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Calculus 3 : Double Integration in Polar Coordinates

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Example Question #1551 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{50}{(3\cdot (x^{2} + y^{2}))})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.22\text{ and }1.31\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.613,0.790)\text{ and }\overrightarrow{u_2}=(-0.454,-0.891)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{50}{(3\cdot (x^{2} + y^{2}))})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.22\text{ and }1.31\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.613,0.790)\text{ and }\overrightarrow{u_2}=(-0.454,-0.891)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

119.58 $\displaystyle 119.58$

-358.73 $\displaystyle -358.73$

-59.79 $\displaystyle -59.79$

29.89 $\displaystyle 29.89$

Correct answer:

-59.79 $\displaystyle -59.79$

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.790}{-0.613})=0.71\pi;\theta_2=arctan(\frac{-0.891}{-0.454})=1.35\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int_{0.71\pi}^{1.35\pi}\int_{0.22}^{1.31}(-\frac{50}{(3\cdot r)})drd\theta=(-\frac{(50ln(r))}{3})d\theta|_{0.22}^{1.31}\\&\int_{0.71\pi}^{1.35\pi}(-29.74)d\theta=(-29.74\theta)|_{0.71\pi}^{1.35\pi}=-59.79\end{align*}$ $\displaystyle \begin{align*}&\theta_1=arctan(\frac{0.790}{-0.613})=0.71\pi;\theta_2=arctan(\frac{-0.891}{-0.454})=1.35\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int_{0.71\pi}^{1.35\pi}\int_{0.22}^{1.31}(-\frac{50}{(3\cdot r)})drd\theta=(-\frac{(50ln(r))}{3})d\theta|_{0.22}^{1.31}\\&\int_{0.71\pi}^{1.35\pi}(-29.74)d\theta=(-29.74\theta)|_{0.71\pi}^{1.35\pi}=-59.79\end{align*}$

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Example Question #1552 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(19sin(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}{2}-\frac{ (\frac{47}{(\frac{2}{5})^{(x^{2} + y^{2})}})}{4})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.41\text{ and }1.71\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.790,0.613)\text{ and }\overrightarrow{u_2}=(1.000,-0.031)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(19sin(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}{2}-\frac{ (\frac{47}{(\frac{2}{5})^{(x^{2} + y^{2})}})}{4})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.41\text{ and }1.71\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.790,0.613)\text{ and }\overrightarrow{u_2}=(1.000,-0.031)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

1753.86 $\displaystyle 1753.86$

-58.46 $\displaystyle -58.46$

58.46 $\displaystyle 58.46$

-292.31 $\displaystyle -292.31$

Correct answer:

-292.31 $\displaystyle -292.31$

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(19sin(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}{2}-\frac{ (\frac{47}{(\frac{2}{5})^{(x^{2} + y^{2})}})}{4})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(19\cdot rsin(\frac{r^{2}}{2}))}{2}-\frac{ (47\cdot r)}{(4\cdot (\frac{2}{5})^{(r^{2})})})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$ $\displaystyle \begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(19sin(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}{2}-\frac{ (\frac{47}{(\frac{2}{5})^{(x^{2} + y^{2})}})}{4})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(19\cdot rsin(\frac{r^{2}}{2}))}{2}-\frac{ (47\cdot r)}{(4\cdot (\frac{2}{5})^{(r^{2})})})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.613}{-0.790})=0.79\pi;\theta_2=arctan(\frac{-0.031}{1.000})=1.99\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0.79\pi}^{1.99\pi}\int_{0.41}^{1.71}(\frac{(19\cdot rsin(\frac{r^{2}}{2}))}{2}-\frac{ (47\cdot r)}{(4\cdot (\frac{2}{5})^{(r^{2})})})drd\theta=(\frac{47}{(8\cdot (\frac{2}{5})^{(r^{2})}ln(\frac{2}{5}))}-\frac{ (19cos(\frac{r^{2}}{2}))}{2})d\theta|_{0.41}^{1.71}\\&\int_{0.79\pi}^{1.99\pi}(-77.54)d\theta=(-77.54\theta)|_{0.79\pi}^{1.99\pi}=-292.31\end{align*}$ $\displaystyle \begin{align*}&\theta_1=arctan(\frac{0.613}{-0.790})=0.79\pi;\theta_2=arctan(\frac{-0.031}{1.000})=1.99\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0.79\pi}^{1.99\pi}\int_{0.41}^{1.71}(\frac{(19\cdot rsin(\frac{r^{2}}{2}))}{2}-\frac{ (47\cdot r)}{(4\cdot (\frac{2}{5})^{(r^{2})})})drd\theta=(\frac{47}{(8\cdot (\frac{2}{5})^{(r^{2})}ln(\frac{2}{5}))}-\frac{ (19cos(\frac{r^{2}}{2}))}{2})d\theta|_{0.41}^{1.71}\\&\int_{0.79\pi}^{1.99\pi}(-77.54)d\theta=(-77.54\theta)|_{0.79\pi}^{1.99\pi}=-292.31\end{align*}$

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Calculus 3 : Double Integration in Polar Coordinates

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Example Question #1551 : Calculus 3

Example Question #1552 : Calculus 3

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