Multiple Integration - AP Calculus BC

Card 0 of 3445

Question

Evaluate $\int \int_R 3x+2y \ dA$ , where is the trapezoidal region with vertices given by , , $(\frac{5}{2}, \frac{5}{2})$ , and $(\frac{5}{2}, -\frac{5}{2})$ ,

using the transformation , and .

Answer

The first thing we have to do is figure out the general equations for the lines that create the trapezoid.

Now we have the general equations for out trapezoid, now we need to plug in our transformations into these equations.

$-6v=0\rightarrow v=0$

$4u=0\rightarrow u=0$

$4u=5\rightarrow u=\frac{5}{4}$

$-6v=-5\rightarrow v=\frac{5}{6}$

So our region is a rectangle given by $0 \leq u \leq \frac{5}{4}$ , $0 \leq v \leq \frac{5}{6}$

Next we need to calculate the Jacobian.

$\begin{vmatrix} 2 & 3\ 2 & -3 \end{vmatrix} =-6-6=-12$

Now we can put the integral together.

$\int_{0}^{\frac{5}{6}} \int_{0}^{\frac{5}{4}} (3(2u+3v)+2(2u-3v))) \cdot \left | -12\right | \ du\ dv$

$=\int_{0}^{\frac{5}{6}} \int_{0}^{\frac{5}{4}} 120u+36v \ du\ dv$

$=\int_{0}^{\frac{5}{6}} 60u^2+36uv \Big|_{0}^{\frac{5}{4}}\ dv$

$=\int_{0}^{\frac{5}{6}} 60( \frac{5}{4})^2+36(\frac{5}{4})v-0 \ dv$

$=\int_{0}^{\frac{5}{6}} \frac{375}{4}+45v \ dv$

$=\frac{375}{4}v+\frac{45}{2}v^2\Big|_{0}^{\frac{5}{6}} \ dv$

$=\frac{375}{4}\cdot\frac{5}{6}+\frac{45}{2}\cdot\Big(\frac{5}{6}\Big)^2 -0=\frac{375}{4}$

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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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