Double Integration in Polar Coordinates

Help Questions

AP Calculus BC › Double Integration in Polar Coordinates

Questions 1 - 10

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(6\cdot (\frac{1}{4})^{(x^{2} + y^{2})})}{19}-\frac{ (23e^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})}{3})dA\text{, where D is}\&\text{the region defined between two circles centered on the origin with radii }\&0.27\text{ and }1.18\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(-1.000,0.031)\text{ and }\overrightarrow{u_2}=(0.992,-0.125)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Explanation

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to polar form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(\frac{(6\cdot (\frac{1}{4})^{(x^{2} + y^{2})})}{19}-\frac{ (23e^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})}{3})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(6\cdot (\frac{1}{4})^{(r^{2})}\cdot r)}{19}-\frac{ (23\cdot re^{(\frac{(2\cdot r^{2})}{3})})}{3})drd\theta\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.031}{-1.000})=0.99\pi;\theta_2=arctan(\frac{-0.125}{0.992})=1.96\pi\&\text{Now, utilizing integral rules:}\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\&\int_{0.99\pi}^{1.96\pi}\int_{0.27}^{1.18}(\frac{(6\cdot (\frac{1}{4})^{(r^{2})}\cdot r)}{19}-\frac{ (23\cdot re^{(\frac{(2\cdot r^{2})}{3})})}{3})drd\theta=(-\frac{ (23e^{(\frac{(2\cdot r^{2})}{3})})}{4}-\frac{ 3}{(19\cdot 4^{(r^{2})}ln(4))})d\theta|_{0.27}^{1.18}\&\int_{0.99\pi}^{1.96\pi}(-8.425)d\theta=(-8.425\cdot \theta)|_{0.99\pi}^{1.96\pi}=-25.68\end{align*}$

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(47sin(x^{2} + y^{2}))}{2}-\frac{ e^{(x^{2} + y^{2})}}{4})dxdy\text{, where D is}\&\text{the region of a circle centered on the origin with a radius of }\&1.14\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.969,0.249)\text{ and }\overrightarrow{u_2}=(-0.988,0.156)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Explanation

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

$\begin{align*}&\theta_1=arctan(\frac{0.249}{0.969})=0.08\pi;\theta_2=arctan(\frac{0.156}{-0.988})=0.95\pi\&\text{Now, utilizing integral rules:}\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\&\int_{0.08\pi}^{0.95\pi}\int_{0}^{1.14}(\frac{(47\cdot rsin(r^{2}))}{2}-\frac{ (re^{(r^{2})})}{4})drd\theta=(-\frac{ (47cos(r^{2}))}{4}-\frac{ e^{(r^{2})}}{8})d\theta|_{0}^{1.14}\&\int_{0.08\pi}^{0.95\pi}(8.269)d\theta=(8.269\cdot \theta)|_{0.08\pi}^{0.95\pi}=22.6\end{align*}$

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(3sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{23}- 32e^{(- 2x^{2} - 2y^{2})})dA\text{, where D is}\&\text{the region of a circle centered on the origin with a radius of }\&1.26\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.685,0.729)\text{ and }\overrightarrow{u_2}=(-0.969,0.249)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Explanation

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to polar form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(\frac{(3sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{23}- 32e^{(- 2x^{2} - 2y^{2})})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(3\cdot rsin(\frac{(2\cdot r^{2})}{3}))}{23}- 32\cdot re^{(-2\cdot r^{2})})drd\theta\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.729}{0.685})=0.26\pi;\theta_2=arctan(\frac{0.249}{-0.969})=0.92\pi\&\text{Now, utilizing integral rules:}\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\&\int_{0.26\pi}^{0.92\pi}\int_{0}^{1.26}(\frac{(3\cdot rsin(\frac{(2\cdot r^{2})}{3}))}{23}- 32\cdot re^{(-2\cdot r^{2})})drd\theta=(8e^{(-2\cdot r^{2})} -\frac{ (9cos(\frac{(2\cdot r^{2})}{3}))}{92})d\theta|_{0}^{1.26}\&\int_{0.26\pi}^{0.92\pi}(-7.616)d\theta=(-7.616\theta)|_{0.26\pi}^{0.92\pi}=-15.79\end{align*}$

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(12sin(2x^{2} + 2y^{2}) -\frac{ (5e^{(- 2x^{2} - 2y^{2})})}{41})dA\text{, where D is}\&\text{the region of a circle centered on the origin with a radius of }\&0.9\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(-0.827,0.562)\text{ and }\overrightarrow{u_2}=(0.661,-0.750)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Explanation

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to polar form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(12sin(2x^{2} + 2y^{2}) -\frac{ (5e^{(- 2x^{2} - 2y^{2})})}{41})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(12\cdot rsin(2\cdot r^{2}) -\frac{ (5\cdot re^{(-2\cdot r^{2})})}{41})drd\theta\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.562}{-0.827})=0.81\pi;\theta_2=arctan(\frac{-0.750}{0.661})=1.73\pi\&\text{Now, utilizing integral rules:}\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\&\int_{0.81\pi}^{1.73\pi}\int_{0}^{0.9}(12\cdot rsin(2\cdot r^{2}) -\frac{ (5\cdot re^{(-2\cdot r^{2})})}{41})drd\theta=(\frac{(5e^{(-2\cdot r^{2})})}{164}- 6cos(r^{2})^{2})d\theta|_{0}^{0.9}\&\int_{0.81\pi}^{1.73\pi}(3.123)d\theta=(3.123\cdot \theta)|_{0.81\pi}^{1.73\pi}=9.03\end{align*}$

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(- 2e^{(- x^{2} - y^{2})} -\frac{ 40}{3})dA\text{, where D is}\&\text{the region of a circle centered on the origin with a radius of }\&1.54\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(-0.339,0.941)\text{ and }\overrightarrow{u_2}=(0.218,-0.976)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Explanation

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to polar form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(- 2e^{(- x^{2} - y^{2})} -\frac{ 40}{3})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (40\cdot r)}{3}- 2\cdot re^{(-r^{2})})drd\theta\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.941}{-0.339})=0.61\pi;\theta_2=arctan(\frac{-0.976}{0.218})=1.57\pi\&\text{Now, utilizing integral rules:}\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\&\int_{0.61\pi}^{1.57\pi}\int_{0}^{1.54}(-\frac{ (40\cdot r)}{3}- 2\cdot re^{(-r^{2})})drd\theta=(e^{(-r^{2})} -\frac{ (20\cdot r^{2})}{3})d\theta|_{0}^{1.54}\&\int_{0.61\pi}^{1.57\pi}(-16.72)d\theta=(-16.72\cdot \theta)|_{0.61\pi}^{1.57\pi}=-50.42\end{align*}$

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.

$4\pi$

$\frac{8}{3}\pi$

$2\pi$

$8\pi$

$\frac{5}{4}\pi$

Explanation

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:

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

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

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.

Explanation

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 .

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

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

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

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

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

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

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

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

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

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

Explanation

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

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

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

Explanation

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

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

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}{7}+\frac{ (\frac{5}{4^{(x^{2} + y^{2})}})}{13})dA\text{, where D is}\&\text{the region of a circle centered on the origin with a radius of }\&1.73\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.368,0.930)\text{ and }\overrightarrow{u_2}=(-0.750,-0.661)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Explanation

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to polar form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(\frac{sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}{7}+\frac{ (\frac{5}{4^{(x^{2} + y^{2})}})}{13})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(5\cdot r)}{(13\cdot 4^{(r^{2})})}+\frac{ (rsin(\frac{(2\cdot r^{2})}{3}))}{7})drd\theta\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.930}{0.368})=0.38\pi;\theta_2=arctan(\frac{-0.661}{-0.750})=1.23\pi\&\text{Now, utilizing integral rules:}\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\&\int_{0.38\pi}^{1.23\pi}\int_{0}^{1.73}(\frac{(5\cdot r)}{(13\cdot 4^{(r^{2})})}+\frac{ (rsin(\frac{(2\cdot r^{2})}{3}))}{7})drd\theta=(-\frac{ (3cos(\frac{(2\cdot r^{2})}{3}))}{28}-\frac{ 5}{(52\cdot 2^{(2\cdot r^{2})}ln(2))})d\theta|_{0}^{1.73}\&\int_{0.38\pi}^{1.23\pi}(0.2878)d\theta=(0.2878\theta)|_{0.38\pi}^{1.23\pi}=0.77\end{align*}$

Page 1 of 19