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Calculus 3 : Triple Integration in Cartesian Coordinates

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Example Questions

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:

$\frac{1}{8}$

$\frac{65}{8}$

$\frac{3465}{8}$

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 #3 : Triple Integration In Cartesian Coordinates

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 #4 : Triple Integration In Cartesian Coordinates

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:

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

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

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

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

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

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 #132 : Triple Integrals

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 #7 : Triple Integration In Cartesian Coordinates

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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Example Question #1 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{-13}^{2}\int_{-13}^{-7}\int_{6}^{9}(3x + y + 2z)dxdydz\end{align*}$

Possible Answers:

405

Correct answer:

405

Explanation:

$\begin{align*}&\text{When performing a triple integral, it is worth noting that the order}\\&\text{in which the integration performed does not entirely matter, since each variable will be}\\&\text{integrated independently, with the others being treated as constants for the}\\&\text{puposes of integration. For example:}\\&\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx\\&\text{Considering our integral:}\\&\int_{-13}^{2}\int_{-13}^{-7}\int_{6}^{9}(3x + y + 2z)dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-13}^{2}\int_{-13}^{-7}\int_{6}^{9}(3x + y + 2z)dxdydz=\int_{-13}^{2}\int_{-13}^{-7}(x{(y + 2z)} + {(3x^2)}/2)dydz|_{6}^{9}\\&\int_{-13}^{2}\int_{-13}^{-7}(3y + 6z + 135/2)dydz=\int_{-13}^{2}({(3y{(y + 4z + 45)})}/2)dz|_{-13}^{-7}\\&\int_{-13}^{2}(36z + 225)dz=9z{(2z + 25)}dz|_{-13}^{2}=405\end{align*}$

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Example Question #1 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{8}^{20}\int_{-7}^{-2}\int_{-11}^{18}(6xy - 2z)dxdydz\end{align*}$

Possible Answers:

527220

Correct answer:

Explanation:

$\begin{align*}&\text{When performing a triple integral, it is worth noting that the order}\\&\text{in which the integration performed does not entirely matter, since each variable will be}\\&\text{integrated independently, with the others being treated as constants for the}\\&\text{puposes of integration. For example:}\\&\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx\\&\text{Considering our integral:}\\&\int_{8}^{20}\int_{-7}^{-2}\int_{-11}^{18}(6xy - 2z)dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{8}^{20}\int_{-7}^{-2}\int_{-11}^{18}(6xy - 2z)dxdydz=\int_{8}^{20}\int_{-7}^{-2}(3x^2y - 2xz)dydz|_{-11}^{18}\\&\int_{8}^{20}\int_{-7}^{-2}(609y - 58z)dydz=\int_{8}^{20}({(29y{(21y - 4z)})}/2)dz|_{-7}^{-2}\\&\int_{8}^{20}(- 290z - 27405/2)dz=-{(145z{(2z + 189)})}/2dz|_{8}^{20}=-213150\end{align*}$

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Example Question #10 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{-4}^{3}\int_{-8}^{-2}\int_{4}^{6}(24xyz^2)dxdydz\end{align*}$

Possible Answers:

127680

Correct answer:

Explanation:

$\begin{align*}&\text{When performing a triple integral, it is worth noting that the order}\\&\text{in which the integration performed does not entirely matter, since each variable will be}\\&\text{integrated independently, with the others being treated as constants for the}\\&\text{puposes of integration. For example:}\\&\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx\\&\text{Considering our integral:}\\&\int_{-4}^{3}\int_{-8}^{-2}\int_{4}^{6}(24xyz^2)dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-4}^{3}\int_{-8}^{-2}\int_{4}^{6}(24xyz^2)dxdydz=\int_{-4}^{3}\int_{-8}^{-2}(12x^2yz^2)dydz|_{4}^{6}\\&\int_{-4}^{3}\int_{-8}^{-2}(240yz^2)dydz=\int_{-4}^{3}(120y^2z^2)dz|_{-8}^{-2}\\&\int_{-4}^{3}(-7200z^2)dz=-2400z^3dz|_{-4}^{3}=-218400\end{align*}$

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Calculus 3 : Triple Integration in Cartesian Coordinates

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1 : Triple Integration In Cartesian Coordinates

Example Question #1 : Triple Integration In Cartesian Coordinates

Example Question #3 : Triple Integration In Cartesian Coordinates

Example Question #4 : Triple Integration In Cartesian Coordinates

Example Question #1 : Triple Integration In Cartesian Coordinates

Example Question #132 : Triple Integrals

Example Question #7 : Triple Integration In Cartesian Coordinates

Example Question #1 : Triple Integration In Cartesian Coordinates

Example Question #1 : Triple Integration In Cartesian Coordinates

Example Question #10 : Triple Integration In Cartesian Coordinates

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