-5.75

-17.26

23.01

Correct answer:

-5.75

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(e^{(z)}sin(x^{2} + y^{2}))}{10}-\frac{ (9cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{2})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(rsin(r^{2})e^{(z)})}{10}-\frac{ (9\cdot rcos(\frac{(2\cdot r^{2})}{3}))}{2})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.309}{0.951})=0.1\pi;\theta_2=arctan(\frac{0.707}{-0.707})=0.75\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.17}\int_{0.1\pi}^{0.75\pi}\int_{0}^{1.11}(\frac{(rsin(r^{2})e^{(z)})}{10}-\frac{ (9\cdot rcos(\frac{(2\cdot r^{2})}{3}))}{2})drd\theta dz=(-\frac{ (27sin(\frac{(2\cdot r^{2})}{3}))}{8}-\frac{ (cos(r^{2})e^{(z)})}{20})d\theta dz|_{0}^{1.11}\\&\int_{0}^{1.17}\int_{0.1\pi}^{0.75\pi}(0.03339e^{(z)} - 2.471)d\theta dz=(1.0\theta\cdot (0.03339e^{(z)} - 2.471))dz|_{0.1\pi}^{0.75\pi}\\&\int_{0}^{1.17}(0.06818e^{(z)} - 5.046)dz=(0.06818e^{(z)} - 5.046z)|_{0}^{1.17}=-5.75\end{align*}$

Report an Error

1748630261 temp image20250530 88155 1ejplln

View Calculus 3 Tutors

Christopher
Certified Tutor

Cornell University, Bachelor of Science, Applied and Engineering Physics. Delft University of Technology, Master of Science, ...

1748630223 temp image20250530 88155 1dxkcdu

View Calculus 3 Tutors

Zachary
Certified Tutor

Brown University, Bachelors, Mathematics.

1748630205 temp image20250530 88155 1xu8b4q

View Calculus 3 Tutors

Taylor
Certified Tutor

United States Military Academy, Current Undergrad, Physics, Nuclear Engineering.

All Calculus 3 Resources

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

Popular Subjects

Physics Tutors in Phoenix, SSAT Tutors in Seattle, Spanish Tutors in Houston, LSAT Tutors in San Francisco-Bay Area, Chemistry Tutors in Houston, Biology Tutors in Miami, SSAT Tutors in Philadelphia, English Tutors in Phoenix, GRE Tutors in Atlanta, Chemistry Tutors in San Francisco-Bay Area

Popular Courses & Classes

MCAT Courses & Classes in Boston, ISEE Courses & Classes in Atlanta, GMAT Courses & Classes in Miami, Spanish Courses & Classes in Boston, MCAT Courses & Classes in Phoenix, GRE Courses & Classes in Chicago, ISEE Courses & Classes in Philadelphia, ACT Courses & Classes in Washington DC, ISEE Courses & Classes in Denver, GMAT Courses & Classes in San Francisco-Bay Area

Popular Test Prep

SAT Test Prep in San Diego, GMAT Test Prep in New York City, SAT Test Prep in Los Angeles, ISEE Test Prep in Chicago, GRE Test Prep in Miami, SSAT Test Prep in Chicago, GRE Test Prep in New York City, GMAT Test Prep in Miami, GRE Test Prep in San Francisco-Bay Area, GRE Test Prep in Dallas Fort Worth

DMCA Complaint

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.

Please follow these steps to file a notice:

You must include the following:

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.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

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

* Please describe the copyrighted work so that it may be easily identified

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.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Calculus 3 : Multiple Integration

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #11 : Triple Integrals

Example Question #11 : Multiple Integration

Example Question #461 : Calculus 3

Example Question #11 : Triple Integrals

Example Question #11 : Multiple Integration

Example Question #12 : Multiple Integration

Example Question #472 : Calculus 3

Example Question #18 : Multiple Integration

Example Question #15 : Triple Integrals

Example Question #473 : Calculus 3

All Calculus 3 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: