We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 3 : Triple Integration in Cylindrical Coordinates

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 #1 : Triple Integration In Cylindrical Coordinates

Evaluate $\int \int \int_D y \ dV$ , where is the region below the plane z=x+1 , above the plane and between the cylinders x^2+y^2=1 , and x^2+y^2=9 .

Possible Answers:

$\frac{\pi}{2}$

$\pi$

$\frac{\pi}{4}$

$2 \pi$

Correct answer:

Explanation:

We need to figure out our boundaries for our integral.

We need to convert everything into cylindrical coordinates. Remeber we are above the plane, this means we are above z=0 .

$0\leq z \leq x+1 \rightarrow 0\leq z \leq r\cos(\theta)+1$

The region is between two circles x^2+y^2=1 , and x^2+y^2=9 .

This means that

$0 \leq \theta \leq 2\pi$

$1\leq r \leq 3$

$\int \int \int_D y \ dV=\int_{0}^{2\pi} \int_{1}^{3} \int_{0}^{r\cos(\theta)+1} r\sin(\theta) r\ dz \ dr \ d\theta$

$=\int_{0}^{2\pi} \int_{1}^{3} \int_{0}^{r\cos(\theta)+1} r^2\sin(\theta) \ dz \ dr \ d\theta$

$=\int_{0}^{2\pi} \int_{1}^{3} zr^2\sin(\theta) \Big|_{0}^{r\cos(\theta)+1} \ dr \ d\theta$

$=\int_{0}^{2\pi} \int_{1}^{3} (r\cos(\theta)+1)r^2\sin(\theta)-0 \ dr \ d\theta$

$=\int_{0}^{2\pi} \int_{1}^{3} r^3\sin(\theta)\cos(\theta)+r^2\sin(\theta) \ dr \ d\theta$

$=\int_{0}^{2\pi} \frac{1}{4}r^4\sin(\theta)\cos(\theta)+\frac{1}{3}r^3\sin(\theta) \Big|_{1}^{3}\ d\theta$

$=\int_{0}^{2\pi} (\frac{1}{4}(3)^4\sin(\theta)\cos(\theta)+\frac{1}{3}(3)^3\sin(\theta))-(\frac{1}{4}(1)^4\sin(\theta)\cos(\theta)+\frac{1}{3}(1)^3\sin(\theta)) \ d\theta$

$=\int_{0}^{2\pi} (\frac{81}{4}\sin(\theta)\cos(\theta)+9\sin(\theta))-(\frac{1}{4}\sin(\theta)\cos(\theta)+\frac{1}{3}\sin(\theta)) \ d\theta$

$=\int_{0}^{2\pi} 20\sin(\theta)\cos(\theta)+\frac{26}{3}\sin(\theta) \ d\theta$

$= -\frac{1}{2}\cdot20\cos^2(\theta)-\frac{26}{3}\cos(\theta)\Big|_{0}^{2\pi}$

$= -\frac{1}{2}\cdot20\cos^2(2\pi)-\frac{26}{3}\cos(2\pi)-(-\frac{1}{2}\cdot20\cos^2(0)-\frac{26}{3}\cos(0))$

$= -\frac{1}{2}\cdot20(1)-\frac{26}{3}(1)-(-\frac{1}{2}\cdot20(1)-\frac{26}{3}(1))$

$= -10-\frac{26}{3}-(-10-\frac{26}{3})=0$

Report an Error

Example Question #2 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (3cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))}{41}-\frac{ (sin(z + 1)cos(x^{2} + y^{2}))}{25})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin with radii }\\&0.23\text{ and }1.56\\&\text{and length }1.97\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.094,0.996)\text{ and }\overrightarrow{u_2}=(0.637,-0.771)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.03

0.01

-0.07

-0.02

Correct answer:

0.03

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{ (3cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))}{41}-\frac{ (sin(z + 1)cos(x^{2} + y^{2}))}{25})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (3\cdot rcos(\frac{(3\cdot r^{2})}{2}))}{41}-\frac{ (rcos(r^{2})sin(z + 1))}{25})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.996}{-0.094})=0.53\pi;\theta_2=arctan(\frac{-0.771}{0.637})=1.72\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\end{align*}$

$\begin{align*}\\&\int[sin(az)]=-\frac{cos(az)}{a}\\&\int_{0}^{1.97}\int_{0.53\pi}^{1.72\pi}\int_{0.23}^{1.56}(-\frac{ (3\cdot rcos(\frac{(3\cdot r^{2})}{2}))}{41}-\frac{ (rcos(r^{2})sin(z + 1))}{25})drd\theta dz=(-\frac{ sin(\frac{(3\cdot r^{2})}{2})}{41}-\frac{ (sin(r^{2})sin(z + 1))}{50})d\theta dz|_{0.23}^{1.56}\\&\int_{0}^{1.97}\int_{0.53\pi}^{1.72\pi}(0.01381 - 0.01195sin(z + 1))d\theta dz=(-1.0\theta\cdot (0.01195sin(z + 1) - 0.01381))dz|_{0.53\pi}^{1.72\pi}\\&\int_{0}^{1.97}(0.05165 - 0.04467sin(z + 1))dz=0.03\end{align*}$

Report an Error

Example Question #1 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})}}{4}-\frac{ (49e^{(20z)}e^{(x^{2} + y^{2})})}{5})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.92\\&\text{and length }1.65\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.218,0.976)\text{ and }\overrightarrow{u_2}=(0.661,-0.750)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

$-2.08871\cdot10^{15}$

$1.39247\cdot10^{15}$

$-8.35483\cdot10^{15}$

$8.35483\cdot10^{15}$

Correct answer:

$-8.35483\cdot10^{15}$

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})}}{4}-\frac{ (49e^{(20z)}e^{(x^{2} + y^{2})})}{5})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(re^{(-\frac{(2\cdot r^{2})}{3})})}{4}-\frac{ (49\cdot re^{(r^{2})}e^{(20z)})}{5})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.976}{0.218})=0.43\pi;\theta_2=arctan(\frac{-0.750}{0.661})=1.73\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.65}\int_{0.43\pi}^{1.73\pi}\int_{0}^{1.92}(\frac{(re^{(-\frac{(2\cdot r^{2})}{3})})}{4}-\frac{ (49\cdot re^{(r^{2})}e^{(20z)})}{5})drd\theta dz=(-\frac{ (49e^{(20z + r^{2})})}{10}-\frac{ (3e^{(-\frac{(2\cdot r^{2})}{3})})}{16})d\theta dz|_{0}^{1.92}\\&\int_{0}^{1.65}\int_{0.43\pi}^{1.73\pi}(0.1714 - 190.6e^{(20z)})d\theta dz=(-1.0\theta\cdot (190.6e^{(20z)} - 0.1714))dz|_{0.43\pi}^{1.73\pi}\\&\int_{0}^{1.65}(0.7002 - 778.5e^{(20z)})dz=(0.7002z - 38.92e^{(20z)})|_{0}^{1.65}=-8.35483\cdot10^{15}\end{align*}$

Report an Error

Example Question #3 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(5cos(x^{2} + y^{2}) +\frac{ (3\cdot 2^ze^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})})}{13})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin with radii }\\&0.31\text{ and }1.89\\&\text{and length }1.31\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.988,0.156)\text{ and }\overrightarrow{u_2}=(-1.000,0.031)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-4.55

2.27

27.28

-27.28

Correct answer:

-4.55

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(5cos(x^{2} + y^{2}) +\frac{ (3\cdot 2^ze^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})})}{13})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(5\cdot rcos(r^{2}) +\frac{ (3\cdot 2^z\cdot re^{(-\frac{(3\cdot r^{2})}{2})})}{13})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.156}{0.988})=0.05\pi;\theta_2=arctan(\frac{0.031}{-1.000})=0.99\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{1.31}\int_{0.05\pi}^{0.99\pi}\int_{0.31}^{1.89}(5\cdot rcos(r^{2}) +\frac{ (3\cdot 2^z\cdot re^{(-\frac{(3\cdot r^{2})}{2})})}{13})drd\theta dz=(\frac{(5sin(r^{2}))}{2}-\frac{ (2^ze^{(-\frac{(3\cdot r^{2})}{2})})}{13})d\theta dz|_{0.31}^{1.89}\\&\int_{0}^{1.31}\int_{0.05\pi}^{0.99\pi}(0.06623\cdot 2^z - 1.283)d\theta dz=(\theta\cdot (0.06623\cdot 2^z - 1.283))dz|_{0.05\pi}^{0.99\pi}\\&\int_{0}^{1.31}(0.1956\cdot 2^z - 3.789)dz=(0.2822\cdot 2^z - 3.789z)|_{0}^{1.31}=-4.55\end{align*}$

Report an Error

Example Question #4 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(- 16e^{(x^{2} + y^{2})} -\frac{ (5e^{(2z)})}{(19\cdot (x^{2} + y^{2}))})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin with radii }\\&0.42\text{ and }1.42\\&\text{and length }1.72\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.368,0.930)\text{ and }\overrightarrow{u_2}=(-0.156,-0.988)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-1196.57

39.89

239.31

-239.31

Correct answer:

-239.31

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(- 16e^{(x^{2} + y^{2})} -\frac{ (5e^{(2z)})}{(19\cdot (x^{2} + y^{2}))})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(- 16\cdot re^{(r^{2})} -\frac{ (5e^{(2z)})}{(19\cdot r)})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.930}{-0.368})=0.62\pi;\theta_2=arctan(\frac{-0.988}{-0.156})=1.45\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.72}\int_{0.62\pi}^{1.45\pi}\int_{0.42}^{1.42}(- 16\cdot re^{(r^{2})} -\frac{ (5e^{(2z)})}{(19\cdot r)})drd\theta dz=(- 8e^{(r^{2})} -\frac{ (5e^{(2z)}ln(r))}{19})d\theta dz|_{0.42}^{1.42}\\&\int_{0}^{1.72}\int_{0.62\pi}^{1.45\pi}(- 0.3206e^{(2z)} - 50.55)d\theta dz=(-1.0\theta\cdot (0.3206e^{(2z)} + 50.55))dz|_{0.62\pi}^{1.45\pi}\\&\int_{0}^{1.72}(- 0.8359e^{(2z)} - 131.8)dz=(- 131.8z - 0.4179e^{(2z)})|_{0}^{1.72}=-239.31\end{align*}$

Report an Error

Example Question #1 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(2\cdot 4^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})})}{7}-\frac{ (7\cdot 3^zcos(x^{2} + y^{2}))}{2})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin with radii }\\&0.27\text{ and }1.86\\&\text{and length }1.09\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.876,0.482)\text{ and }\overrightarrow{u_2}=(0.368,-0.930)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

16.76

-4.19

100.58

-83.82

Correct answer:

16.76

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(2\cdot 4^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})})}{7}-\frac{ (7\cdot 3^zcos(x^{2} + y^{2}))}{2})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(2\cdot 4^{(\frac{r^{2}}{2})}\cdot r)}{7}-\frac{ (7\cdot 3^z\cdot rcos(r^{2}))}{2})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.482}{0.876})=0.16\pi;\theta_2=arctan(\frac{-0.930}{0.368})=1.62\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{1.09}\int_{0.16\pi}^{1.62\pi}\int_{0.27}^{1.86}(\frac{(2\cdot 4^{(\frac{r^{2}}{2})}\cdot r)}{7}-\frac{ (7\cdot 3^z\cdot rcos(r^{2}))}{2})drd\theta dz=(\frac{2^{(r^{2})}}{(7ln(2))}-\frac{ (7\cdot 3^zsin(r^{2}))}{4})d\theta dz|_{0.27}^{1.86}\\&\int_{0}^{1.09}\int_{0.16\pi}^{1.62\pi}(0.6746\cdot 3^z + 2.051)d\theta dz=(\theta\cdot (0.6746\cdot 3^z + 2.051))dz|_{0.16\pi}^{1.62\pi}\\&\int_{0}^{1.09}(3.094\cdot 3^z + 9.405)dz=(9.405z + 2.817\cdot 3^z)|_{0}^{1.09}=16.76\end{align*}$

Report an Error

Example Question #1 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(3e^{(- x^{2} - y^{2})})}{11}+\frac{ (49e^{(2z)}cos(x^{2} + y^{2}))}{6})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.87\\&\text{and length }1.45\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.844,0.536)\text{ and }\overrightarrow{u_2}=(0.982,-0.187)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

211.22

-7.04

-42.24

7.04

Correct answer:

-42.24

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(3e^{(- x^{2} - y^{2})})}{11}+\frac{ (49e^{(2z)}cos(x^{2} + y^{2}))}{6})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(3\cdot re^{(-r^{2})})}{11}+\frac{ (49\cdot rcos(r^{2})e^{(2z)})}{6})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.536}{-0.844})=0.82\pi;\theta_2=arctan(\frac{-0.187}{0.982})=1.94\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.45}\int_{0.82\pi}^{1.94\pi}\int_{0}^{1.87}(\frac{(3\cdot re^{(-r^{2})})}{11}+\frac{ (49\cdot rcos(r^{2})e^{(2z)})}{6})drd\theta dz=(\frac{(49sin(r^{2})e^{(2z)})}{12}-\frac{ (3e^{(-r^{2})})}{22})d\theta dz|_{0}^{1.87}\\&\int_{0}^{1.45}\int_{0.82\pi}^{1.94\pi}(0.1322 - 1.421e^{(2z)})d\theta dz=(-\theta\cdot (1.421e^{(2z)} - 0.1322))dz|_{0.82\pi}^{1.94\pi}\\&\int_{0}^{1.45}(0.4653 - 4.998e^{(2z)})dz=(0.4653z - 2.499e^{(2z)})|_{0}^{1.45}=-42.24\end{align*}$

Report an Error

Example Question #2 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(48cos(3z)sin(2x^{2} + 2y^{2}) - 20sin(x^{2} + y^{2}))dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin with radii }\\&0.17\text{ and }1.61\\&\text{and length }0.54\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.707,0.707)\text{ and }\overrightarrow{u_2}=(0.125,-0.992)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-6.48

-19.43

3.89

77.73

Correct answer:

-19.43

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(48cos(3z)sin(2x^{2} + 2y^{2}) - 20sin(x^{2} + y^{2}))dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(48\cdot rcos(3z)sin(2\cdot r^{2}) - 20\cdot rsin(r^{2}))drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.707}{-0.707})=0.75\pi;\theta_2=arctan(\frac{-0.992}{0.125})=1.54\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[cos(az)]=\frac{sin(az)}{a}\\&\int_{0}^{0.54}\int_{0.75\pi}^{1.54\pi}\int_{0.17}^{1.61}(48\cdot rcos(3z)sin(2\cdot r^{2}) - 20\cdot rsin(r^{2}))drd\theta dz=(10cos(r^{2}) - 12cos(3z)cos(2\cdot r^{2}))d\theta dz|_{0.17}^{1.61}\\&\int_{0}^{0.54}\int_{0.75\pi}^{1.54\pi}(6.526cos(3z) - 18.52)d\theta dz=(\theta\cdot (6.526cos(3z) - 18.52))dz|_{0.75\pi}^{1.54\pi}\\&\int_{0}^{0.54}(16.2cos(3z) - 45.97)dz=(10.8cos(1.5z)sin(1.5z) - 45.97z)|_{0}^{0.54}=-19.43\end{align*}$

Report an Error

Example Question #6 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(13cos(2x^{2} + 2y^{2}))}{6}-\frac{ (2e^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}e^{(-2z)})}{15})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.94\\&\text{and length }1.01\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.876,0.482)\text{ and }\overrightarrow{u_2}=(-0.844,0.536)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-20.39

3.4

-10.2

40.78

Correct answer:

-10.2

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(13cos(2x^{2} + 2y^{2}))}{6}-\frac{ (2e^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}e^{(-2z)})}{15})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(13\cdot rcos(2\cdot r^{2}))}{6}-\frac{ (2\cdot re^{(-2z)}e^{(\frac{(3\cdot r^{2})}{2})})}{15})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.482}{0.876})=0.16\pi;\theta_2=arctan(\frac{0.536}{-0.844})=0.82\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.01}\int_{0.16\pi}^{0.82\pi}\int_{0}^{1.94}(\frac{(13\cdot rcos(2\cdot r^{2}))}{6}-\frac{ (2\cdot re^{(-2z)}e^{(\frac{(3\cdot r^{2})}{2})})}{15})drd\theta dz=(\frac{(13sin(2\cdot r^{2}))}{24}-\frac{ (2e^{(\frac{(3\cdot r^{2})}{2}- 2z)})}{45})d\theta dz|_{0}^{1.94}\\&\int_{0}^{1.01}\int_{0.16\pi}^{0.82\pi}(0.513 - 12.53e^{(-2z)})d\theta dz=(-\theta\cdot (12.53e^{(-2z)} - 0.513))dz|_{0.16\pi}^{0.82\pi}\\&\int_{0}^{1.01}(1.064 - 25.99e^{(-2z)})dz=(1.064z + 12.99e^{(-2z)})|_{0}^{1.01}=-10.2\end{align*}$

Report an Error

Example Question #7 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{29}{(3\cdot (2x^{2} + 2y^{2}))}-\frac{ (4sin(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})e^{(-z)})}{3})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin with radii }\\&0.33\text{ and }1.47\\&\text{and length }1.88\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.094,0.996)\text{ and }\overrightarrow{u_2}=(-0.509,-0.861)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-138.64

138.64

34.66

-11.55

Correct answer:

34.66

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{29}{(3\cdot (2x^{2} + 2y^{2}))}-\frac{ (4sin(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})e^{(-z)})}{3})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{29}{(6\cdot r)}-\frac{ (4\cdot re^{(-z)}sin(\frac{(3\cdot r^{2})}{2}))}{3})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.996}{0.094})=0.47\pi;\theta_2=arctan(\frac{-0.861}{-0.509})=1.33\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.88}\int_{0.47\pi}^{1.33\pi}\int_{0.33}^{1.47}(\frac{29}{(6\cdot r)}-\frac{ (4\cdot re^{(-z)}sin(\frac{(3\cdot r^{2})}{2}))}{3})drd\theta dz=(\frac{(29ln(r))}{6}+\frac{ (4e^{(-z)}cos(\frac{(3\cdot r^{2})}{2}))}{9})d\theta dz|_{0.33}^{1.47}\\&\int_{0}^{1.88}\int_{0.47\pi}^{1.33\pi}(7.221 - 0.8808e^{(-z)})d\theta dz=(-\theta\cdot (0.8808e^{(-z)} - 7.221))dz|_{0.47\pi}^{1.33\pi}\\&\int_{0}^{1.88}(19.51 - 2.38e^{(-z)})dz=(19.51z + 2.38e^{(-z)})|_{0}^{1.88}=34.66\end{align*}$

Report an Error

View Calculus 3 Tutors

Joseph
Certified Tutor

Ferris State University, Bachelor of Science, Applied Mathematics. University of North Carolina at Chapel Hill, Master of Sci...

View Calculus 3 Tutors

Sayantan
Certified Tutor

Western Michigan University, Bachelor of Science, Applied Mathematics.

View Calculus 3 Tutors

Ryanto
Certified Tutor

Excelsior College, Albany NY, Bachelor, Business with Math Minor. University of Houston, Master's/Graduate, Accounting.

All Calculus 3 Resources

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

Popular Subjects

French Tutors in New York City, French Tutors in Seattle, MCAT Tutors in Philadelphia, Physics Tutors in San Francisco-Bay Area, SAT Tutors in Washington DC, Calculus Tutors in Denver, English Tutors in Phoenix, LSAT Tutors in New York City, ACT Tutors in New York City, Reading Tutors in Chicago

Popular Courses & Classes

ISEE Courses & Classes in Washington DC, SAT Courses & Classes in Atlanta, SAT Courses & Classes in San Francisco-Bay Area, ISEE Courses & Classes in Boston, Spanish Courses & Classes in Atlanta, MCAT Courses & Classes in Denver, SSAT Courses & Classes in Washington DC, Spanish Courses & Classes in San Francisco-Bay Area, GMAT Courses & Classes in Philadelphia, SSAT Courses & Classes in Los Angeles

Popular Test Prep

SAT Test Prep in Los Angeles, MCAT Test Prep in Boston, GMAT Test Prep in Philadelphia, LSAT Test Prep in Boston, GRE Test Prep in Dallas Fort Worth, SAT Test Prep in Houston, LSAT Test Prep in New York City, SAT Test Prep in San Diego, LSAT Test Prep in Houston, GMAT Test Prep in Washington DC

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 : Triple Integration in Cylindrical Coordinates

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1 : Triple Integration In Cylindrical Coordinates

Example Question #2 : Triple Integrals

Example Question #1 : Triple Integration In Cylindrical Coordinates

Example Question #3 : Triple Integrals

Example Question #4 : Triple Integrals

Example Question #1 : Multiple Integration

Example Question #1 : Triple Integrals

Example Question #2 : Triple Integrals

Example Question #6 : Triple Integrals

Example Question #7 : Triple Integrals

All Calculus 3 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: