Double Integrals - AP Calculus BC

Card 0 of 1030

Question

Evaluate the following integral by converting into Polar Coordinates.

$\int \int_D 10xy \ dA$ , where is the portion between the circles of radius and and lies in first quadrant.

Answer

We have to remember how to convert cartesian coordinates into polar coordinates.

$x=r\cos(\theta)$

$y=r\sin(\theta)$

Lets write the ranges of our variables and $\theta$ .

$4\leq r \leq 10$

$0\leq \theta \leq \frac{\pi}{2}$

Now lets setup our double integral, don't forgot the extra .

$\int_{0}^{\frac{\pi}{2}} \int_{4}^{10}(r\cos(\theta))(r\sin(\theta)) r\ dr\ d\theta$

$=\int_{0}^{\frac{\pi}{2}} \int_{4}^{10}r^3\cos(\theta)\sin(\theta)\ dr\ d\theta$

$=\int_{0}^{\frac{\pi}{2}} \frac{1}{4}r^4\cos(\theta)\sin(\theta)\Big|_{4}^{10}\ d\theta$

$=\int_{0}^{\frac{\pi}{2}} \frac{1}{4}(10)^4\cos(\theta)\sin(\theta)-\frac{1}{4}(4)^4\cos(\theta)\sin(\theta) d\theta$

$=\int_{0}^{\frac{\pi}{2}} 2500\cos(\theta)\sin(\theta)-64\cos(\theta)\sin(\theta) d\theta$

$=\int_{0}^{\frac{\pi}{2}} 2436\cos(\theta)\sin(\theta) d\theta$

$=-\frac{1}{2}\cdot 2436\cos^2(\theta) \Big|_{0}^{\frac{\pi}{2}}$

$=-\frac{1}{2}\cdot 2436\cos^2\Big(\Big(\frac{\pi}{2}\Big)\Big)+\frac{1}{2}\cdot 2436\cos^2\Big(\Big(0\Big)\Big)$

$=\frac{1}{2}\cdot 2436(1)$

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Question

Evaluate the integral

$\int\int_D(x^2+y^2)dxdy$

where D is the region above the x-axis and within a circle centered at the origin of radius 2.

Answer

The conversions for Cartesian into polar coordinates is:

$x=r\cos \theta, ;y=r\sin \theta, ; dxdy=rdrd\theta$

The condition that the region is above the x-axis says:

$0\leq\theta\leq\pi$

And the condition that the region is within a circle of radius two says:

$0\leq r \leq 2$

With these conditions and conversions, the integral becomes:

$\int^\pi_0 \int^2_0(r^2\cos^2(\theta)+r^2\sin^2(\theta))rdrd\theta=\int^\pi_0 \int^2_0r^3drd\theta$

$=\int^\pi_0 (\frac{1}{4}r^4|^2_0)d\theta=\int^\pi_0 4d\theta=4\theta|^\pi_0=4\pi$

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Question

$\begin{align}&\text{Evaluate the integral, }\iint_{D}(10cos(x^{2} + y^{2}))dxdy\text{, where D is}\&\text{the region of a circle centered on the origin with a radius of }\&1.49\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.063,0.998)\text{ and }\overrightarrow{u_2}=(-0.729,-0.685)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align}$

Answer

$\begin{align}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to polar form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dxdy=rdrd\theta\&\text{With this we can find: }\iint_{D}(10cos(x^{2} + y^{2}))dxdy\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}10\cdot rcos(r^{2})drd\theta\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align}$

$\begin{align}&\theta_1=arctan(\frac{0.998}{0.063})=0.48\pi;\theta_2=arctan(\frac{-0.685}{-0.729})=1.24\pi\&\text{Now, utilizing integral rules:}\&\int[sin(ax)]=-\frac{cos(ax)}{a}\&\int_{0.48\pi}^{1.24\pi}\int_{0}^{1.49}10\cdot rcos(r^{2}))drd\theta=(5sin(r^{2}))d\theta|_{0}^{1.49}\&\int_{0.48\pi}^{1.24\pi}(3.983)d\theta=(3.983\cdot \theta)|_{0.48\pi}^{1.24\pi}=9.51\end{align}$

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Question

$\begin{align}&\text{Evaluate the integral, }\iint_{D}(-\frac{(25e^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})})}{2})dxdy\text{, where D is}\&\text{the region of a circle centered on the origin with a radius of }\&1.77\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.960,0.279)\text{ and }\overrightarrow{u_2}=(-0.218,-0.976)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align}$

Answer

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

$\begin{align}&\theta_1=arctan(\frac{0.279}{0.960})=0.09\pi;\theta_2=arctan(\frac{-0.976}{-0.218})=1.43\pi\&\text{Now, utilizing integral rules:}\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\&\int_{0.09\pi}^{1.43\pi}\int_{0}^{1.77}-\frac{(25\cdot re^{(-\frac{(3\cdot r^{2})}{2})})}{2})drd\theta=(\frac{(25e^{(-\frac{(3\cdot r^{2})}{2})})}{6})d\theta|_{0}^{1.77}\&\int_{0.09\pi}^{1.43\pi}(-4.129)d\theta=(-4.129\cdot \theta)|_{0.09\pi}^{1.43\pi}=-17.38\end{align}$

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Question

$\begin{align}&\text{Evaluate the integral, }\iint_{D}(43cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))dxdy\text{, where D is}\&\text{the region of a circle centered on the origin with a radius of }\&1.94\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.397,0.918)\text{ and }\overrightarrow{u_2}=(-0.951,-0.309)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align}$

Answer

$\begin{align}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to polar form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dxdy=rdrd\theta\&\text{With this we can find: }\iint_{D}(43cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))dxdy\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}43\cdot rcos(\frac{r^{2}}{2})drd\theta\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align}$

$\begin{align}&\theta_1=arctan(\frac{0.918}{0.397})=0.37\pi;\theta_2=arctan(\frac{-0.309}{-0.951})=1.1\pi\&\text{Now, utilizing integral rules:}\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}\&\int_{0.37\pi}^{1.1\pi}\int_{0}^{1.94}43\cdot rcos(\frac{r^{2}}{2}))drd\theta=(43sin(\frac{r^{2}}{2}))d\theta|_{0}^{1.94}\&\int_{0.37\pi}^{1.1\pi}(40.94)d\theta=(40.94\cdot \theta)|_{0.37\pi}^{1.1\pi}=93.88\end{align}$

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Question

$\begin{align}&\text{Evaluate the integral, }\iint_{D}(-41\cdot (\frac{3}{2})^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})dxdy\text{, where D is}\&\text{the region of a circle centered on the origin with a radius of }\&1.73\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.482,0.876)\text{ and }\overrightarrow{u_2}=(-0.536,-0.844)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align}$

Answer

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

$\begin{align}&\theta_1=arctan(\frac{0.876}{0.482})=0.34\pi;\theta_2=arctan(\frac{-0.844}{-0.536})=1.32\pi\&\text{Now, utilizing integral rules:}\&\int[rb^{ar^2}]=\frac{e^{ar^2}}{2aln(b)}\&\int_{0.34\pi}^{1.32\pi}\int_{0}^{1.73}-41\cdot (\frac{3}{2})^{(\frac{(2\cdot r^{2})}{3})}\cdot r)drd\theta=(-\frac{(123\cdot 3^{(\frac{(2\cdot r^{2})}{3})})}{(4\cdot 2^{(\frac{(2\cdot r^{2})}{3})}ln(\frac{3}{2}))})d\theta|_{0}^{1.73}\&\int_{0.34\pi}^{1.32\pi}(-94.47)d\theta=(-94.47\cdot \theta)|_{0.34\pi}^{1.32\pi}=-290.85\end{align}$

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Question

$\begin{align}&\text{Evaluate the integral, }\iint_{D}(-50e^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})dxdy\text{, where D is}\&\text{the region of a circle centered on the origin with a radius of }\&1.31\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.750,0.661)\text{ and }\overrightarrow{u_2}=(-0.562,-0.827)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align}$

Answer

$\begin{align}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to polar form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dxdy=rdrd\theta\&\text{With this we can find: }\iint_{D}(-50e^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})dxdy\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-50\cdot re^{(\frac{(2\cdot r^{2})}{3})})drd\theta\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align}$

$\begin{align}&\theta_1=arctan(\frac{0.661}{0.750})=0.23\pi;\theta_2=arctan(\frac{-0.827}{-0.562})=1.31\pi\&\text{Now, utilizing integral rules:}\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\&\int_{0.23\pi}^{1.31\pi}\int_{0}^{1.31}(-50\cdot re^{(\frac{(2\cdot r^{2})}{3})})drd\theta=(-\frac{(75e^{(\frac{(2\cdot r^{2})}{3})})}{2})d\theta|_{0}^{1.31}\&\int_{0.23\pi}^{1.31\pi}(-80.23)d\theta=(-80.23\cdot \theta)|_{0.23\pi}^{1.31\pi}=-272.22\end{align}$

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Question

$\begin{align}&\text{Evaluate the integral, }\iint_{D}(20cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))dxdy\text{, where D is}\&\text{the region defined between two circles centered on the origin with radii }\&0.5\text{ and }1.35\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(-0.876,0.482)\text{ and }\overrightarrow{u_2}=(-0.156,-0.988)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align}$

Answer

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

$\begin{align}&\theta_1=arctan(\frac{0.482}{-0.876})=0.84\pi;\theta_2=arctan(\frac{-0.988}{-0.156})=1.45\pi\&\text{Now, utilizing integral rules:}\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}\&\int_{0.84\pi}^{1.45\pi}\int_{0.5}^{1.35}(20\cdot rcos(\frac{(2\cdot r^{2})}{3}))drd\theta=(15sin(\frac{(2\cdot r^{2})}{3}))d\theta|_{0.5}^{1.35}\&\int_{0.84\pi}^{1.45\pi}(11.57)d\theta=(11.57\cdot \theta)|_{0.84\pi}^{1.45\pi}=22.18\end{align}$

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Question

$\begin{align}&\text{Evaluate the integral, }\iint_{D}(16sin(x^{2} + y^{2}))dxdy\text{, where D is}\&\text{the region defined between two circles centered on the origin with radii }\&0.33\text{ and }1.76\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(-0.982,0.187)\text{ and }\overrightarrow{u_2}=(0.918,-0.397)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align}$

Answer

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

$\begin{align}&\theta_1=arctan(\frac{0.187}{-0.982})=0.94\pi;\theta_2=arctan(\frac{-0.397}{0.918})=1.87\pi\&\text{Now, utilizing integral rules:}\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\&\int_{0.94\pi}^{1.87\pi}\int_{0.33}^{1.76}(16\cdot rsin(r^{2}))drd\theta=(-8cos(r^{2}))d\theta|_{0.33}^{1.76}\&\int_{0.94\pi}^{1.87\pi}(15.94)d\theta=(15.94\cdot \theta)|_{0.94\pi}^{1.87\pi}=46.59\end{align}$

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Question

$\begin{align}&\text{Evaluate the integral, }\iint_{D}(\frac{49}{(\frac{2}{3})^{(x^{2}+ y^{2})}})dxdy\text{, where D is}\&\text{the region defined between two circles centered on the origin with radii }\&0.16\text{ and }0.97\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.482,0.876)\text{ and }\overrightarrow{u_2}=(0.750,-0.661)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align}$

Answer

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

$\begin{align}&\theta_1=arctan(\frac{0.876}{0.482})=0.34\pi;\theta_2=arctan(\frac{-0.661}{0.750})=1.77\pi\&\text{Now, utilizing integral rules:}\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\&\int_{0.34\pi}^{1.77\pi}\int_{0.16}^{0.97}(\frac{(49\cdot r)}{(\frac{2}{3})^{(r^{2})}})drd\theta=(-\frac{(49\cdot (\frac{3}{2})^{(r^{2})})}{(2ln(\frac{2}{3}))})d\theta|_{0.16}^{0.97}\&\int_{0.34\pi}^{1.77\pi}(27.44)d\theta=(27.44\cdot \theta)|_{0.34\pi}^{1.77\pi}=123.25\end{align}$

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Question

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

Answer

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

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

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Question

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

Answer

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

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

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Question

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

Answer

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

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

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Question

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

Answer

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

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

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Question

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

Answer

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

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

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Question

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

Answer

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

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

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Question

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

Answer

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

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

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Question

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

Answer

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

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

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Question

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

Answer

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

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

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Question

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

Answer

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

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

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