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Calculus 3 : Calculus 3

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

Example Questions

Example Question #71 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(6e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})}e^{(-z)})}{29})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.25\text{ and }1.23\\&\text{and length }1.8\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.891,0.454)\text{ and }\overrightarrow{u_2}=(0.930,-0.368)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.12

-0.04

-0.25

0.5

Correct answer:

-0.25

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.454}{-0.891})=0.85\pi;\theta_2=arctan(\frac{-0.368}{0.930})=1.88\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.8}\int_{0.85\pi}^{1.88\pi}\int_{0.25}^{1.23}(-\frac{(6\cdot re^{(-z)}e^{(-\frac{(2\cdot r^{2})}{3})})}{29})drd\theta dz=(\frac{(9e^{(- z -\frac{ (2\cdot r^{2})}{3})})}{58})d\theta dz|_{0.25}^{1.23}\\&\int_{0}^{1.8}\int_{0.85\pi}^{1.88\pi}(-0.09224e^{(-z)})d\theta dz=(-0.09224\theta e^{(-z)})dz|_{0.85\pi}^{1.88\pi}\\&\int_{0}^{1.8}(-0.2985e^{(-z)})dz=(0.2985e^{(-z)})|_{0}^{1.8}=-0.25\end{align*}$

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Example Question #72 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(32e^{(-\frac{ x^{2}}{2}-\frac{ y^{2}}{2})}e^{(-2z)})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.33\text{ and }1.49\\&\text{and length }0.76\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.905,0.426)\text{ and }\overrightarrow{u_2}=(0.988,-0.156)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-105.73

4.41

-13.22

26.43

Correct answer:

26.43

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(32e^{(-\frac{ x^{2}}{2}-\frac{ y^{2}}{2})}e^{(-2z)})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(32\cdot re^{(-2z)}e^{(-\frac{r^{2}}{2})})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.426}{-0.905})=0.86\pi;\theta_2=arctan(\frac{-0.156}{0.988})=1.95\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{0.76}\int_{0.86\pi}^{1.95\pi}\int_{0.33}^{1.49}(32\cdot re^{(-2z)}e^{(-\frac{r^{2}}{2})})drd\theta dz=(-32e^{(- 2z -\frac{ r^{2}}{2})})d\theta dz|_{0.33}^{1.49}\\&\int_{0}^{0.76}\int_{0.86\pi}^{1.95\pi}(19.76e^{(-2z)})d\theta dz=(19.76\theta e^{(-2z)})dz|_{0.86\pi}^{1.95\pi}\\&\int_{0}^{0.76}(67.66e^{(-2z)})dz=(-33.83e^{(-2z)})|_{0}^{0.76}=26.43\end{align*}$

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Example Question #76 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(4\cdot 2^z)}{(3\cdot (2x^{2} + 2y^{2}))})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.31\text{ and }0.99\\&\text{and length }1.07\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.536,0.844)\text{ and }\overrightarrow{u_2}=(0.790,-0.613)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-4.28

8.56

-12.84

0.71

Correct answer:

-4.28

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.844}{-0.536})=0.68\pi;\theta_2=arctan(\frac{-0.613}{0.790})=1.79\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{1.07}\int_{0.68\pi}^{1.79\pi}\int_{0.31}^{0.99}(-\frac{(2\cdot 2^z)}{(3\cdot r)})drd\theta dz=(-\frac{(2\cdot 2^zln(r))}{3})d\theta dz|_{0.31}^{0.99}\\&\int_{0}^{1.07}\int_{0.68\pi}^{1.79\pi}(-0.7741\cdot 2^z)d\theta dz=(-0.7741\cdot 2^z\theta)dz|_{0.68\pi}^{1.79\pi}\\&\int_{0}^{1.07}(-2.699\cdot 2^z)dz=(-3.894\cdot 2^z)|_{0}^{1.07}=-4.28\end{align*}$

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Example Question #77 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(4e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})}e^{(-2z)})}{47})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.17\\&\text{and length }0.86\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.613,0.790)\text{ and }\overrightarrow{u_2}=(-1.000,-0.031)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.11

0.02

-0.01

0.04

Correct answer:

0.04

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(4e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})}e^{(-2z)})}{47})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(4\cdot re^{(-2z)}e^{(-\frac{(2\cdot r^{2})}{3})})}{47})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.790}{0.613})=0.29\pi;\theta_2=arctan(\frac{-0.031}{-1.000})=1.01\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{0.86}\int_{0.29\pi}^{1.01\pi}\int_{0}^{1.17}(\frac{(4\cdot re^{(-2z)}e^{(-\frac{(2\cdot r^{2})}{3})})}{47})drd\theta dz=(-\frac{(3e^{(- 2z -\frac{ (2\cdot r^{2})}{3})})}{47})d\theta dz|_{0}^{1.17}\\&\int_{0}^{0.86}\int_{0.29\pi}^{1.01\pi}(0.0382e^{(-2z)})d\theta dz=(0.0382\theta e^{(-2z)})dz|_{0.29\pi}^{1.01\pi}\\&\int_{0}^{0.86}(0.08641e^{(-2z)})dz=(-0.04321e^{(-2z)})|_{0}^{0.86}=0.04\end{align*}$

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Example Question #78 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(\frac{6}{(\frac{2}{5})^{(x^{2} + y^{2})}e^{(2z)}})}{31})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&0.84\\&\text{and length }1.56\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.969,0.249)\text{ and }\overrightarrow{u_2}=(-0.930,0.368)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

15.67

-5.22

-2.61

0.87

Correct answer:

-2.61

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(\frac{6}{(\frac{2}{5})^{(x^{2} + y^{2})}e^{(2z)}})}{31})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(6\cdot re^{(2z)})}{(31\cdot (\frac{2}{5})^{(r^{2})})})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.249}{0.969})=0.08\pi;\theta_2=arctan(\frac{0.368}{-0.930})=0.88\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.56}\int_{0.08\pi}^{0.88\pi}\int_{0}^{0.84}(-\frac{(6\cdot re^{(2z)})}{(31\cdot (\frac{2}{5})^{(r^{2})})})drd\theta dz=(\frac{(3\cdot (\frac{5}{2})^{(r^{2})}e^{(2z)})}{(31ln(\frac{2}{5}))})d\theta dz|_{0}^{0.84}\\&\int_{0}^{1.56}\int_{0.08\pi}^{0.88\pi}(-0.096e^{(2z)})d\theta dz=(-0.096\theta e^{(2z)})dz|_{0.08\pi}^{0.88\pi}\\&\int_{0}^{1.56}(-0.2413e^{(2z)})dz=(-0.1206e^{(2z)})|_{0}^{1.56}=-2.61\end{align*}$

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Example Question #531 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(18\cdot (\frac{7}{2})^{(x^{2} + y^{2})}cos(z + 2))}{5})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.66\\&\text{and length }1.48\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.187,0.982)\text{ and }\overrightarrow{u_2}=(-0.000,-1.000)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-643.97

804.96

-160.99

80.5

Correct answer:

-160.99

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(18\cdot (\frac{7}{2})^{(x^{2} + y^{2})}cos(z + 2))}{5})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(18\cdot (\frac{7}{2})^{(r^{2})}\cdot rcos(z + 2))}{5})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.982}{-0.187})=0.56\pi;\theta_2=arctan(\frac{-1.000}{-0.000})=1.5\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[cos(az)]=\frac{sin(az)}{a}\\&\int_{0}^{1.48}\int_{0.56\pi}^{1.5\pi}\int_{0}^{1.66}(\frac{(18\cdot (\frac{7}{2})^{(r^{2})}\cdot rcos(z + 2))}{5})drd\theta dz=(\frac{(9\cdot (\frac{7}{2})^{(r^{2})}cos(z + 2))}{(5ln(\frac{7}{2}))})d\theta dz|_{0}^{1.66}\\&\int_{0}^{1.48}\int_{0.56\pi}^{1.5\pi}(43.92cos(z + 2))d\theta dz=(43.92\theta cos(z + 2))dz|_{0.56\pi}^{1.5\pi}\\&\int_{0}^{1.48}(129.7cos(z + 2))dz=(129.7sin(z + 2))|_{0}^{1.48}=-160.99\end{align*}$

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Example Question #73 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(4sin(x^{2} + y^{2}))}{(29\cdot (z + 1)^{2})})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.54\\&\text{and length }0.82\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.844,0.536)\text{ and }\overrightarrow{u_2}=(-0.941,0.339)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.36

-0.12

0.02

-0.6

Correct answer:

-0.12

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(4sin(x^{2} + y^{2}))}{(29\cdot (z + 1)^{2})})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(4\cdot rsin(r^{2}))}{(29\cdot (z + 1)^{2})})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.536}{0.844})=0.18\pi;\theta_2=arctan(\frac{0.339}{-0.941})=0.89\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0}^{0.82}\int_{0.18\pi}^{0.89\pi}\int_{0}^{1.54}(-\frac{(4\cdot rsin(r^{2}))}{(29\cdot (z + 1)^{2})})drd\theta dz=(\frac{(2cos(r^{2}))}{(58z + 29z^{2} + 29)})d\theta dz|_{0}^{1.54}\\&\int_{0}^{0.82}\int_{0.18\pi}^{0.89\pi}(-\frac{0.1185}{(z + 1)^{2}})d\theta dz=(-\frac{(0.1185\theta )}{(z + 1)^{2}})dz|_{0.18\pi}^{0.89\pi}\\&\int_{0}^{0.82}(-\frac{0.2643}{(z + 1)^{2}})dz=(\frac{0.2643}{(z + 1)})|_{0}^{0.82}=-0.12\end{align*}$

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Example Question #81 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{46}{3^{(x^{2} + y^{2})}e^{(-z)}})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.35\text{ and }1.64\\&\text{and length }1.46\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.891,0.454)\text{ and }\overrightarrow{u_2}=(-0.536,-0.844)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

48.56

-24.28

9.71

-97.13

Correct answer:

48.56

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{46}{3^{(x^{2} + y^{2})}e^{(-z)}})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(46\cdot re^{(-z)})}{3^{(r^{2})}})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.454}{0.891})=0.15\pi;\theta_2=arctan(\frac{-0.844}{-0.536})=1.32\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.46}\int_{0.15\pi}^{1.32\pi}\int_{0.35}^{1.64}(\frac{(46\cdot re^{(-z)})}{3^{(r^{2})}})drd\theta dz=(-\frac{(23e^{(-z)})}{(3^{(r^{2})}ln(3))})d\theta dz|_{0.35}^{1.64}\\&\int_{0}^{1.46}\int_{0.15\pi}^{1.32\pi}(17.21e^{(-z)})d\theta dz=(17.21\theta e^{(-z)})dz|_{0.15\pi}^{1.32\pi}\\&\int_{0}^{1.46}(63.25e^{(-z)})dz=(-63.25e^{(-z)})|_{0}^{1.46}=48.56\end{align*}$

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Example Question #81 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(22cos(2x^{2} + 2y^{2})sin(z + 2))}{3})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&0.8\\&\text{and length }1.94\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.844,0.536)\text{ and }\overrightarrow{u_2}=(-0.951,0.309)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

6.71

-1.12

0.56

-2.24

Correct answer:

-1.12

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(22cos(2x^{2} + 2y^{2})sin(z + 2))}{3})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(22\cdot rsin(z + 2)cos(2\cdot r^{2}))}{3})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.536}{0.844})=0.18\pi;\theta_2=arctan(\frac{0.309}{-0.951})=0.9\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[sin(az)]=-\frac{cos(az)}{a}\\&\int_{0}^{1.94}\int_{0.18\pi}^{0.9\pi}\int_{0}^{0.8}(-\frac{(22\cdot rsin(z + 2)cos(2\cdot r^{2}))}{3})drd\theta dz=(-\frac{(11sin(z + 2)sin(2\cdot r^{2}))}{6})d\theta dz|_{0}^{0.8}\\&\int_{0}^{1.94}\int_{0.18\pi}^{0.9\pi}(-1.756sin(z + 2))d\theta dz=(-1.756\theta sin(z + 2))dz|_{0.18\pi}^{0.9\pi}\\&\int_{0}^{1.94}(-3.973sin(z + 2))dz=(3.973cos(z + 2))|_{0}^{1.94}=-1.12\end{align*}$

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Example Question #83 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(\frac{(2z)}{47}+\frac{ 2}{47})}{(2x^{2} + 2y^{2})})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.45\text{ and }1.91\\&\text{and length }1.77\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-1.000,0.000)\text{ and }\overrightarrow{u_2}=(-0.000,-1.000)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.03

-0.97

0.16

0.04

Correct answer:

0.16

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.000}{-1.000})=1\pi;\theta_2=arctan(\frac{-1.000}{-0.000})=1.5\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int_{0}^{1.77}\int_{1\pi}^{1.5\pi}\int_{0.45}^{1.91}(\frac{(\frac{(2z)}{47}+\frac{ 2}{47})}{(2\cdot r)})drd\theta dz=(ln(r)\cdot (\frac{z}{47}+\frac{ 1}{47}))d\theta dz|_{0.45}^{1.91}\\&\int_{0}^{1.77}\int_{1\pi}^{1.5\pi}(0.03076z + 0.03076)d\theta dz=(0.03076\theta \cdot (z + 1))dz|_{1\pi}^{1.5\pi}\\&\int_{0}^{1.77}(0.04831z + 0.04831)dz=(0.02416\cdot (z + 1)^{2})|_{0}^{1.77}=0.16\end{align*}$

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Calculus 3 : Calculus 3

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Example Question #71 : Triple Integration In Cylindrical Coordinates

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