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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 #19 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-6e^{(- 2x^{2} - 2y^{2})})dxdy\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.85\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.969,0.249)\text{ and }\overrightarrow{u_2}=(-0.031,-1.000)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

10.73

-0.89

-2.68

-16.1

Correct answer:

-2.68

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.249}{-0.969})=0.92\pi;\theta_2=arctan(\frac{-1.000}{-0.031})=1.49\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.92\pi}^{1.49\pi}\int_{0}^{1.85}(-6\cdot re^{(-2\cdot r^{2})})drd\theta=(\frac{(3e^{(-2\cdot r^{2})})}{2})d\theta|_{0}^{1.85}\\&\int_{0.92\pi}^{1.49\pi}(-1.498)d\theta=(-1.498\cdot \theta)|_{0.92\pi}^{1.49\pi}=-2.68\end{align*}$

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Example Question #20 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(5\cdot (\frac{7}{2})^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})})}{6})dxdy\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&0.53\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.982,0.187)\text{ and }\overrightarrow{u_2}=(-0.125,-0.992)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.63

-0.21

1.25

-0.07

Correct answer:

-0.21

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dxdy=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(5\cdot (\frac{7}{2})^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})})}{6})dxdy\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(5\cdot (\frac{7}{2})^{(\frac{r^{2}}{2})}\cdot r)}{6})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.187}{-0.982})=0.94\pi;\theta_2=arctan(\frac{-0.992}{-0.125})=1.46\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.94\pi}^{1.46\pi}\int_{0}^{0.53}(-\frac{(5\cdot (\frac{7}{2})^{(\frac{r^{2}}{2})}\cdot r)}{6})drd\theta=(-\frac{(5\cdot 7^{(\frac{r^{2}}{2})})}{(6\cdot 2^{(\frac{r^{2}}{2})}ln(\frac{7}{2}))})d\theta|_{0}^{0.53}\\&\int_{0.94\pi}^{1.46\pi}(-0.128)d\theta=(-0.128\cdot \theta)|_{0.94\pi}^{1.46\pi}=-0.21\end{align*}$

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Example Question #21 : Double Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(24sin(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))dxdy\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.26\text{ and }1.93\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.729,0.685)\text{ and }\overrightarrow{u_2}=(0.976,-0.218)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-567.65

37.84

28.38

113.53

Correct answer:

113.53

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dxdy=rdrd\theta\\&\text{With this we can find: }\iint_{D}(24sin(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))dxdy\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(24\cdot rsin(\frac{r^{2}}{2}))drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.685}{-0.729})=0.76\pi;\theta_2=arctan(\frac{-0.218}{0.976})=1.93\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0.76\pi}^{1.93\pi}\int_{0.26}^{1.93}(24\cdot rsin(\frac{r^{2}}{2}))drd\theta=(-24cos(\frac{r^{2}}{2}))d\theta|_{0.26}^{1.93}\\&\int_{0.76\pi}^{1.93\pi}(30.89)d\theta=(30.89\cdot \theta)|_{0.76\pi}^{1.93\pi}=113.53\end{align*}$

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Example Question #22 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-7cos(x^{2} + y^{2}))dxdy\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.19\text{ and }1.48\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.000,1.000)\text{ and }\overrightarrow{u_2}=(0.454,-0.891)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

9.84

-2.46

-59.03

-9.84

Correct answer:

-9.84

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{1.000}{0.000})=0.5\pi;\theta_2=arctan(\frac{-0.891}{0.454})=1.65\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0.5\pi}^{1.65\pi}\int_{0.19}^{1.48}(-7\cdot rcos(r^{2}))drd\theta=(-\frac{(7sin(r^{2}))}{2})d\theta|_{0.19}^{1.48}\\&\int_{0.5\pi}^{1.65\pi}(-2.723)d\theta=(-2.723\cdot \theta)|_{0.5\pi}^{1.65\pi}=-9.84\end{align*}$

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Example Question #501 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(21\cdot (\frac{3}{2})^{(x^{2} + y^{2})})}{2})dxdy\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&0.97\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.588,0.809)\text{ and }\overrightarrow{u_2}=(-0.661,-0.750)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

6.11

-18.33

18.33

4.58

Correct answer:

18.33

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.809}{0.588})=0.3\pi;\theta_2=arctan(\frac{-0.750}{-0.661})=1.27\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.3\pi}^{1.27\pi}\int_{0}^{0.97}(\frac{(21\cdot (\frac{3}{2})^{(r^{2})}\cdot r)}{2})drd\theta=(\frac{(21\cdot (\frac{3}{2})^{(r^{2})})}{(4ln(\frac{3}{2}))})d\theta|_{0}^{0.97}\\&\int_{0.3\pi}^{1.27\pi}(6.014)d\theta=(6.014\cdot \theta)|_{0.3\pi}^{1.27\pi}=18.33\end{align*}$

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Example Question #502 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(4e^{(-\frac{ x^{2}}{2}-\frac{ y^{2}}{2})})}{37})dxdy\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.24\text{ and }0.84\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.996,0.094)\text{ and }\overrightarrow{u_2}=(-0.992,0.125)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.01

-0.51

0.42

0.08

Correct answer:

0.08

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dxdy=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(4e^{(-\frac{ x^{2}}{2}-\frac{ y^{2}}{2})})}{37})dxdy\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(4\cdot re^{(-\frac{r^{2}}{2})})}{37})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.094}{0.996})=0.03\pi;\theta_2=arctan(\frac{0.125}{-0.992})=0.96\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.03\pi}^{0.96\pi}\int_{0.24}^{0.84}(\frac{(4\cdot re^{(-\frac{r^{2}}{2})})}{37})drd\theta=(-\frac{(4e^{(-\frac{r^{2}}{2})})}{37})d\theta|_{0.24}^{0.84}\\&\int_{0.03\pi}^{0.96\pi}(0.02907)d\theta=(0.02907\cdot \theta)|_{0.03\pi}^{0.96\pi}=0.08\end{align*}$

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Example Question #25 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(2sin(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))}{37})dxdy\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.41\text{ and }1.85\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.063,0.998)\text{ and }\overrightarrow{u_2}=(0.156,-0.988)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.03

-0.16

0.2

0.08

Correct answer:

0.03

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dxdy=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(2sin(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))}{37})dxdy\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(2\cdot rsin(\frac{(3\cdot r^{2})}{2}))}{37})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.998}{-0.063})=0.52\pi;\theta_2=arctan(\frac{-0.988}{0.156})=1.55\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0.52\pi}^{1.55\pi}\int_{0.41}^{1.85}(\frac{(2\cdot rsin(\frac{(3\cdot r^{2})}{2}))}{37})drd\theta=(-\frac{(2cos(\frac{(3\cdot r^{2})}{2}))}{111})d\theta|_{0.41}^{1.85}\\&\int_{0.52\pi}^{1.55\pi}(0.01008)d\theta=(0.01008\cdot \theta)|_{0.52\pi}^{1.55\pi}=0.03\end{align*}$

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Example Question #26 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-4sin(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))dxdy\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.97\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.685,0.729)\text{ and }\overrightarrow{u_2}=(0.218,-0.976)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

85.19

-14.2

-3.55

-70.99

Correct answer:

-14.2

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.729}{-0.685})=0.74\pi;\theta_2=arctan(\frac{-0.976}{0.218})=1.57\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0.74\pi}^{1.57\pi}\int_{0}^{1.97}(-4\cdot rsin(\frac{r^{2}}{2}))drd\theta=(4cos(\frac{r^{2}}{2}))d\theta|_{0}^{1.97}\\&\int_{0.74\pi}^{1.57\pi}(-5.445)d\theta=(-5.445\cdot \theta)|_{0.74\pi}^{1.57\pi}=-14.2\end{align*}$

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Example Question #22 : Double Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(47sin(x^{2} + y^{2}))}{2}-\frac{ e^{(x^{2} + y^{2})}}{4})dxdy\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.14\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.969,0.249)\text{ and }\overrightarrow{u_2}=(-0.988,0.156)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-11.3

67.8

22.6

-90.4

Correct answer:

22.6

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.249}{0.969})=0.08\pi;\theta_2=arctan(\frac{0.156}{-0.988})=0.95\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.08\pi}^{0.95\pi}\int_{0}^{1.14}(\frac{(47\cdot rsin(r^{2}))}{2}-\frac{ (re^{(r^{2})})}{4})drd\theta=(-\frac{ (47cos(r^{2}))}{4}-\frac{ e^{(r^{2})}}{8})d\theta|_{0}^{1.14}\\&\int_{0.08\pi}^{0.95\pi}(8.269)d\theta=(8.269\cdot \theta)|_{0.08\pi}^{0.95\pi}=22.6\end{align*}$

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Example Question #28 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(5e^{(-\frac{ x^{2}}{2}-\frac{ y^{2}}{2})})}{44}- 8e^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})})dxdy\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.62\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.930,0.368)\text{ and }\overrightarrow{u_2}=(0.509,-0.861)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

8.95

-53.69

161.06

-10.74

Correct answer:

-53.69

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dxdy=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(5e^{(-\frac{ x^{2}}{2}-\frac{ y^{2}}{2})})}{44}- 8e^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})})dxdy\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(5\cdot re^{(-\frac{r^{2}}{2})})}{44}- 8\cdot re^{(\frac{r^{2}}{2})})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.368}{-0.930})=0.88\pi;\theta_2=arctan(\frac{-0.861}{0.509})=1.67\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.88\pi}^{1.67\pi}\int_{0}^{1.62}(\frac{(5\cdot re^{(-\frac{r^{2}}{2})})}{44}- 8\cdot re^{(\frac{r^{2}}{2})})drd\theta=(-\frac{(e^{(-\frac{r^{2}}{2})}\cdot (352e^{(r^{2})} + 5))}{44})d\theta|_{0}^{1.62}\\&\int_{0.88\pi}^{1.67\pi}(-21.63)d\theta=(-21.63\cdot \theta)|_{0.88\pi}^{1.67\pi}=-53.69\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #19 : Double Integration In Polar Coordinates

Example Question #20 : Double Integration In Polar Coordinates

Example Question #21 : Double Integrals

Example Question #22 : Double Integration In Polar Coordinates

Example Question #501 : Multiple Integration

Example Question #502 : Multiple Integration

Example Question #25 : Double Integration In Polar Coordinates

Example Question #26 : Double Integration In Polar Coordinates

Example Question #22 : Double Integrals

Example Question #28 : Double Integration In Polar Coordinates

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