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Calculus 3 : Triple Integrals

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Example Questions

Example Question #131 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(4cos(z + 2)sin(x^{2} + y^{2}))}{15})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.12\text{ and }1.92\\&\text{and length }0.63\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.309,0.951)\text{ and }\overrightarrow{u_2}=(-0.536,-0.844)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.12

-0.06

-0.23

0.94

Correct answer:

-0.23

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(4cos(z + 2)sin(x^{2} + y^{2}))}{15})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(4\cdot rcos(z + 2)sin(r^{2}))}{15})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.951}{-0.309})=0.6\pi;\theta_2=arctan(\frac{-0.844}{-0.536})=1.32\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[cos(az)]=\frac{sin(az)}{a}\\&\int_{0}^{0.63}\int_{0.6\pi}^{1.32\pi}\int_{0.12}^{1.92}(\frac{(4\cdot rcos(z + 2)sin(r^{2}))}{15})drd\theta dz=(-\frac{(2cos(r^{2})cos(z + 2))}{15})d\theta dz|_{0.12}^{1.92}\\&\int_{0}^{0.63}\int_{0.6\pi}^{1.32\pi}(0.2473cos(z + 2))d\theta dz=(0.2473\theta cos(z + 2))dz|_{0.6\pi}^{1.32\pi}\\&\int_{0}^{0.63}(0.5595cos(z + 2))dz=(0.5595sin(z + 2))|_{0}^{0.63}=-0.23\end{align*}$

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Example Question #132 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(\frac{4}{(\frac{7}{2})^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}e^{(-2z)}})}{27})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.65\\&\text{and length }1.77\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.536,0.844)\text{ and }\overrightarrow{u_2}=(-0.844,-0.536)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.01

-0.03

0.01

0.03

Correct answer:

-0.03

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(\frac{4}{(\frac{7}{2})^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}e^{(-2z)}})}{27})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(4\cdot re^{(-2z)})}{(27\cdot (\frac{7}{2})^{(\frac{(3\cdot r^{2})}{2})})})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.844}{-0.536})=0.68\pi;\theta_2=arctan(\frac{-0.536}{-0.844})=1.18\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.77}\int_{0.68\pi}^{1.18\pi}\int_{0}^{1.65}(-\frac{(4\cdot re^{(-2z)})}{(27\cdot (\frac{7}{2})^{(\frac{(3\cdot r^{2})}{2})})})drd\theta dz=(\frac{(4\cdot 2^{(\frac{(3\cdot r^{2})}{2})}e^{(-2z)})}{(81\cdot 7^{(\frac{(3\cdot r^{2})}{2})}ln(\frac{7}{2}))})d\theta dz|_{0}^{1.65}\\&\int_{0}^{1.77}\int_{0.68\pi}^{1.18\pi}(-0.03918e^{(-2z)})d\theta dz=(-0.03918\theta e^{(-2z)})dz|_{0.68\pi}^{1.18\pi}\\&\int_{0}^{1.77}(-0.06155e^{(-2z)})dz=(0.03077e^{(-2z)})|_{0}^{1.77}=-0.03\end{align*}$

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Example Question #133 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(3sin(z + 1)e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})})}{28})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&0.86\\&\text{and length }0.89\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.790,0.613)\text{ and }\overrightarrow{u_2}=(-0.960,-0.279)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.07

-0.02

-0.37

0.02

Correct answer:

0.07

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(3sin(z + 1)e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})})}{28})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(3\cdot rsin(z + 1)e^{(-\frac{(2\cdot r^{2})}{3})})}{28})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.613}{0.790})=0.21\pi;\theta_2=arctan(\frac{-0.279}{-0.960})=1.09\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[sin(az)]=-\frac{cos(az)}{a}\\&\int_{0}^{0.89}\int_{0.21\pi}^{1.09\pi}\int_{0}^{0.86}(\frac{(3\cdot rsin(z + 1)e^{(-\frac{(2\cdot r^{2})}{3})})}{28})drd\theta dz=(-\frac{(9sin(z + 1)e^{(-\frac{(2\cdot r^{2})}{3})})}{112})d\theta dz|_{0}^{0.86}\\&\int_{0}^{0.89}\int_{0.21\pi}^{1.09\pi}(0.03128sin(z + 1))d\theta dz=(0.03128\theta sin(z + 1))dz|_{0.21\pi}^{1.09\pi}\\&\int_{0}^{0.89}(0.08647sin(z + 1))dz=(-0.08647cos(z + 1))|_{0}^{0.89}=0.07\end{align*}$

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Example Question #1 : Triple Integration In Cartesian Coordinates

Evaluate the following integral

$\int \int \int_D 15xyz \ dV$ , where $D=[1,2]\times[3,4]\times [5,6]$

Possible Answers:

$\frac{65}{8}$

$\frac{3465}{8}$

$\frac{1}{8}$

Correct answer:

$\frac{3465}{8}$

Explanation:

$\int_{1}^{2} \int_{3}^{4} \int_{5}^{6} 15xyz \ dx\ dy\ dz$

$=\int_{1}^{2} \int_{3}^{4} \frac{15}{2}x^2yz\Big|_{5}^{6} dy\ dz$

$=\int_{1}^{2} \int_{3}^{4} (\frac{15}{2}(6)^2yz)-(\frac{15}{2}(5)^2yz) dy\ dz$

$=\int_{1}^{2} \int_{3}^{4} \frac{165}{2}yz\ dy\ dz$

$=\int_{1}^{2} \frac{165}{4}y^2z\Big|_{3}^{4}\ dz$

$=\int_{1}^{2} \frac{165}{4}(4)^2z-\frac{165}{4}(3)^2z\ dz$

$=\int_{1}^{2} \frac{1155}{4}z\ dz$

$=\frac{1155}{8}z^2\Big|_{1}^{2}$

$=\frac{1155}{8}(2)^2-\frac{1155}{8}(1)^2=\frac{3465}{8}$

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Example Question #1 : Triple Integration In Cartesian Coordinates

Find the value of the triple integral $\int \int_{B} \int yz^2dxdydz$ , $B=[0,1]\times[0,2]\times[0,3]$

Possible Answers:

Correct answer:

Explanation:

An integral expression of the form

$\int \int_{B} \int f(x,y,z)dxdydz$ , $B=[a,b]\times[c,d]\times[s,t]$

is equivalent to an integral of the form

$\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz$

Where the innermost integral corresponds to the innermost derivative variable, and the outermost integral corresponds to the outermost variable.

It is worth noting that the order the integration is performed does not matter, since each variable will be integrated independently, with the others being treated as constants for the purposes of integration. For example

$\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx$

Considering our integral:

$\int \int_{B} \int yz^2dxdydz$ , $B=[0,1]\times[0,2]\times[0,3]$

The approach is simply to take it step by step:

$\int_{0}^{3}\int_{0}^{2}\int_{0}^{1}yz^2dxdydz=\int_{0}^{3}\int_{0}^{2}xyz^2|_{0}^{1}dydz$

$\int_{0}^{3}\int_{0}^{2}yz^2dydz=\int_{0}^{3}\frac{y^2z^2}{2}|_{0}^{2}dz$

$\int_{0}^{3}2z^2dz=\frac{2z^3}{3}|_{0}^{3}=18$

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Example Question #131 : Multiple Integration

Find the value of the triple integral $\int \int_{B} \int (x+y+z)dxdydz$ , $B=[0,1]\times[0,2]\times[0,3]$

Possible Answers:

Correct answer:

Explanation:

An integral expression of the form

$\int \int_{B} \int f(x,y,z)dxdydz$ , $B=[a,b]\times[c,d]\times[s,t]$

is equivalent to an integral of the form

$\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz$

Where the innermost integral corresponds to the innermost derivative variable, and the outermost integral corresponds to the outermost variable.

$\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx$

Considering our integral:

$\int \int_{B} \int (x+y+z)dxdydz$ , $B=[0,1]\times[0,2]\times[0,3]$

The approach is simply to take it step by step:

$\int_{0}^{3}\int_{0}^{2}\int_{0}^{1}(x+y+z)dxdydz=\int_{0}^{3}\int_{0}^{2}(\frac{x^2}{2}+xy+xz)|_{0}^{1}dydz$

$\int_{0}^{3}\int_{0}^{2}(y+z+\frac{1}{2})dydz=\int_{0}^{3}(\frac{y^2}{2}+yz+\frac{y}{2})|_{0}^{2}dz$

$\int_{0}^{3}(2z+3)dz=(z^2+3z)|_{0}^{3}=18$

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Example Question #132 : Multiple Integration

Find the value of the triple integral $\int \int_{B} \int (x^2-yz)dxdydz$ , $B=[0,3]\times[-4,0]\times[-1,1]$

Possible Answers:

Correct answer:

Explanation:

An integral expression of the form

$\int \int_{B} \int f(x,y,z)dxdydz$ , $B=[a,b]\times[c,d]\times[s,t]$

is equivalent to an integral of the form

$\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz$

Where the innermost integral corresponds to the innermost derivative variable, and the outermost integral corresponds to the outermost variable.

$\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx$

Considering our integral:

$\int \int_{B} \int (x^2-yz)dxdydz$ , $B=[0,3]\times[-4,0]\times[-1,1]$

The approach is simply to take it step by step:

$\int_{-1}^{1}\int_{-4}^{0}\int_{0}^{3}(x^2-yz)dxdydz=\int_{-1}^{1}\int_{-4}^{0}(\frac{x^3}{3}-xyz)|_{0}^{3}dydz$

$\int_{-1}^{1}\int_{-4}^{0}(9-3yz)dydz=\int_{-1}^{1}(9y-\frac{3y^2z}{2})|_{-4}^{0}dz$

$\int_{-1}^{1}(36+24z)dz=(36z+12z^2)|_{-1}^{1}=72$

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Example Question #1 : Triple Integration In Cartesian Coordinates

Find the value of the triple integral $\int \int_{B} \int (3x^2z^3e^{xyz^2})dxdydz$ , $B=[0,1]\times[0,2]\times[0,3]$

Possible Answers:

$\frac{e^{18}-179}{72}$

$\frac{e^{18}-179}{24}$

$\frac{e^{18}-181}{24}$

$\frac{e^{18}-181}{72}$

$\frac{e^{18}+179}{72}$

Correct answer:

$\frac{e^{18}-181}{24}$

Explanation:

An integral expression of the form

$\int \int_{B} \int f(x,y,z)dxdydz$ , $B=[a,b]\times[c,d]\times[s,t]$

is equivalent to an integral of the form

$\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz$

Where the innermost integral corresponds to the innermost derivative variable, and the outermost integral corresponds to the outermost variable.

$\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx$

Considering our integral:

$\int \int_{B} \int (3x^2z^3e^{xyz^2})dxdydz$ , $B=[0,1]\times[0,2]\times[0,3]$

The approach is simply to take it step by step. But first, we should consider the order in which we wish to integrate; those and terms in front of the exponent are daunting. Let's see if we can eliminate some by integrating with respect to initially:

${\int_{0}^{3}\int_{0}^{1}\int_{0}^{2}(3x^2z^3e^{xyz^2})dydxdz=\int_{0}^{3}\int_{0}^{1}(3xze^{xyz^2})|_{0}^{2}dxdz=\int_{0}^{3}\int_{0}^{1}(3xze^{2xz^2}-3xz)dxdz}$

This is already looking more manageable. Now, however, we have both an and in front of our exponent, and we must decide which one to choose for the next integration. Recall that $\frac{d}{du}e^{u^2}=2ue^{u^2}$ . Integrating with respect to would be a better way of eliminating the terms in front of the exponent:

$\int_{0}^{1}\int_{0}^{3}(3xze^{2xz^2}-3xz)dzdx=\int_{0}^{1}(\frac{3}{4}e^{2xz^2}-\frac{3}{2}xz^2)|_{0}^{3}dx=\frac{3}{4}e^{18x}-\frac{27}{2}x-\frac{3}{4}$

Finally, we're left with a (relatively) simpler integral:

$\int_{0}^{1}(\frac{3}{4}e^{18x}-\frac{27}{2}x-\frac{3}{4})dx=(\frac{e^{18x}}{24}-\frac{27}{4}x^2-\frac{3}{4}x)|_{0}^{1}=\frac{e^{18}-181}{24}$

Being careful about the order of integration can make the work much easier (relatively).

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Example Question #2 : Triple Integration In Cartesian Coordinates

Find the value of the triple integral $\int \int_{B} \int (-xy^2sin(xyz))dxdydz$ , $B=[0,\pi]\times[0,\frac{1}{2}]\times[0,1]$

Possible Answers:

$\frac{8+\pi^2}{8}$

$\frac{8\pi+\pi^2}{8}$

$\frac{8-\pi^2}{8\pi}$

$\frac{8-\pi}{8\pi}$

$\frac{8-\pi^2}{8}$

Correct answer:

$\frac{8-\pi^2}{8\pi}$

Explanation:

An integral expression of the form

$\int \int_{B} \int f(x,y,z)dxdydz$ , $B=[a,b]\times[c,d]\times[s,t]$

is equivalent to an integral of the form

$\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz$

Where the innermost integral corresponds to the innermost derivative variable, and the outermost integral corresponds to the outermost variable.

$\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx$

Considering our integral:

$\int \int_{B} \int (-xy^2sin(xyz))dxdydz$ , $B=[0,\pi]\times[0,\frac{1}{2}]\times[0,1]$

The approach is simply to take it step by step; however, to make the job of integration easier, it might be prudent to begin by integrating with respect to :

$\int_{0}^{\pi}\int_{0}^{\frac{1}{2}}\int_{0}^{1}(-xy^2sin(xyz))dzdydx=\int_{0}^{\pi}\int_{0}^{\frac{1}{2}}ycos(xyz)|_{0}^{1}dydx$

That worked out pretty well in terms of getting rid of some terms in the front; let's use that trick again and take the integral with respect to next:

$\int_{0}^{\frac{1}{2}}\int_{0}^{\pi}(ycos(xy)-y)dxdy=\int_{0}^{\frac{1}{2}}(sin(xy)-xy)|_{0}^{\pi}dy$

This leaves a relatively straightforward integral with respect to :

$\int_{0}^{\frac{1}{2}}(sin(\pi y)-\pi y)dy=(\frac{-cos(\pi y)}{\pi}-\frac{\pi y^2}{2})|_{0}^{\frac{1}{2}}=\frac{8-\pi^2}{8\pi}$

Being careful about the order of integration can make a complicated problem a bit simpler.

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Example Question #134 : Multiple Integration

Find the value of the triple integral $\int \int_{B} \int xydxdydz$ , $B=[0,3]\times[1,2]\times[-1,3]$

Possible Answers:

$\frac{83}{4}$

Correct answer:

Explanation:

An integral expression of the form

$\int \int_{B} \int f(x,y,z)dxdydz$ , $B=[a,b]\times[c,d]\times[s,t]$

is equivalent to an integral of the form

$\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz$

Where the innermost integral corresponds to the innermost derivative variable, and the outermost integral corresponds to the outermost variable.

$\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx$

Considering our integral:

$\int \int_{B} \int xydxdydz$ , $B=[0,3]\times[1,2]\times[-1,3]$

The approach is simply to take it step by step:

$\int_{-1}^{3}\int_{1}^{2}\int_{0}^{3}xydxdydz=\int_{-1}^{3}\int_{1}^{2}\frac{x^2y}{2}|_{0}^{3}dydz$

$\int_{-1}^{3}\int_{1}^{2}\frac{9y}{2}dydz=\int_{-1}^{3}\frac{9y^2}{4}|_{1}^{2}dz$

$\int_{-1}^{3}dz=\frac{27}{4}z|_{-1}^{3}=27$

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Calculus 3 : Triple Integrals

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #131 : Multiple Integration

Example Question #132 : Multiple Integration

Example Question #133 : Multiple Integration

Example Question #1 : Triple Integration In Cartesian Coordinates

Example Question #1 : Triple Integration In Cartesian Coordinates

Example Question #131 : Multiple Integration

Example Question #132 : Multiple Integration

Example Question #1 : Triple Integration In Cartesian Coordinates

Example Question #2 : Triple Integration In Cartesian Coordinates

Example Question #134 : Multiple Integration

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