Multiple Integration

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AP Calculus BC › Multiple Integration

Questions 1 - 10

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

Explanation

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

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

$\begin{align*}&\text{Evaluate the triple integral}\int_{10}^{14}\int_{-6}^{-1}\int_{7}^{11}(\frac{(11cos(y + 1)sin(z + 1))}{(2x)})dxdydz\end{align*}$

Explanation

$\begin{align*}&\text{Performing a triple integral, the order of integration does}\&\text{not entirely matter.}\&\text{For example:}\&\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx\&\text{Considering our integral, and keeping savvy about }\&\text{utilizing integral rules:}\&\int[cos(ax)]=\frac{sin(ax)}{a}\&\int[sin(ax)]=-\frac{cos(ax)}{a}\end{align*}$

$\begin{align*}&\int_{10}^{14}\int_{-6}^{-1}\int_{7}^{11}(\frac{(11cos(y + 1)sin(z + 1))}{(2x)})dxdydz\&\text{The approach is simply to take it step by step:}\&\int_{10}^{14}\int_{-6}^{-1}\int_{7}^{11}(\frac{(11cos(y + 1)sin(z + 1))}{(2x)})dxdydz=\int_{10}^{14}\int_{-6}^{-1}(\frac{(11cos(y + 1)sin(z + 1)ln(x))}{2})dydz|_{7}^{11}\&\int_{10}^{14}\int_{-6}^{-1}(\frac{(11cos(y + 1)sin(z + 1)ln(\frac{11}{7}))}{2})dydz=\int_{10}^{14}(\frac{(11sin(y + 1)sin(z + 1)ln(\frac{11}{7}))}{2})dz|_{-6}^{-1}\&\int_{10}^{14}(\frac{(11sin(z + 1)ln(\frac{11}{7})sin(5))}{2})dz=-\frac{(11cos(z + 1)ln(\frac{11}{7})sin(5))}{2}|_{10}^{14}=-1.82\end{align*}$

$\begin{align*}&\text{Evaluate the triple integral}\int_{5}^{10}\int_{-2}^{3}\int_{1}^{5}(2x + 4y - 3x^2)dxdydz\end{align*}$

Explanation

$\begin{align*}&\text{When performing a triple integral, it is worth noting that the order}\&\text{in which the integration performed does not entirely matter, since each variable will be}\&\text{integrated independently, with the others being treated as constants for the}\&\text{puposes of integration. For example:}\&\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx\&\text{Considering our integral:}\&\int_{5}^{10}\int_{-2}^{3}\int_{1}^{5}(2x + 4y - 3x^2)dxdydz\&\text{The approach is simply to take it step by step:}\&\int_{5}^{10}\int_{-2}^{3}\int_{1}^{5}(2x + 4y - 3x^2)dxdydz=\int_{5}^{10}\int_{-2}^{3}(x{(x + 4y - x^2)})dydz|_{1}^{5}\&\int_{5}^{10}\int_{-2}^{3}(16y - 100)dydz=\int_{5}^{10}(4y{(2y - 25)})dz|_{-2}^{3}\&\int_{5}^{10}(-460)dz=-460zdz|_{5}^{10}=-2300\end{align*}$

$\begin{align*}&\text{Evaluate the triple integral}\int_{3}^{5.5}\int_{3}^{7.5}\int_{-4.5}^{-1}(8e^{(-2y)}e^{(-2z)}e^{(x)})dxdydz\end{align*}$

$4.35\cdot10^{-6}$

$-1.45\cdot10^{-6}$

$1.45\cdot10^{-6}$

$-1.31\cdot10^{-5}$

Explanation

$\begin{align*}&\int_{3}^{5.5}\int_{3}^{7.5}\int_{-4.5}^{-1}(8e^{(-2y)}e^{(-2z)}e^{(x)})dxdydz\&\text{The approach is simply to take it step by step:}\&\int_{3}^{5.5}\int_{3}^{7.5}\int_{-4.5}^{-1}(8e^{(-2y)}e^{(-2z)}e^{(x)})dxdydz=\int_{3}^{5.5}\int_{3}^{7.5}(8e^{(-2y)}e^{(-2z)}e^{(x)})dydz|_{-4.5}^{-1}\&\int_{3}^{5.5}\int_{3}^{7.5}(e^{(- 2y - 2z)}\cdot (8e^{(-1)} - 8e^{(-\frac{9}{2})}))dydz=\int_{3}^{5.5}(-4e^{(-\frac{9}{2})}e^{(- 2y - 2z)}\cdot (e^{(\frac{7}{2})} - 1))dz|_{3}^{7.5}\&\int_{3}^{5.5}(4e^{(-2z)}\cdot (e^{(-16)} - e^{(-\frac{39}{2})})\cdot (e^{(9)} - 1))dz=-2e^{(-2z)}\cdot (e^{(-16)} - e^{(-\frac{39}{2})})\cdot (e^{(9)} - 1)|_{3}^{5.5}=4.35\cdot10^{-6}\end{align*}$

$\begin{align*}&\text{Evaluate the triple integral}\int_{10}^{13.5}\int_{5}^{7}\int_{-7}^{-2}(\frac{(8\cdot 3^ysin(3x))}{(31z)})dxdydz\end{align*}$

Explanation

$\begin{align*}&\text{Performing a triple integral, the order of integration does}\&\text{not entirely matter.}\&\text{For example:}\&\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx\&\text{Considering our integral, and keeping savvy about }\&\text{utilizing integral rules:}\&\int[sin(ax)]=-\frac{cos(ax)}{a}\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\end{align*}$

$\begin{align*}&\int_{10}^{13.5}\int_{5}^{7}\int_{-7}^{-2}(\frac{(8\cdot 3^ysin(3x))}{(31z)})dxdydz\&\text{The approach is simply to take it step by step:}\&\int_{10}^{13.5}\int_{5}^{7}\int_{-7}^{-2}(\frac{(8\cdot 3^ysin(3x))}{(31z)})dxdydz=\int_{10}^{13.5}\int_{5}^{7}(-\frac{(8\cdot 3^ycos(3x))}{(93z)})dydz|_{-7}^{-2}\&\int_{10}^{13.5}\int_{5}^{7}(-\frac{(8\cdot 3^y\cdot (\frac{cos(6)}{3}-\frac{ cos(21)}{3}))}{(31z)})dydz=\int_{10}^{13.5}(-\frac{(8\cdot 3^y\cdot (cos(6) - cos(21)))}{(93zln(3))})dz|_{5}^{7}\&\int_{10}^{13.5}(-\frac{(5184\cdot (cos(6) - cos(21)))}{(31zln(3))})dz=-\frac{(ln(z)\cdot (5184cos(6) - 5184cos(21)))}{(31ln(3))}|_{10}^{13.5}=-68.88\end{align*}$

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

Explanation

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

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

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

Explanation

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

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

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(6e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})}e^{(-z)})}{29})dA\text{, where D is}\&\text{the region of a cylindrical annulus centered on the origin,}\&\text{with inner/outer radii }\&0.25\text{ and }1.23\&\text{and length }1.8\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(-0.891,0.454)\text{ and }\overrightarrow{u_2}=(0.930,-0.368)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Explanation

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

$\begin{align*}&\theta_1=arctan(\frac{0.454}{-0.891})=0.85\pi;\theta_2=arctan(\frac{-0.368}{0.930})=1.88\pi\&\text{Now, utilizing integral rules:}\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\&\int[e^{az}]=\frac{e^{az}}{a}\&\int_{0}^{1.8}\int_{0.85\pi}^{1.88\pi}\int_{0.25}^{1.23}(-\frac{(6\cdot re^{(-z)}e^{(-\frac{(2\cdot r^{2})}{3})})}{29})drd\theta dz=(\frac{(9e^{(- z -\frac{ (2\cdot r^{2})}{3})})}{58})d\theta dz|_{0.25}^{1.23}\&\int_{0}^{1.8}\int_{0.85\pi}^{1.88\pi}(-0.09224e^{(-z)})d\theta dz=(-0.09224\theta e^{(-z)})dz|_{0.85\pi}^{1.88\pi}\&\int_{0}^{1.8}(-0.2985e^{(-z)})dz=(0.2985e^{(-z)})|_{0}^{1.8}=-0.25\end{align*}$

$\begin{align*}&\text{Evaluate the triple integral}\int_{9}^{14}\int_{-6}^{-1}\int_{4.5}^{9.5}(\frac{(sin(3y)e^{(2x)})}{(6z)})dxdydz\end{align*}$

$3.615\cdot10^{6}$

$1.807\cdot10^{6}$

$-2.169\cdot10^{7}$

Explanation

$\begin{align*}&\text{Performing a triple integral, the order of integration does}\&\text{not entirely matter.}\&\text{For example:}\&\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx\&\text{Considering our integral, and keeping savvy about }\&\text{utilizing integral rules:}\&\int[e^{ax}]=\frac{e^{ax}}{a}\&\int[sin(ax)]=-\frac{cos(ax)}{a}\end{align*}$

$\begin{align*}&\int_{9}^{14}\int_{-6}^{-1}\int_{4.5}^{9.5}(\frac{(sin(3y)e^{(2x)})}{(6z)})dxdydz\&\text{The approach is simply to take it step by step:}\&\int_{9}^{14}\int_{-6}^{-1}\int_{4.5}^{9.5}(\frac{(sin(3y)e^{(2x)})}{(6z)})dxdydz=\int_{9}^{14}\int_{-6}^{-1}(\frac{(sin(3y)e^{(2x)})}{(12z)})dydz|_{4.5}^{9.5}\&\int_{9}^{14}\int_{-6}^{-1}(\frac{(sin(3y)e^{(9)}\cdot(e^{(10)} - 1))}{(12z)})dydz=\int_{9}^{14}(-\frac{(cos(3y)e^{(9)}\cdot(e^{(10)} - 1))}{(36z)})dz|_{-6}^{-1}\&\int_{9}^{14}(-\frac{(e^{(9)}\cdot(cos(3) - cos(18))\cdot(e^{(10)} - 1))}{(36z)})dz=ln(z)\cdot(\frac{(cos(3)e^{(9)})}{36}-\frac{ (cos(3)e^{(19)})}{36}-\frac{ (cos(18)e^{(9)})}{36}+\frac{ (cos(18)e^{(19)})}{36})|_{9}^{14}=3.615\cdot10^{6}\end{align*}$

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

Explanation

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

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

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