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Calculus 3 : Double Integrals

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Example Questions

Example Question #11 : Double Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-46e^{(2x^{2} + 2y^{2})})dxdy\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.99\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.941,0.339)\text{ and }\overrightarrow{u_2}=(0.031,-1.000)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

12325.7

-12325.7

-61628.4

123257

Correct answer:

-61628.4

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.339}{-0.941})=0.89\pi;\theta_2=arctan(\frac{-1.000}{0.031})=1.51\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.89\pi}^{1.51\pi}\int_{0}^{1.99}(-46\cdot re^{(2\cdot r^{2})})drd\theta=(-\frac{(23e^{(2\cdot r^{2})})}{2})d\theta|_{0}^{1.99}\\&\int_{0.89\pi}^{1.51\pi}(-31644.0)d\theta=(-31644\cdot \theta)|_{0.89\pi}^{1.51\pi}=-61628.4\end{align*}$

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Example Question #12 : Double Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(45e^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})})dxdy\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.55\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.996,0.094)\text{ and }\overrightarrow{u_2}=(0.729,-0.685)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

86.53

259.58

86.53

-778.75

Correct answer:

259.58

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dxdy=rdrd\theta\\&\text{With this we can find: }\iint_{D}(45e^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})})dxdy\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(45\cdot re^{(\frac{r^{2}}{2})})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.094}{-0.996})=0.97\pi;\theta_2=arctan(\frac{-0.685}{0.729})=1.76\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.97\pi}^{1.76\pi}\int_{0}^{1.55}(45\cdot re^{(\frac{r^{2}}{2})})drd\theta=(45e^{(\frac{r^{2}}{2})})d\theta|_{0}^{1.55}\\&\int_{0.97\pi}^{1.76\pi}(104.6)d\theta=(104.6\cdot \theta)|_{0.97\pi}^{1.76\pi}=259.58\end{align*}$

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Example Question #1381 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{24}{(x^{2} + y^{2})})dxdy\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.14\text{ and }1.98\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.729,0.685)\text{ and }\overrightarrow{u_2}=(0.613,-0.790)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

94.88

-189.76

189.76

759.03

Correct answer:

189.76

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.685}{-0.729})=0.76\pi;\theta_2=arctan(\frac{-0.790}{0.613})=1.71\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int_{0.76\pi}^{1.71\pi}\int_{0.14}^{1.98}(\frac{24}{r^{2}}r)drd\theta=(24ln(r))d\theta|_{0.14}^{1.98}\\&\int_{0.76\pi}^{1.71\pi}(63.58)d\theta=(63.58\cdot \theta)|_{0.76\pi}^{1.71\pi}=189.76\end{align*}$

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Example Question #11 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{11}{(x^{2} + y^{2})})dxdy\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.43\text{ and }1.68\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.125,0.992)\text{ and }\overrightarrow{u_2}=(-0.218,-0.976)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-91.36

-45.68

91.36

-11.42

Correct answer:

-45.68

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.992}{0.125})=0.46\pi;\theta_2=arctan(\frac{-0.976}{-0.218})=1.43\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int_{0.46\pi}^{1.43\pi}\int_{0.43}^{1.68}(-\frac{11}{r^{2}}r)drd\theta=(-11ln(r))d\theta|_{0.43}^{1.68}\\&\int_{0.46\pi}^{1.43\pi}(-14.99)d\theta=(-14.99\cdot \theta)|_{0.46\pi}^{1.43\pi}=-45.68\end{align*}$

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Example Question #1381 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{33}{3^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})}})dxdy\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.1\text{ and }1.76\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.707,0.707)\text{ and }\overrightarrow{u_2}=(0.063,-0.998)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-29.51

236.04

-14.75

-59.01

Correct answer:

-59.01

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dxdy=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{33}{3^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})}})dxdy\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(33\cdot r)}{3^{(\frac{r^{2}}{2})}})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.707}{-0.707})=0.75\pi;\theta_2=arctan(\frac{-0.998}{0.063})=1.52\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.75\pi}^{1.52\pi}\int_{0.1}^{1.76}(-\frac{(33\cdot r)}{3^{(\frac{r^{2}}{2})}})drd\theta=(\frac{33}{(3^{(\frac{r^{2}}{2})}ln(3))})d\theta|_{0.1}^{1.76}\\&\int_{0.75\pi}^{1.52\pi}(-24.39)d\theta=(-24.39\cdot \theta)|_{0.75\pi}^{1.52\pi}=-59.01\end{align*}$

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Example Question #11 : Double Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-6sin(2x^{2} + 2y^{2}))dxdy\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.31\text{ and }1.64\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.918,0.397)\text{ and }\overrightarrow{u_2}=(-0.309,-0.951)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

5.44

-0.91

-0.18

-4.53

Correct answer:

-0.91

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.397}{-0.918})=0.87\pi;\theta_2=arctan(\frac{-0.951}{-0.309})=1.4\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0.87\pi}^{1.4\pi}\int_{0.31}^{1.64}(-6\cdot rsin(2\cdot r^{2}))drd\theta=(3cos(r^{2})^{2})d\theta|_{0.31}^{1.64}\\&\int_{0.87\pi}^{1.4\pi}(-0.5447)d\theta=(-0.5447\cdot \theta)|_{0.87\pi}^{1.4\pi}=-0.91\end{align*}$

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Example Question #1382 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(3e^{(- 2x^{2} - 2y^{2})})}{17})dxdy\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.28\text{ and }1.32\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.637,0.771)\text{ and }\overrightarrow{u_2}=(0.397,-0.918)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.31

0.15

-0.15

0.08

Correct answer:

0.15

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.771}{0.637})=0.28\pi;\theta_2=arctan(\frac{-0.918}{0.397})=1.63\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.28\pi}^{1.63\pi}\int_{0.28}^{1.32}(\frac{(3\cdot re^{(-2\cdot r^{2})})}{17})drd\theta=(-\frac{(3e^{(-2\cdot r^{2})})}{68})d\theta|_{0.28}^{1.32}\\&\int_{0.28\pi}^{1.63\pi}(0.03636)d\theta=(0.03636\cdot \theta)|_{0.28\pi}^{1.63\pi}=0.15\end{align*}$

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Example Question #17 : Double Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(45e^{(-\frac{ x^{2}}{2}-\frac{ y^{2}}{2})})}{4})dxdy\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.39\text{ and }1.92\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.279,0.960)\text{ and }\overrightarrow{u_2}=(-0.891,-0.454)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-45.63

-2.53

15.21

-15.21

Correct answer:

-15.21

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dxdy=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(45e^{(-\frac{ x^{2}}{2}-\frac{ y^{2}}{2})})}{4})dxdy\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(45\cdot re^{(-\frac{r^{2}}{2})})}{4})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.960}{-0.279})=0.59\pi;\theta_2=arctan(\frac{-0.454}{-0.891})=1.15\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.59\pi}^{1.15\pi}\int_{0.39}^{1.92}(-\frac{(45\cdot re^{(-\frac{r^{2}}{2})})}{4})drd\theta=(\frac{(45e^{(-\frac{r^{2}}{2})})}{4})d\theta|_{0.39}^{1.92}\\&\int_{0.59\pi}^{1.15\pi}(-8.645)d\theta=(-8.645\cdot \theta)|_{0.59\pi}^{1.15\pi}=-15.21\end{align*}$

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Example Question #1391 : Calculus 3

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

Possible Answers:

10.73

-2.68

-16.1

-0.89

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 #11 : 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:

1.25

-0.63

-0.07

-0.21

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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Calculus 3 : Double Integrals

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #11 : Double Integrals

Example Question #12 : Double Integrals

Example Question #1381 : Calculus 3

Example Question #11 : Double Integration In Polar Coordinates

Example Question #1381 : Calculus 3

Example Question #11 : Double Integrals

Example Question #1382 : Calculus 3

Example Question #17 : Double Integrals

Example Question #1391 : Calculus 3

Example Question #11 : Double Integration In Polar Coordinates

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