Triple Integration in Cylindrical Coordinates - AP Calculus BC

Card 0 of 665

Question

Evaluate $\int \int \int_D y \ dV$ , where is the region below the plane , above the plane and between the cylinders , and .

Answer

We need to figure out our boundaries for our integral.

We need to convert everything into cylindrical coordinates. Remeber we are above the plane, this means we are above .

$0\leq z \leq x+1 \rightarrow 0\leq z \leq r\cos(\theta)+1$

The region is between two circles , and .

This means that

$0 \leq \theta \leq 2\pi$

$1\leq r \leq 3$

$\int \int \int_D y \ dV=\int_{0}^{2\pi} \int_{1}^{3} \int_{0}^{r\cos(\theta)+1} r\sin(\theta) r\ dz \ dr \ d\theta$

$=\int_{0}^{2\pi} \int_{1}^{3} \int_{0}^{r\cos(\theta)+1} r^2\sin(\theta) \ dz \ dr \ d\theta$

$=\int_{0}^{2\pi} \int_{1}^{3} zr^2\sin(\theta) \Big|_{0}^{r\cos(\theta)+1} \ dr \ d\theta$

$=\int_{0}^{2\pi} \int_{1}^{3} (r\cos(\theta)+1)r^2\sin(\theta)-0 \ dr \ d\theta$

$=\int_{0}^{2\pi} \int_{1}^{3} r^3\sin(\theta)\cos(\theta)+r^2\sin(\theta) \ dr \ d\theta$

$=\int_{0}^{2\pi} \frac{1}{4}r^4\sin(\theta)\cos(\theta)+\frac{1}{3}r^3\sin(\theta) \Big|_{1}^{3}\ d\theta$

$=\int_{0}^{2\pi} (\frac{1}{4}(3)^4\sin(\theta)\cos(\theta)+\frac{1}{3}(3)^3\sin(\theta))-(\frac{1}{4}(1)^4\sin(\theta)\cos(\theta)+\frac{1}{3}(1)^3\sin(\theta)) \ d\theta$

$=\int_{0}^{2\pi} (\frac{81}{4}\sin(\theta)\cos(\theta)+9\sin(\theta))-(\frac{1}{4}\sin(\theta)\cos(\theta)+\frac{1}{3}\sin(\theta)) \ d\theta$

$=\int_{0}^{2\pi} 20\sin(\theta)\cos(\theta)+\frac{26}{3}\sin(\theta) \ d\theta$

$= -\frac{1}{2}\cdot20\cos^2(\theta)-\frac{26}{3}\cos(\theta)\Big|_{0}^{2\pi}$

$= -\frac{1}{2}\cdot20\cos^2(2\pi)-\frac{26}{3}\cos(2\pi)-(-\frac{1}{2}\cdot20\cos^2(0)-\frac{26}{3}\cos(0))$

$= -\frac{1}{2}\cdot20(1)-\frac{26}{3}(1)-(-\frac{1}{2}\cdot20(1)-\frac{26}{3}(1))$

$= -10-\frac{26}{3}-(-10-\frac{26}{3})=0$

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Question

$\begin{align}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (3cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))}{41}-\frac{ (sin(z + 1)cos(x^{2} + y^{2}))}{25})dA\text{, where D is}\&\text{the region of a cylindrical annulus centered on the origin with radii }\&0.23\text{ and }1.56\&\text{and length }1.97\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(-0.094,0.996)\text{ and }\overrightarrow{u_2}=(0.637,-0.771)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align}$

Answer

$\begin{align}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(-\frac{ (3cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))}{41}-\frac{ (sin(z + 1)cos(x^{2} + y^{2}))}{25})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (3\cdot rcos(\frac{(3\cdot r^{2})}{2}))}{41}-\frac{ (rcos(r^{2})sin(z + 1))}{25})drd\theta dz\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align}$

$\begin{align}&\theta_1=arctan(\frac{0.996}{-0.094})=0.53\pi;\theta_2=arctan(\frac{-0.771}{0.637})=1.72\pi\&\text{Now, utilizing integral rules:}\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\end{align}$

$\begin{align}\&\int[sin(az)]=-\frac{cos(az)}{a}\&\int_{0}^{1.97}\int_{0.53\pi}^{1.72\pi}\int_{0.23}^{1.56}(-\frac{ (3\cdot rcos(\frac{(3\cdot r^{2})}{2}))}{41}-\frac{ (rcos(r^{2})sin(z + 1))}{25})drd\theta dz=(-\frac{ sin(\frac{(3\cdot r^{2})}{2})}{41}-\frac{ (sin(r^{2})sin(z + 1))}{50})d\theta dz|_{0.23}^{1.56}\&\int_{0}^{1.97}\int_{0.53\pi}^{1.72\pi}(0.01381 - 0.01195sin(z + 1))d\theta dz=(-1.0\theta\cdot (0.01195sin(z + 1) - 0.01381))dz|_{0.53\pi}^{1.72\pi}\&\int_{0}^{1.97}(0.05165 - 0.04467sin(z + 1))dz=0.03\end{align}$

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Question

$\begin{align}&\text{Evaluate the integral, }\iint_{D}(\frac{e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})}}{4}-\frac{ (49e^{(20z)}e^{(x^{2} + y^{2})})}{5})dA\text{, where D is}\&\text{the region of a cylinder centered on the origin with a radius of }\&1.92\&\text{and length }1.65\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.218,0.976)\text{ and }\overrightarrow{u_2}=(0.661,-0.750)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align}$

Answer

$\begin{align}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(\frac{e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})}}{4}-\frac{ (49e^{(20z)}e^{(x^{2} + y^{2})})}{5})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(re^{(-\frac{(2\cdot r^{2})}{3})})}{4}-\frac{ (49\cdot re^{(r^{2})}e^{(20z)})}{5})drd\theta dz\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align}$

$\begin{align}&\theta_1=arctan(\frac{0.976}{0.218})=0.43\pi;\theta_2=arctan(\frac{-0.750}{0.661})=1.73\pi\&\text{Now, utilizing integral rules:}\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\&\int[e^{az}]=\frac{e^{az}}{a}\&\int_{0}^{1.65}\int_{0.43\pi}^{1.73\pi}\int_{0}^{1.92}(\frac{(re^{(-\frac{(2\cdot r^{2})}{3})})}{4}-\frac{ (49\cdot re^{(r^{2})}e^{(20z)})}{5})drd\theta dz=(-\frac{ (49e^{(20z + r^{2})})}{10}-\frac{ (3e^{(-\frac{(2\cdot r^{2})}{3})})}{16})d\theta dz|_{0}^{1.92}\&\int_{0}^{1.65}\int_{0.43\pi}^{1.73\pi}(0.1714 - 190.6e^{(20z)})d\theta dz=(-1.0\theta\cdot (190.6e^{(20z)} - 0.1714))dz|_{0.43\pi}^{1.73\pi}\&\int_{0}^{1.65}(0.7002 - 778.5e^{(20z)})dz=(0.7002z - 38.92e^{(20z)})|_{0}^{1.65}=-8.35483\cdot10^{15}\end{align}$

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Question

$\begin{align}&\text{Evaluate the integral, }\iint_{D}(5cos(x^{2} + y^{2}) +\frac{ (3\cdot 2^ze^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})})}{13})dA\text{, where D is}\&\text{the region of a cylindrical annulus centered on the origin with radii }\&0.31\text{ and }1.89\&\text{and length }1.31\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.988,0.156)\text{ and }\overrightarrow{u_2}=(-1.000,0.031)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align}$

Answer

$\begin{align}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(5cos(x^{2} + y^{2}) +\frac{ (3\cdot 2^ze^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})})}{13})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(5\cdot rcos(r^{2}) +\frac{ (3\cdot 2^z\cdot re^{(-\frac{(3\cdot r^{2})}{2})})}{13})drd\theta dz\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align}$

$\begin{align}&\theta_1=arctan(\frac{0.156}{0.988})=0.05\pi;\theta_2=arctan(\frac{0.031}{-1.000})=0.99\pi\&\text{Now, utilizing integral rules:}\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\&\int_{0}^{1.31}\int_{0.05\pi}^{0.99\pi}\int_{0.31}^{1.89}(5\cdot rcos(r^{2}) +\frac{ (3\cdot 2^z\cdot re^{(-\frac{(3\cdot r^{2})}{2})})}{13})drd\theta dz=(\frac{(5sin(r^{2}))}{2}-\frac{ (2^ze^{(-\frac{(3\cdot r^{2})}{2})})}{13})d\theta dz|_{0.31}^{1.89}\&\int_{0}^{1.31}\int_{0.05\pi}^{0.99\pi}(0.06623\cdot 2^z - 1.283)d\theta dz=(\theta\cdot (0.06623\cdot 2^z - 1.283))dz|_{0.05\pi}^{0.99\pi}\&\int_{0}^{1.31}(0.1956\cdot 2^z - 3.789)dz=(0.2822\cdot 2^z - 3.789z)|_{0}^{1.31}=-4.55\end{align}$

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Question

$\begin{align}&\text{Evaluate the integral, }\iint_{D}(- 16e^{(x^{2} + y^{2})} -\frac{ (5e^{(2z)})}{(19\cdot (x^{2} + y^{2}))})dA\text{, where D is}\&\text{the region of a cylindrical annulus centered on the origin with radii }\&0.42\text{ and }1.42\&\text{and length }1.72\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(-0.368,0.930)\text{ and }\overrightarrow{u_2}=(-0.156,-0.988)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align}$

Answer

$\begin{align}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(- 16e^{(x^{2} + y^{2})} -\frac{ (5e^{(2z)})}{(19\cdot (x^{2} + y^{2}))})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(- 16\cdot re^{(r^{2})} -\frac{ (5e^{(2z)})}{(19\cdot r)})drd\theta dz\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align}$

$\begin{align}&\theta_1=arctan(\frac{0.930}{-0.368})=0.62\pi;\theta_2=arctan(\frac{-0.988}{-0.156})=1.45\pi\&\text{Now, utilizing integral rules:}\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\&\int[e^{az}]=\frac{e^{az}}{a}\&\int_{0}^{1.72}\int_{0.62\pi}^{1.45\pi}\int_{0.42}^{1.42}(- 16\cdot re^{(r^{2})} -\frac{ (5e^{(2z)})}{(19\cdot r)})drd\theta dz=(- 8e^{(r^{2})} -\frac{ (5e^{(2z)}ln(r))}{19})d\theta dz|_{0.42}^{1.42}\&\int_{0}^{1.72}\int_{0.62\pi}^{1.45\pi}(- 0.3206e^{(2z)} - 50.55)d\theta dz=(-1.0\theta\cdot (0.3206e^{(2z)} + 50.55))dz|_{0.62\pi}^{1.45\pi}\&\int_{0}^{1.72}(- 0.8359e^{(2z)} - 131.8)dz=(- 131.8z - 0.4179e^{(2z)})|_{0}^{1.72}=-239.31\end{align}$

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Question

$\begin{align}&\text{Evaluate the integral, }\iint_{D}(\frac{(2\cdot 4^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})})}{7}-\frac{ (7\cdot 3^zcos(x^{2} + y^{2}))}{2})dA\text{, where D is}\&\text{the region of a cylindrical annulus centered on the origin with radii }\&0.27\text{ and }1.86\&\text{and length }1.09\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.876,0.482)\text{ and }\overrightarrow{u_2}=(0.368,-0.930)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align}$

Answer

$\begin{align}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(\frac{(2\cdot 4^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})})}{7}-\frac{ (7\cdot 3^zcos(x^{2} + y^{2}))}{2})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(2\cdot 4^{(\frac{r^{2}}{2})}\cdot r)}{7}-\frac{ (7\cdot 3^z\cdot rcos(r^{2}))}{2})drd\theta dz\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align}$

$\begin{align}&\theta_1=arctan(\frac{0.482}{0.876})=0.16\pi;\theta_2=arctan(\frac{-0.930}{0.368})=1.62\pi\&\text{Now, utilizing integral rules:}\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\&\int_{0}^{1.09}\int_{0.16\pi}^{1.62\pi}\int_{0.27}^{1.86}(\frac{(2\cdot 4^{(\frac{r^{2}}{2})}\cdot r)}{7}-\frac{ (7\cdot 3^z\cdot rcos(r^{2}))}{2})drd\theta dz=(\frac{2^{(r^{2})}}{(7ln(2))}-\frac{ (7\cdot 3^zsin(r^{2}))}{4})d\theta dz|_{0.27}^{1.86}\&\int_{0}^{1.09}\int_{0.16\pi}^{1.62\pi}(0.6746\cdot 3^z + 2.051)d\theta dz=(\theta\cdot (0.6746\cdot 3^z + 2.051))dz|_{0.16\pi}^{1.62\pi}\&\int_{0}^{1.09}(3.094\cdot 3^z + 9.405)dz=(9.405z + 2.817\cdot 3^z)|_{0}^{1.09}=16.76\end{align}$

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Question

$\begin{align}&\text{Evaluate the integral, }\iint_{D}(\frac{(3e^{(- x^{2} - y^{2})})}{11}+\frac{ (49e^{(2z)}cos(x^{2} + y^{2}))}{6})dA\text{, where D is}\&\text{the region of a cylinder centered on the origin with a radius of }\&1.87\&\text{and length }1.45\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(-0.844,0.536)\text{ and }\overrightarrow{u_2}=(0.982,-0.187)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align}$

Answer

$\begin{align}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(\frac{(3e^{(- x^{2} - y^{2})})}{11}+\frac{ (49e^{(2z)}cos(x^{2} + y^{2}))}{6})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(3\cdot re^{(-r^{2})})}{11}+\frac{ (49\cdot rcos(r^{2})e^{(2z)})}{6})drd\theta dz\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align}$

$\begin{align}&\theta_1=arctan(\frac{0.536}{-0.844})=0.82\pi;\theta_2=arctan(\frac{-0.187}{0.982})=1.94\pi\&\text{Now, utilizing integral rules:}\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\&\int[e^{az}]=\frac{e^{az}}{a}\&\int_{0}^{1.45}\int_{0.82\pi}^{1.94\pi}\int_{0}^{1.87}(\frac{(3\cdot re^{(-r^{2})})}{11}+\frac{ (49\cdot rcos(r^{2})e^{(2z)})}{6})drd\theta dz=(\frac{(49sin(r^{2})e^{(2z)})}{12}-\frac{ (3e^{(-r^{2})})}{22})d\theta dz|_{0}^{1.87}\&\int_{0}^{1.45}\int_{0.82\pi}^{1.94\pi}(0.1322 - 1.421e^{(2z)})d\theta dz=(-\theta\cdot (1.421e^{(2z)} - 0.1322))dz|_{0.82\pi}^{1.94\pi}\&\int_{0}^{1.45}(0.4653 - 4.998e^{(2z)})dz=(0.4653z - 2.499e^{(2z)})|_{0}^{1.45}=-42.24\end{align}$

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Question

$\begin{align}&\text{Evaluate the integral, }\iint_{D}(48cos(3z)sin(2x^{2} + 2y^{2}) - 20sin(x^{2} + y^{2}))dA\text{, where D is}\&\text{the region of a cylindrical annulus centered on the origin with radii }\&0.17\text{ and }1.61\&\text{and length }0.54\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(-0.707,0.707)\text{ and }\overrightarrow{u_2}=(0.125,-0.992)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align}$

Answer

$\begin{align}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(48cos(3z)sin(2x^{2} + 2y^{2}) - 20sin(x^{2} + y^{2}))dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(48\cdot rcos(3z)sin(2\cdot r^{2}) - 20\cdot rsin(r^{2}))drd\theta dz\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align}$

$\begin{align}&\theta_1=arctan(\frac{0.707}{-0.707})=0.75\pi;\theta_2=arctan(\frac{-0.992}{0.125})=1.54\pi\&\text{Now, utilizing integral rules:}\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\&\int[cos(az)]=\frac{sin(az)}{a}\&\int_{0}^{0.54}\int_{0.75\pi}^{1.54\pi}\int_{0.17}^{1.61}(48\cdot rcos(3z)sin(2\cdot r^{2}) - 20\cdot rsin(r^{2}))drd\theta dz=(10cos(r^{2}) - 12cos(3z)cos(2\cdot r^{2}))d\theta dz|_{0.17}^{1.61}\&\int_{0}^{0.54}\int_{0.75\pi}^{1.54\pi}(6.526cos(3z) - 18.52)d\theta dz=(\theta\cdot (6.526cos(3z) - 18.52))dz|_{0.75\pi}^{1.54\pi}\&\int_{0}^{0.54}(16.2cos(3z) - 45.97)dz=(10.8cos(1.5z)sin(1.5z) - 45.97z)|_{0}^{0.54}=-19.43\end{align}$

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Question

$\begin{align}&\text{Evaluate the integral, }\iint_{D}(\frac{(13cos(2x^{2} + 2y^{2}))}{6}-\frac{ (2e^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}e^{(-2z)})}{15})dA\text{, where D is}\&\text{the region of a cylinder centered on the origin with a radius of }\&1.94\&\text{and length }1.01\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.876,0.482)\text{ and }\overrightarrow{u_2}=(-0.844,0.536)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align}$

Answer

$\begin{align}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(\frac{(13cos(2x^{2} + 2y^{2}))}{6}-\frac{ (2e^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}e^{(-2z)})}{15})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(13\cdot rcos(2\cdot r^{2}))}{6}-\frac{ (2\cdot re^{(-2z)}e^{(\frac{(3\cdot r^{2})}{2})})}{15})drd\theta dz\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align}$

$\begin{align}&\theta_1=arctan(\frac{0.482}{0.876})=0.16\pi;\theta_2=arctan(\frac{0.536}{-0.844})=0.82\pi\&\text{Now, utilizing integral rules:}\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\&\int[e^{az}]=\frac{e^{az}}{a}\&\int_{0}^{1.01}\int_{0.16\pi}^{0.82\pi}\int_{0}^{1.94}(\frac{(13\cdot rcos(2\cdot r^{2}))}{6}-\frac{ (2\cdot re^{(-2z)}e^{(\frac{(3\cdot r^{2})}{2})})}{15})drd\theta dz=(\frac{(13sin(2\cdot r^{2}))}{24}-\frac{ (2e^{(\frac{(3\cdot r^{2})}{2}- 2z)})}{45})d\theta dz|_{0}^{1.94}\&\int_{0}^{1.01}\int_{0.16\pi}^{0.82\pi}(0.513 - 12.53e^{(-2z)})d\theta dz=(-\theta\cdot (12.53e^{(-2z)} - 0.513))dz|_{0.16\pi}^{0.82\pi}\&\int_{0}^{1.01}(1.064 - 25.99e^{(-2z)})dz=(1.064z + 12.99e^{(-2z)})|_{0}^{1.01}=-10.2\end{align}$

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Question

$\begin{align}&\text{Evaluate the integral, }\iint_{D}(\frac{29}{(3\cdot (2x^{2} + 2y^{2}))}-\frac{ (4sin(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})e^{(-z)})}{3})dA\text{, where D is}\&\text{the region of a cylindrical annulus centered on the origin with radii }\&0.33\text{ and }1.47\&\text{and length }1.88\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.094,0.996)\text{ and }\overrightarrow{u_2}=(-0.509,-0.861)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align}$

Answer

$\begin{align}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(\frac{29}{(3\cdot (2x^{2} + 2y^{2}))}-\frac{ (4sin(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})e^{(-z)})}{3})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{29}{(6\cdot r)}-\frac{ (4\cdot re^{(-z)}sin(\frac{(3\cdot r^{2})}{2}))}{3})drd\theta dz\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align}$

$\begin{align}&\theta_1=arctan(\frac{0.996}{0.094})=0.47\pi;\theta_2=arctan(\frac{-0.861}{-0.509})=1.33\pi\&\text{Now, utilizing integral rules:}\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\&\int[e^{az}]=\frac{e^{az}}{a}\&\int_{0}^{1.88}\int_{0.47\pi}^{1.33\pi}\int_{0.33}^{1.47}(\frac{29}{(6\cdot r)}-\frac{ (4\cdot re^{(-z)}sin(\frac{(3\cdot r^{2})}{2}))}{3})drd\theta dz=(\frac{(29ln(r))}{6}+\frac{ (4e^{(-z)}cos(\frac{(3\cdot r^{2})}{2}))}{9})d\theta dz|_{0.33}^{1.47}\&\int_{0}^{1.88}\int_{0.47\pi}^{1.33\pi}(7.221 - 0.8808e^{(-z)})d\theta dz=(-\theta\cdot (0.8808e^{(-z)} - 7.221))dz|_{0.47\pi}^{1.33\pi}\&\int_{0}^{1.88}(19.51 - 2.38e^{(-z)})dz=(19.51z + 2.38e^{(-z)})|_{0}^{1.88}=34.66\end{align}$

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Question

$\begin{align}&\text{Evaluate the integral, }\iint_{D}(\frac{5}{(39\cdot (x^{2} + y^{2}))}+\frac{ (4sin(2x^{2} + 2y^{2})e^{(-2z)})}{3})dA\text{, where D is}\&\text{the region of a cylindrical annulus centered on the origin with radii }\&0.4\text{ and }1.4\&\text{and length }0.62\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.685,0.729)\text{ and }\overrightarrow{u_2}=(0.218,-0.976)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align}$

Answer

$\begin{align}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(\frac{5}{(39\cdot (x^{2} + y^{2}))}+\frac{ (4sin(2x^{2} + 2y^{2})e^{(-2z)})}{3})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{5}{(39\cdot r)}+\frac{ (4\cdot re^{(-2z)}sin(2\cdot r^{2}))}{3})drd\theta dz\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align}$

$\begin{align}&\theta_1=arctan(\frac{0.729}{0.685})=0.26\pi;\theta_2=arctan(\frac{-0.976}{0.218})=1.57\pi\&\text{Now, utilizing integral rules:}\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\&\int[e^{az}]=\frac{e^{az}}{a}\&\int_{0}^{0.62}\int_{0.26\pi}^{1.57\pi}\int_{0.4}^{1.4}(\frac{5}{(39\cdot r)}+\frac{ (4\cdot re^{(-2z)}sin(2\cdot r^{2}))}{3})drd\theta dz=(\frac{(5ln(r))}{39}-\frac{ (2cos(r^{2})^{2}e^{(-2z)})}{3})d\theta dz|_{0.4}^{1.4}\&\int_{0}^{0.62}\int_{0.26\pi}^{1.57\pi}(0.5538e^{(-2z)} + 0.1606)d\theta dz=(\theta\cdot (0.5538e^{(-2z)} + 0.1606))dz|_{0.26\pi}^{1.57\pi}\&\int_{0}^{0.62}(2.279e^{(-2z)} + 0.661)dz=(0.661z - 1.139e^{(-2z)})|_{0}^{0.62}=1.22\end{align}$

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$\begin{align}&\text{Evaluate the integral, }\iint_{D}(4cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}) - 21cos(z + 2)e^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})dA\text{, where D is}\&\text{the region of a cylinder centered on the origin with a radius of }\&1.81\&\text{and length }1.37\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.685,0.729)\text{ and }\overrightarrow{u_2}=(-0.982,0.187)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align}$

Answer

$\begin{align}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(4cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}) - 21cos(z + 2)e^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(4\cdot rcos(\frac{(3\cdot r^{2})}{2}) - 21\cdot rcos(z + 2)e^{(\frac{(2\cdot r^{2})}{3})})drd\theta dz\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align}$

$\begin{align}&\theta_1=arctan(\frac{0.729}{0.685})=0.26\pi;\theta_2=arctan(\frac{0.187}{-0.982})=0.94\pi\&\text{Now, utilizing integral rules:}\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\&\int[cos(az)]=\frac{sin(az)}{a}\&\int_{0}^{1.37}\int_{0.26\pi}^{0.94\pi}\int_{0}^{1.81}(4\cdot rcos(\frac{(3\cdot r^{2})}{2}) - 21\cdot rcos(z + 2)e^{(\frac{(2\cdot r^{2})}{3})})drd\theta dz=(\frac{(4sin(\frac{(3\cdot r^{2})}{2}))}{3}-\frac{ (63cos(z + 2)e^{(\frac{(2\cdot r^{2})}{3})})}{4})d\theta dz|_{0}^{1.81}\&\int_{0}^{1.37}\int_{0.26\pi}^{0.94\pi}(- 124.1cos(z + 2) - 1.306)d\theta dz=(-\theta\cdot (124.1cos(z + 2) + 1.306))dz|_{0.26\pi}^{0.94\pi}\&\int_{0}^{1.37}(- 265.2cos(z + 2) - 2.791)dz=(- 2.791z - 265.2sin(z + 2))|_{0}^{1.37}=297.39\end{align}$

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$\begin{align}&\text{Evaluate the integral, }\iint_{D}(\frac{(4cos(x^{2} + y^{2}))}{9}+\frac{ (cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})e^{(-z)})}{4})dA\text{, where D is}\&\text{the region of a cylinder centered on the origin with a radius of }\&0.86\&\text{and length }1.1\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.941,0.339)\text{ and }\overrightarrow{u_2}=(0.588,-0.809)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align}$

Answer

$\begin{align}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(\frac{(4cos(x^{2} + y^{2}))}{9}+\frac{ (cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})e^{(-z)})}{4})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(4\cdot rcos(r^{2}))}{9}+\frac{ (re^{(-z)}cos(\frac{(2\cdot r^{2})}{3}))}{4})drd\theta dz\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align}$

$\begin{align}&\theta_1=arctan(\frac{0.339}{0.941})=0.11\pi;\theta_2=arctan(\frac{-0.809}{0.588})=1.7\pi\&\text{Now, utilizing integral rules:}\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\&\int[e^{az}]=\frac{e^{az}}{a}\&\int_{0}^{1.1}\int_{0.11\pi}^{1.7\pi}\int_{0}^{0.86}(\frac{(4\cdot rcos(r^{2}))}{9}+\frac{ (re^{(-z)}cos(\frac{(2\cdot r^{2})}{3}))}{4})drd\theta dz=(\frac{(2sin(r^{2}))}{9}+\frac{ (3e^{(-z)}sin(\frac{(2\cdot r^{2})}{3}))}{16})d\theta dz|_{0}^{0.86}\&\int_{0}^{1.1}\int_{0.11\pi}^{1.7\pi}(0.08875e^{(-z)} + 0.1498)d\theta dz=(\theta\cdot (0.08875e^{(-z)} + 0.1498))dz|_{0.11\pi}^{1.7\pi}\&\int_{0}^{1.1}(0.4433e^{(-z)} + 0.7482)dz=(0.7482z - 0.4433e^{(-z)})|_{0}^{1.1}=1.12\end{align}$

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Question

$\begin{align}&\text{Evaluate the integral, }\iint_{D}(3e^{(- 2x^{2} - 2y^{2})} -\frac{ cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2})}{(11\cdot 2^{(\frac{z}{2})})})dA\text{, where D is}\&\text{the region of a cylinder centered on the origin with a radius of }\&0.84\&\text{and length }1.77\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.809,0.588)\text{ and }\overrightarrow{u_2}=(-0.960,-0.279)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align}$

Answer

$\begin{align}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(3e^{(- 2x^{2} - 2y^{2})} -\frac{ cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2})}{(11\cdot 2^{(\frac{z}{2})})})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(3\cdot re^{(-2\cdot r^{2})} -\frac{ (rcos(\frac{r^{2}}{2}))}{(11\cdot 2^{(\frac{z}{2})})})drd\theta dz\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align}$

$\begin{align}&\theta_1=arctan(\frac{0.588}{0.809})=0.2\pi;\theta_2=arctan(\frac{-0.279}{-0.960})=1.09\pi\&\text{Now, utilizing integral rules:}\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\&\int_{0}^{1.77}\int_{0.2\pi}^{1.09\pi}\int_{0}^{0.84}(3\cdot re^{(-2\cdot r^{2})} -\frac{ (rcos(\frac{r^{2}}{2}))}{(11\cdot 2^{(\frac{z}{2})})})drd\theta dz=(-\frac{ (3e^{(-2\cdot r^{2})})}{4}-\frac{ sin(\frac{r^{2}}{2})}{(11\cdot 2^{(\frac{z}{2})})})d\theta dz|_{0}^{0.84}\&\int_{0}^{1.77}\int_{0.2\pi}^{1.09\pi}(0.5671 -\frac{ 0.03141}{2^{(0.5z)}})d\theta dz=(-1.0\theta\cdot (\frac{0.03141}{2^{(0.5z)}}- 0.5671))dz|_{0.2\pi}^{1.09\pi}\&\int_{0}^{1.77}(1.586 -\frac{ 0.08783}{2^{(0.5z)}})dz=(1.586z +\frac{ 0.2534}{2^{(0.5z)}})|_{0}^{1.77}=2.69\end{align}$

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Question

$\begin{align}&\text{Evaluate the integral, }\iint_{D}(\frac{(4e^{(x^{2} + y^{2})})}{25}+\frac{ (6cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})e^{(-2z)})}{23})dA\text{, where D is}\&\text{the region of a cylindrical annulus centered on the origin with radii }\&0.13\text{ and }0.99\&\text{and length }1.7\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.536,0.844)\text{ and }\overrightarrow{u_2}=(-0.536,-0.844)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align}$

Answer

$\begin{align}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(\frac{(4e^{(x^{2} + y^{2})})}{25}+\frac{ (6cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})e^{(-2z)})}{23})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(4\cdot re^{(r^{2})})}{25}+\frac{ (6\cdot re^{(-2z)}cos(\frac{(3\cdot r^{2})}{2}))}{23})drd\theta dz\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align}$

$\begin{align}&\theta_1=arctan(\frac{0.844}{0.536})=0.32\pi;\theta_2=arctan(\frac{-0.844}{-0.536})=1.32\pi\&\text{Now, utilizing integral rules:}\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\&\int[e^{az}]=\frac{e^{az}}{a}\&\int_{0}^{1.7}\int_{0.32\pi}^{1.32\pi}\int_{0.13}^{0.99}(\frac{(4\cdot re^{(r^{2})})}{25}+\frac{ (6\cdot re^{(-2z)}cos(\frac{(3\cdot r^{2})}{2}))}{23})drd\theta dz=(\frac{(2e^{(r^{2})})}{25}+\frac{ (2e^{(-2z)}sin(\frac{(3\cdot r^{2})}{2}))}{23})d\theta dz|_{0.13}^{0.99}\&\int_{0}^{1.7}\int_{0.32\pi}^{1.32\pi}(0.08431e^{(-2z)} + 0.1318)d\theta dz=(\theta\cdot (0.08431e^{(-2z)} + 0.1318))dz|_{0.32\pi}^{1.32\pi}\&\int_{0}^{1.7}(0.2649e^{(-2z)} + 0.4141)dz=(0.4141z - 0.1324e^{(-2z)})|_{0}^{1.7}=0.83\end{align}$

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Question

$\begin{align}&\text{Evaluate the integral, }\iint_{D}(\frac{(19zsin(2x^{2} + 2y^{2}))}{2}-\frac{ sin(x^{2} + y^{2})}{47})dA\text{, where D is}\&\text{the region of a cylindrical annulus centered on the origin with radii }\&0.27\text{ and }1.58\&\text{and length }1.44\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.982,0.187)\text{ and }\overrightarrow{u_2}=(-0.918,-0.397)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align}$

Answer

$\begin{align}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(\frac{(19zsin(2x^{2} + 2y^{2}))}{2}-\frac{ sin(x^{2} + y^{2})}{47})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(19\cdot rzsin(2\cdot r^{2}))}{2}-\frac{ (rsin(r^{2}))}{47})drd\theta dz\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align}$

$\begin{align}&\theta_1=arctan(\frac{0.187}{0.982})=0.06\pi;\theta_2=arctan(\frac{-0.397}{-0.918})=1.13\pi\&\text{Now, utilizing integral rules:}\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\&\int_{0}^{1.44}\int_{0.06\pi}^{1.13\pi}\int_{0.27}^{1.58}(\frac{(19\cdot rzsin(2\cdot r^{2}))}{2}-\frac{ (rsin(r^{2}))}{47})drd\theta dz=(-\frac{(cos(r^{2})\cdot (893zcos(r^{2}) - 2))}{188})d\theta dz|_{0.27}^{1.58}\&\int_{0}^{1.44}\int_{0.06\pi}^{1.13\pi}(1.693z - 0.01911)d\theta dz=(\theta\cdot (1.693z - 0.01911))dz|_{0.06\pi}^{1.13\pi}\&\int_{0}^{1.44}(5.689z - 0.06424)dz=(0.993\cdot (1.693z - 0.01911)^{2})|_{0}^{1.44}=5.81\end{align}$

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Question

$\begin{align}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (3e^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})})}{4}- 17cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2})e^{(-z)})dA\text{, where D is}\&\text{the region of a cylindrical annulus centered on the origin with radii }\&0.38\text{ and }1.63\&\text{and length }0.76\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(-0.588,0.809)\text{ and }\overrightarrow{u_2}=(0.661,-0.750)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align}$

Answer

$\begin{align}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(-\frac{ (3e^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})})}{4}- 17cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2})e^{(-z)})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (3\cdot re^{(\frac{r^{2}}{2})})}{4}- 17\cdot re^{(-z)}cos(\frac{r^{2}}{2}))drd\theta dz\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align}$

$\begin{align}&\theta_1=arctan(\frac{0.809}{-0.588})=0.7\pi;\theta_2=arctan(\frac{-0.750}{0.661})=1.73\pi\&\text{Now, utilizing integral rules:}\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\&\int[e^{az}]=\frac{e^{az}}{a}\&\int_{0}^{0.76}\int_{0.7\pi}^{1.73\pi}\int_{0.38}^{1.63}(-\frac{ (3\cdot re^{(\frac{r^{2}}{2})})}{4}- 17\cdot re^{(-z)}cos(\frac{r^{2}}{2}))drd\theta dz=(-\frac{ (3e^{(\frac{r^{2}}{2})})}{4}- 17e^{(-z)}sin(\frac{r^{2}}{2}))d\theta dz|_{0.38}^{1.63}\&\int_{0}^{0.76}\int_{0.7\pi}^{1.73\pi}(- 15.28e^{(-z)} - 2.025)d\theta dz=(-\theta\cdot (15.28e^{(-z)} + 2.025))dz|_{0.7\pi}^{1.73\pi}\&\int_{0}^{0.76}(- 49.43e^{(-z)} - 6.553)dz=(49.43e^{(-z)} - 6.553z)|_{0}^{0.76}=-31.3\end{align}$

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Question

$\begin{align}&\text{Evaluate the integral, }\iint_{D}(\frac{(39\cdot (\frac{2}{3})^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}e^{(2z)})}{2}-\frac{ (\frac{2}{(\frac{1}{2})^{(x^{2} + y^{2})}})}{11})dA\text{, where D is}\&\text{the region of a cylinder centered on the origin with a radius of }\&0.73\&\text{and length }1.01\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.861,0.509)\text{ and }\overrightarrow{u_2}=(0.397,-0.918)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align}$

Answer

$\begin{align}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(\frac{(39\cdot (\frac{2}{3})^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}e^{(2z)})}{2}-\frac{ (\frac{2}{(\frac{1}{2})^{(x^{2} + y^{2})}})}{11})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(39\cdot (\frac{2}{3})^{(\frac{(3\cdot r^{2})}{2})}\cdot re^{(2z)})}{2}-\frac{ (2\cdot r)}{(11\cdot (\frac{1}{2})^{(r^{2})})})drd\theta dz\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align}$

$\begin{align}&\theta_1=arctan(\frac{0.509}{0.861})=0.17\pi;\theta_2=arctan(\frac{-0.918}{0.397})=1.63\pi\&\text{Now, utilizing integral rules:}\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\&\int[e^{az}]=\frac{e^{az}}{a}\&\int_{0}^{1.01}\int_{0.17\pi}^{1.63\pi}\int_{0}^{0.73}(\frac{(39\cdot (\frac{2}{3})^{(\frac{(3\cdot r^{2})}{2})}\cdot re^{(2z)})}{2}-\frac{ (2\cdot r)}{(11\cdot (\frac{1}{2})^{(r^{2})})})drd\theta dz=(\frac{(13\cdot (\frac{2}{3})^{(\frac{(3\cdot r^{2})}{2})}e^{(2z)})}{(2ln(\frac{2}{3}))}-\frac{ 2^{(r^{2})}}{(11ln(2))})d\theta dz|_{0}^{0.73}\&\int_{0}^{1.01}\int_{0.17\pi}^{1.63\pi}(4.438e^{(2z)} - 0.0586)d\theta dz=(1.0\theta\cdot (4.438e^{(2z)} - 0.0586))dz|_{0.17\pi}^{1.63\pi}\&\int_{0}^{1.01}(20.36e^{(2z)} - 0.2688)dz=(10.18e^{(2z)} - 0.2688z)|_{0}^{1.01}=66.27\end{align}$

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Question

$\begin{align}&\text{Evaluate the integral, }\iint_{D}(\frac{(22e^{(-2z)})}{(2x^{2} + 2y^{2})}-\frac{ cos(x^{2} + y^{2})}{11})dA\text{, where D is}\&\text{the region of a cylindrical annulus centered on the origin with radii }\&0.24\text{ and }1.58\&\text{and length }1.12\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.125,0.992)\text{ and }\overrightarrow{u_2}=(0.454,-0.891)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align}$

Answer

$\begin{align}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(\frac{(22e^{(-2z)})}{(2x^{2} + 2y^{2})}-\frac{ cos(x^{2} + y^{2})}{11})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(11e^{(-2z)})}{r}-\frac{ (rcos(r^{2}))}{11})drd\theta dz\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align}$

$\begin{align}&\theta_1=arctan(\frac{0.992}{0.125})=0.46\pi;\theta_2=arctan(\frac{-0.891}{0.454})=1.65\pi\&\text{Now, utilizing integral rules:}\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\&\int[e^{az}]=\frac{e^{az}}{a}\&\int_{0}^{1.12}\int_{0.46\pi}^{1.65\pi}\int_{0.24}^{1.58}(\frac{(11e^{(-2z)})}{r}-\frac{ (rcos(r^{2}))}{11})drd\theta dz=(11e^{(-2z)}ln(r) -\frac{ sin(r^{2})}{22})d\theta dz|_{0.24}^{1.58}\&\int_{0}^{1.12}\int_{0.46\pi}^{1.65\pi}(20.73e^{(-2z)} - 0.02472)d\theta dz=(\theta\cdot (20.73e^{(-2z)} - 0.02472))dz|_{0.46\pi}^{1.65\pi}\&\int_{0}^{1.12}(77.5e^{(-2z)} - 0.09241)dz=(- 0.09241z - 38.75e^{(-2z)})|_{0}^{1.12}=34.52\end{align}$

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Question

$\begin{align}&\text{Evaluate the integral, }\iint_{D}(\frac{(e^{(z)}sin(x^{2} + y^{2}))}{10}-\frac{ (9cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{2})dA\text{, where D is}\&\text{the region of a cylinder centered on the origin with a radius of }\&1.11\&\text{and length }1.17\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.951,0.309)\text{ and }\overrightarrow{u_2}=(-0.707,0.707)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align}$

Answer

$\begin{align}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(\frac{(e^{(z)}sin(x^{2} + y^{2}))}{10}-\frac{ (9cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{2})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(rsin(r^{2})e^{(z)})}{10}-\frac{ (9\cdot rcos(\frac{(2\cdot r^{2})}{3}))}{2})drd\theta dz\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align}$

$\begin{align}&\theta_1=arctan(\frac{0.309}{0.951})=0.1\pi;\theta_2=arctan(\frac{0.707}{-0.707})=0.75\pi\&\text{Now, utilizing integral rules:}\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\&\int[e^{az}]=\frac{e^{az}}{a}\&\int_{0}^{1.17}\int_{0.1\pi}^{0.75\pi}\int_{0}^{1.11}(\frac{(rsin(r^{2})e^{(z)})}{10}-\frac{ (9\cdot rcos(\frac{(2\cdot r^{2})}{3}))}{2})drd\theta dz=(-\frac{ (27sin(\frac{(2\cdot r^{2})}{3}))}{8}-\frac{ (cos(r^{2})e^{(z)})}{20})d\theta dz|_{0}^{1.11}\&\int_{0}^{1.17}\int_{0.1\pi}^{0.75\pi}(0.03339e^{(z)} - 2.471)d\theta dz=(1.0\theta\cdot (0.03339e^{(z)} - 2.471))dz|_{0.1\pi}^{0.75\pi}\&\int_{0}^{1.17}(0.06818e^{(z)} - 5.046)dz=(0.06818e^{(z)} - 5.046z)|_{0}^{1.17}=-5.75\end{align}$

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