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Calculus 3 : Triple Integration in Cylindrical Coordinates

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Example Questions

Example Question #131 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(4cos(z + 2)sin(x^{2} + y^{2}))}{15})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.12\text{ and }1.92\\&\text{and length }0.63\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.309,0.951)\text{ and }\overrightarrow{u_2}=(-0.536,-0.844)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.12

-0.06

-0.23

0.94

Correct answer:

-0.23

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(4cos(z + 2)sin(x^{2} + y^{2}))}{15})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(4\cdot rcos(z + 2)sin(r^{2}))}{15})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.951}{-0.309})=0.6\pi;\theta_2=arctan(\frac{-0.844}{-0.536})=1.32\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[cos(az)]=\frac{sin(az)}{a}\\&\int_{0}^{0.63}\int_{0.6\pi}^{1.32\pi}\int_{0.12}^{1.92}(\frac{(4\cdot rcos(z + 2)sin(r^{2}))}{15})drd\theta dz=(-\frac{(2cos(r^{2})cos(z + 2))}{15})d\theta dz|_{0.12}^{1.92}\\&\int_{0}^{0.63}\int_{0.6\pi}^{1.32\pi}(0.2473cos(z + 2))d\theta dz=(0.2473\theta cos(z + 2))dz|_{0.6\pi}^{1.32\pi}\\&\int_{0}^{0.63}(0.5595cos(z + 2))dz=(0.5595sin(z + 2))|_{0}^{0.63}=-0.23\end{align*}$

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Example Question #131 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(\frac{4}{(\frac{7}{2})^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}e^{(-2z)}})}{27})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.65\\&\text{and length }1.77\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.536,0.844)\text{ and }\overrightarrow{u_2}=(-0.844,-0.536)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.01

0.03

-0.03

0.01

Correct answer:

-0.03

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(\frac{4}{(\frac{7}{2})^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}e^{(-2z)}})}{27})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(4\cdot re^{(-2z)})}{(27\cdot (\frac{7}{2})^{(\frac{(3\cdot r^{2})}{2})})})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.844}{-0.536})=0.68\pi;\theta_2=arctan(\frac{-0.536}{-0.844})=1.18\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.77}\int_{0.68\pi}^{1.18\pi}\int_{0}^{1.65}(-\frac{(4\cdot re^{(-2z)})}{(27\cdot (\frac{7}{2})^{(\frac{(3\cdot r^{2})}{2})})})drd\theta dz=(\frac{(4\cdot 2^{(\frac{(3\cdot r^{2})}{2})}e^{(-2z)})}{(81\cdot 7^{(\frac{(3\cdot r^{2})}{2})}ln(\frac{7}{2}))})d\theta dz|_{0}^{1.65}\\&\int_{0}^{1.77}\int_{0.68\pi}^{1.18\pi}(-0.03918e^{(-2z)})d\theta dz=(-0.03918\theta e^{(-2z)})dz|_{0.68\pi}^{1.18\pi}\\&\int_{0}^{1.77}(-0.06155e^{(-2z)})dz=(0.03077e^{(-2z)})|_{0}^{1.77}=-0.03\end{align*}$

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Example Question #133 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(3sin(z + 1)e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})})}{28})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&0.86\\&\text{and length }0.89\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.790,0.613)\text{ and }\overrightarrow{u_2}=(-0.960,-0.279)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.07

-0.02

-0.37

0.02

Correct answer:

0.07

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(3sin(z + 1)e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})})}{28})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(3\cdot rsin(z + 1)e^{(-\frac{(2\cdot r^{2})}{3})})}{28})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.613}{0.790})=0.21\pi;\theta_2=arctan(\frac{-0.279}{-0.960})=1.09\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[sin(az)]=-\frac{cos(az)}{a}\\&\int_{0}^{0.89}\int_{0.21\pi}^{1.09\pi}\int_{0}^{0.86}(\frac{(3\cdot rsin(z + 1)e^{(-\frac{(2\cdot r^{2})}{3})})}{28})drd\theta dz=(-\frac{(9sin(z + 1)e^{(-\frac{(2\cdot r^{2})}{3})})}{112})d\theta dz|_{0}^{0.86}\\&\int_{0}^{0.89}\int_{0.21\pi}^{1.09\pi}(0.03128sin(z + 1))d\theta dz=(0.03128\theta sin(z + 1))dz|_{0.21\pi}^{1.09\pi}\\&\int_{0}^{0.89}(0.08647sin(z + 1))dz=(-0.08647cos(z + 1))|_{0}^{0.89}=0.07\end{align*}$

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Calculus 3 : Triple Integration in Cylindrical Coordinates

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #131 : Multiple Integration

Example Question #131 : Triple Integration In Cylindrical Coordinates

Example Question #133 : Multiple Integration

All Calculus 3 Resources

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