Double Integrals

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AP Calculus BC › Double Integrals

Questions 1 - 10

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

Explanation

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

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

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(3cos(x^{2} + y^{2}))}{2}-\frac{ cos(2x^{2} + 2y^{2})}{8})dA\text{, where D is}\&\text{the region defined between two circles centered on the origin with radii }\&0.25\text{ and }1.93\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.218,0.976)\text{ and }\overrightarrow{u_2}=(-0.941,-0.339)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Explanation

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to polar form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(\frac{(3cos(x^{2} + y^{2}))}{2}-\frac{ cos(2x^{2} + 2y^{2})}{8})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(3\cdot rcos(r^{2}))}{2}-\frac{ (rcos(2\cdot r^{2}))}{8})drd\theta\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.976}{0.218})=0.43\pi;\theta_2=arctan(\frac{-0.339}{-0.941})=1.11\pi\&\text{Now, utilizing integral rules:}\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\&\int_{0.43\pi}^{1.11\pi}\int_{0.25}^{1.93}(\frac{(3\cdot rcos(r^{2}))}{2}-\frac{ (rcos(2\cdot r^{2}))}{8})drd\theta=(-\frac{(sin(r^{2})\cdot (cos(r^{2}) - 12))}{16})d\theta|_{0.25}^{1.93}\&\int_{0.43\pi}^{1.11\pi}(-0.4848)d\theta=(-0.4848\theta)|_{0.43\pi}^{1.11\pi}=-1.04\end{align*}$

Calculate the following Integral.

$\int_{0}^{1}\int_{x}^{x^2} xy^2 dydx$

$-\frac{3}{120}$

$-\frac{1}{120}$

$\frac{1}{120}$

Explanation

$\int_{0}^{1}\int_{x}^{x^2} xy^2 dydx=\int_{0}^{1}\Big(\int_{x}^{x^2} xy^2 dy\Big)dx$

Lets deal with the inner integral first.

$\int_{x}^{x^2}xy^2 dy$

$\int_{x}^{x^2}xy^2 dy=\frac{1}{3}y^3x\Big|_{x}^{x^2}$

$=(\frac{1}{3}(x^2)^3\cdot x)-(\frac{1}{3}x^3\cdot x)$

$=\frac{1}{3}x^7-\frac{1}{3}x^4$

Now we evaluate this expression in the outer integral.

$\int_{0}^{1} \frac{1}{3}x^7-\frac{1}{3}x^4dx$

$=\frac{1}{3}\int_{0}^{1} x^7-x^4dx$

$=\frac{1}{3}\Big( \frac{1}{8}x^8-\frac{1}{5}x^5\Big)\Big|_{0}^{1}$

$=\frac{1}{3}\Big(\frac{1}{8}1^8-\frac{1}{5}1^5\Big)-\frac{1}{3}\Big(\frac{1}{8}0^8-\frac{1}{5}0^5\Big)$

$=\frac{1}{3}\Big(\frac{1}{8}-\frac{1}{5}\Big)-\frac{1}{3}\Big(0-0\Big)$

$=\frac{1}{3}\Big(\frac{5}{40}-\frac{8}{40}\Big)$

$=\frac{1}{3}\Big(-\frac{3}{40}\Big)=-\frac{3}{120}$

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (13\cdot (\frac{2}{7})^{(2x^{2} + 2y^{2})})}{6}-\frac{ (2\cdot 3^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})})}{3})dA\text{, where D is}\&\text{the region defined between two circles centered on the origin with radii }\&0.48\text{ and }0.98\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.279,0.960)\text{ and }\overrightarrow{u_2}=(-0.982,-0.187)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Explanation

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to polar form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(-\frac{ (13\cdot (\frac{2}{7})^{(2x^{2} + 2y^{2})})}{6}-\frac{ (2\cdot 3^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})})}{3})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (2\cdot 3^{(\frac{(3\cdot r^{2})}{2})}\cdot r)}{3}-\frac{ (13\cdot (\frac{2}{7})^{(2\cdot r^{2})}\cdot r)}{6})drd\theta\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.960}{0.279})=0.41\pi;\theta_2=arctan(\frac{-0.187}{-0.982})=1.06\pi\&\text{Now, utilizing integral rules:}\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\&\int_{0.41\pi}^{1.06\pi}\int_{0.48}^{0.98}(-\frac{ (2\cdot 3^{(\frac{(3\cdot r^{2})}{2})}\cdot r)}{3}-\frac{ (13\cdot (\frac{2}{7})^{(2\cdot r^{2})}\cdot r)}{6})drd\theta=(-\frac{ (2\cdot 3^{(\frac{(3\cdot r^{2})}{2})})}{(9ln(3))}-\frac{ (13\cdot (\frac{2}{7})^{(2\cdot r^{2})})}{(24ln(\frac{2}{7}))})d\theta|_{0.48}^{0.98}\&\int_{0.41\pi}^{1.06\pi}(-0.8927)d\theta=(-0.8927\theta)|_{0.41\pi}^{1.06\pi}=-1.82\end{align*}$

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(10e^{(x^{2} + y^{2})} +\frac{ (\frac{48}{(\frac{2}{7})^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}})}{5})dA\text{, where D is}\&\text{the region of a circle centered on the origin with a radius of }\&1.16\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.094,0.996)\text{ and }\overrightarrow{u_2}=(0.941,-0.339)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Explanation

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to polar form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(10e^{(x^{2} + y^{2})} +\frac{ (\frac{48}{(\frac{2}{7})^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}})}{5})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(10\cdot re^{(r^{2})} +\frac{ (48\cdot r)}{(5\cdot (\frac{2}{7})^{(\frac{(2\cdot r^{2})}{3})})})drd\theta\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.996}{0.094})=0.47\pi;\theta_2=arctan(\frac{-0.339}{0.941})=1.89\pi\&\text{Now, utilizing integral rules:}\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\&\int_{0.47\pi}^{1.89\pi}\int_{0}^{1.16}(10\cdot re^{(r^{2})} +\frac{ (48\cdot r)}{(5\cdot (\frac{2}{7})^{(\frac{(2\cdot r^{2})}{3})})})drd\theta=(5e^{(r^{2})} -\frac{ 36}{(5\cdot (\frac{2}{7})^{(\frac{(2\cdot r^{2})}{3})}ln(\frac{2}{7}))})d\theta|_{0}^{1.16}\&\int_{0.47\pi}^{1.89\pi}(26.14)d\theta=(26.14\theta)|_{0.47\pi}^{1.89\pi}=116.6\end{align*}$

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(\frac{2}{(\frac{7}{2})^{(x^{2} + y^{2})}})}{13}+ 46\cdot (\frac{1}{2})^{(x^{2} + y^{2})})dA\text{, where D is}\&\text{the region of a circle centered on the origin with a radius of }\&1.51\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.729,0.685)\text{ and }\overrightarrow{u_2}=(0.279,-0.960)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Explanation

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to polar form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(\frac{(\frac{2}{(\frac{7}{2})^{(x^{2} + y^{2})}})}{13}+ 46\cdot (\frac{1}{2})^{(x^{2} + y^{2})})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(46\cdot (\frac{1}{2})^{(r^{2})}\cdot r +\frac{ (2\cdot r)}{(13\cdot (\frac{7}{2})^{(r^{2})})})drd\theta\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.685}{0.729})=0.24\pi;\theta_2=arctan(\frac{-0.960}{0.279})=1.59\pi\&\text{Now, utilizing integral rules:}\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\&\int_{0.24\pi}^{1.59\pi}\int_{0}^{1.51}(46\cdot (\frac{1}{2})^{(r^{2})}\cdot r +\frac{ (2\cdot r)}{(13\cdot (\frac{7}{2})^{(r^{2})})})drd\theta=(-\frac{ 23}{(2^{(r^{2})}ln(2))}-\frac{ 1}{(13\cdot (\frac{7}{2})^{(r^{2})}ln(\frac{7}{2}))})d\theta|_{0}^{1.51}\&\int_{0.24\pi}^{1.59\pi}(26.41)d\theta=(26.41\theta)|_{0.24\pi}^{1.59\pi}=112\end{align*}$

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(3e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})})}{16}-\frac{ (4cos(x^{2} + y^{2}))}{5})dA\text{, where D is}\&\text{the region of a circle centered on the origin with a radius of }\&0.97\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(-0.960,0.279)\text{ and }\overrightarrow{u_2}=(0.771,-0.637)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Explanation

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to polar form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(\frac{(3e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})})}{16}-\frac{ (4cos(x^{2} + y^{2}))}{5})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(3\cdot re^{(-\frac{(2\cdot r^{2})}{3})})}{16}-\frac{ (4\cdot rcos(r^{2}))}{5})drd\theta\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.279}{-0.960})=0.91\pi;\theta_2=arctan(\frac{-0.637}{0.771})=1.78\pi\&\text{Now, utilizing integral rules:}\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\&\int_{0.91\pi}^{1.78\pi}\int_{0}^{0.97}(\frac{(3\cdot re^{(-\frac{(2\cdot r^{2})}{3})})}{16}-\frac{ (4\cdot rcos(r^{2}))}{5})drd\theta=(-\frac{ (2sin(r^{2}))}{5}-\frac{ (9e^{(-\frac{(2\cdot r^{2})}{3})})}{64})d\theta|_{0}^{0.97}\&\int_{0.91\pi}^{1.78\pi}(-0.2577)d\theta=(-0.2577\theta)|_{0.91\pi}^{1.78\pi}=-0.7\end{align*}$

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(3cos(2x^{2} + 2y^{2}))}{34}+\frac{ (46sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{3})dA\text{, where D is}\&\text{the region of a circle centered on the origin with a radius of }\&0.63\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.930,0.368)\text{ and }\overrightarrow{u_2}=(-0.536,-0.844)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Explanation

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to polar form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(\frac{(3cos(2x^{2} + 2y^{2}))}{34}+\frac{ (46sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{3})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(3\cdot rcos(2\cdot r^{2}))}{34}+\frac{ (46\cdot rsin(\frac{(2\cdot r^{2})}{3}))}{3})drd\theta\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.368}{0.930})=0.12\pi;\theta_2=arctan(\frac{-0.844}{-0.536})=1.32\pi\&\text{Now, utilizing integral rules:}\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\&\int_{0.12\pi}^{1.32\pi}\int_{0}^{0.63}(\frac{(3\cdot rcos(2\cdot r^{2}))}{34}+\frac{ (46\cdot rsin(\frac{(2\cdot r^{2})}{3}))}{3})drd\theta=(\frac{(3sin(2\cdot r^{2}))}{136}-\frac{ (23cos(\frac{(2\cdot r^{2})}{3}))}{2})d\theta|_{0}^{0.63}\&\int_{0.12\pi}^{1.32\pi}(0.416)d\theta=(0.416\theta)|_{0.12\pi}^{1.32\pi}=1.57\end{align*}$

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{5}{(x^{2} + y^{2})}+ 2\cdot (\frac{2}{5})^{(x^{2} + y^{2})})dA\text{, where D is}\&\text{the region defined between two circles centered on the origin with radii }\&0.48\text{ and }1.84\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.891,0.454)\text{ and }\overrightarrow{u_2}=(-0.661,-0.750)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Explanation

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

$\begin{align*}&\theta_1=arctan(\frac{0.454}{0.891})=0.15\pi;\theta_2=arctan(\frac{-0.750}{-0.661})=1.27\pi\&\text{Now, utilizing integral rules:}\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\&\int_{0.15\pi}^{1.27\pi}\int_{0.48}^{1.84}(2\cdot (\frac{2}{5})^{(r^{2})}\cdot r +\frac{ 5}{r})drd\theta=(5ln(r) + (\frac{\frac{2}{5})^{(r^{2})}}{ln(\frac{2}{5})})d\theta|_{0.48}^{1.84}\&\int_{0.15\pi}^{1.27\pi}(7.553)d\theta=(7.553\theta)|_{0.15\pi}^{1.27\pi}=26.58\end{align*}$

Explanation

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to polar form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(-\frac{ (13\cdot (\frac{2}{7})^{(2x^{2} + 2y^{2})})}{6}-\frac{ (2\cdot 3^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})})}{3})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (2\cdot 3^{(\frac{(3\cdot r^{2})}{2})}\cdot r)}{3}-\frac{ (13\cdot (\frac{2}{7})^{(2\cdot r^{2})}\cdot r)}{6})drd\theta\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

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