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Calculus 3 : Multiple Integration

Study concepts, example questions & explanations for Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #281 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{9}^{13.5}\int_{-8}^{-6}\int_{-9}^{-6.5}(3cos(4y)sin(x + 1)sin(z + 2))dxdydz\end{align*}$

Possible Answers:

-0.92

2.75

0.31

-0.23

Correct answer:

-0.92

Explanation:

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

$\begin{align*}&\int_{9}^{13.5}\int_{-8}^{-6}\int_{-9}^{-6.5}(3cos(4y)sin(x + 1)sin(z + 2))dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{9}^{13.5}\int_{-8}^{-6}\int_{-9}^{-6.5}(3cos(4y)sin(x + 1)sin(z + 2))dxdydz=\int_{9}^{13.5}\int_{-8}^{-6}(-3cos(x + 1)cos(4y)sin(z + 2))dydz|_{-9}^{-6.5}\\&\int_{9}^{13.5}\int_{-8}^{-6}(3cos(4y)sin(z + 2)\cdot (cos(8) - cos(\frac{11}{2})))dydz=\int_{9}^{13.5}(\frac{(3sin(4y)sin(z + 2)\cdot (cos(8) - cos(\frac{11}{2})))}{4})dz|_{-8}^{-6}\\&\int_{9}^{13.5}(-\frac{(3sin(z + 2)\cdot (cos(8) - cos(\frac{11}{2}))\cdot (sin(24) - sin(32)))}{4})dz=\frac{(3cos(z + 2)\cdot (cos(8) - cos(\frac{11}{2}))\cdot (sin(24) - sin(32)))}{4}|_{9}^{13.5}=-0.92\end{align*}$

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Example Question #281 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{6}^{9}\int_{-8}^{-4.5}\int_{-7}^{-2}(\frac{(80cos(y + 1)cos(4z))}{(3x^{3})})dxdydz\end{align*}$

Possible Answers:

-0.33

-0.03

0.07

0.2

Correct answer:

0.07

Explanation:

$\begin{align*}&\int_{6}^{9}\int_{-8}^{-4.5}\int_{-7}^{-2}(\frac{(80cos(y + 1)cos(4z))}{(3x^{3})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{6}^{9}\int_{-8}^{-4.5}\int_{-7}^{-2}(\frac{(80cos(y + 1)cos(4z))}{(3x^{3})})dxdydz=\int_{6}^{9}\int_{-8}^{-4.5}(-\frac{(40cos(y + 1)cos(4z))}{(3x^{2})})dydz|_{-7}^{-2}\\&\int_{6}^{9}\int_{-8}^{-4.5}(-\frac{(150cos(y + 1)cos(4z))}{49})dydz=\int_{6}^{9}(-\frac{(150cos(4z)sin(y + 1))}{49})dz|_{-8}^{-4.5}\\&\int_{6}^{9}(-\frac{(150cos(4z)\cdot (sin(7) - sin(\frac{7}{2})))}{49})dz=-\frac{(75sin(4z)\cdot (sin(7) - sin(\frac{7}{2})))}{98}|_{6}^{9}=0.07\end{align*}$

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Example Question #423 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{9}^{14}\int_{4.5}^{8}\int_{8}^{9.5}(\frac{(29\cdot 2^{(\frac{y}{2})}sin(z + 2))}{(5x^{3})})dxdydz\end{align*}$

Possible Answers:

0.21

0.41

-2.47

-0.21

Correct answer:

0.41

Explanation:

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

$\begin{align*}&\int_{9}^{14}\int_{4.5}^{8}\int_{8}^{9.5}(\frac{(29\cdot 2^{(\frac{y}{2})}sin(z + 2))}{(5x^{3})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{9}^{14}\int_{4.5}^{8}\int_{8}^{9.5}(\frac{(29\cdot 2^{(\frac{y}{2})}sin(z + 2))}{(5x^{3})})dxdydz=\int_{9}^{14}\int_{4.5}^{8}(-\frac{(29\cdot 2^{(\frac{y}{2})}sin(z + 2))}{(10x^{2})})dydz|_{8}^{9.5}\\&\int_{9}^{14}\int_{4.5}^{8}(\frac{(609\cdot 2^{(\frac{y}{2})}sin(z + 2))}{46208})dydz=\int_{9}^{14}(\frac{(609\cdot 2^{(\frac{y}{2})}sin(z + 2))}{(23104ln(2))})dz|_{4.5}^{8}\\&\int_{9}^{14}(-\frac{(609sin(z + 2)\cdot (2^{(\frac{1}{4})} - 4))}{(5776ln(2))})dz=\frac{(609cos(z + 2)\cdot (2^{(\frac{1}{4})} - 4))}{(5776ln(2))}|_{9}^{14}=0.41\end{align*}$

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Example Question #424 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{7}^{10}\int_{4}^{8.5}\int_{-9}^{-4.5}(\frac{(2cos(z + 1)e^{(y)})}{61})dxdydz\end{align*}$

Possible Answers:

-356.6

8559

237.8

Correct answer:

Explanation:

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

$\begin{align*}&\int_{7}^{10}\int_{4}^{8.5}\int_{-9}^{-4.5}(\frac{(2cos(z + 1)e^{(y)})}{61})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{7}^{10}\int_{4}^{8.5}\int_{-9}^{-4.5}(\frac{(2cos(z + 1)e^{(y)})}{61})dxdydz=\int_{7}^{10}\int_{4}^{8.5}(\frac{(2xcos(z + 1)e^{(y)})}{61})dydz|_{-9}^{-4.5}\\&\int_{7}^{10}\int_{4}^{8.5}(\frac{(9cos(z + 1)e^{(y)})}{61})dydz=\int_{7}^{10}(\frac{(9cos(z + 1)e^{(y)})}{61})dz|_{4}^{8.5}\\&\int_{7}^{10}(-\frac{(9cos(z + 1)\cdot (e^{(4)} - e^{(\frac{17}{2})}))}{61})dz=\frac{(9sin(z + 1)e^{(4)}\cdot (e^{(\frac{9}{2})} - 1))}{61}|_{7}^{10}=-1427\end{align*}$

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Example Question #425 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{-7}^{-3}\int_{-6}^{-4}\int_{4.5}^{5.5}(\frac{(10cos(y + 1)cos(z + 1)e^{(2x)})}{31})dxdydz\end{align*}$

Possible Answers:

2184

10922

Correct answer:

10922

Explanation:

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

$\begin{align*}&\int_{-7}^{-3}\int_{-6}^{-4}\int_{4.5}^{5.5}(\frac{(10cos(y + 1)cos(z + 1)e^{(2x)})}{31})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-7}^{-3}\int_{-6}^{-4}\int_{4.5}^{5.5}(\frac{(10cos(y + 1)cos(z + 1)e^{(2x)})}{31})dxdydz=\int_{-7}^{-3}\int_{-6}^{-4}(\frac{(5cos(y + 1)cos(z + 1)e^{(2x)})}{31})dydz|_{4.5}^{5.5}\\&\int_{-7}^{-3}\int_{-6}^{-4}(\frac{(5cos(y + 1)cos(z + 1)e^{(9)}\cdot (e^{(2)} - 1))}{31})dydz=\int_{-7}^{-3}(\frac{(5cos(z + 1)sin(y + 1)e^{(9)}\cdot (e^{(2)} - 1))}{31})dz|_{-6}^{-4}\\&\int_{-7}^{-3}(-\frac{(5cos(z + 1)e^{(9)}\cdot (sin(3) - sin(5))\cdot (e^{(2)} - 1))}{31})dz=-\frac{(5sin(z + 1)e^{(9)}\cdot (sin(3) - sin(5))\cdot (e^{(2)} - 1))}{31}|_{-7}^{-3}=10922\end{align*}$

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Example Question #426 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{-3.5}^{1}\int_{3}^{6.5}\int_{7}^{11}(14\cdot 2^{(\frac{y}{2})}\cdot 2^zsin(4x))dxdydz\end{align*}$

Possible Answers:

-365.4

1462

91.35

Correct answer:

-365.4

Explanation:

$\begin{align*}&\int_{-3.5}^{1}\int_{3}^{6.5}\int_{7}^{11}(14\cdot 2^{(\frac{y}{2})}\cdot 2^zsin(4x))dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-3.5}^{1}\int_{3}^{6.5}\int_{7}^{11}(14\cdot 2^{(\frac{y}{2})}\cdot 2^zsin(4x))dxdydz=\int_{-3.5}^{1}\int_{3}^{6.5}(-\frac{(7\cdot 2^{(\frac{y}{2}+ z)}cos(4x))}{2})dydz|_{7}^{11}\\&\int_{-3.5}^{1}\int_{3}^{6.5}(2^{(\frac{y}{2}+ z)}\cdot (\frac{(7cos(28))}{2}-\frac{ (7cos(44))}{2}))dydz=\int_{-3.5}^{1}(\frac{(7\cdot 2^{(\frac{y}{2}+ z)}\cdot (cos(28) - cos(44)))}{ln(2)})dz|_{3}^{6.5}\\&\int_{-3.5}^{1}(-\frac{(14\cdot 2^{(\frac{1}{4})}\cdot 2^z\cdot (cos(28) - cos(44))\cdot (2^{(\frac{1}{4})} - 4))}{ln(2)})dz=-\frac{(14\cdot 2^z\cdot (2^{(\frac{1}{2})}cos(28) - 4\cdot 2^{(\frac{1}{4})}cos(28) - 2^{(\frac{1}{2})}cos(44) + 4\cdot 2^{(\frac{1}{4})}cos(44)))}{ln(2)^{2}}|_{-3.5}^{1}=-365.4\end{align*}$

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Example Question #427 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{-3}^{-2}\int_{8}^{10}\int_{3}^{7.5}(\frac{(3^{(\frac{x}{3})}\cdot 3^{(\frac{z}{2})})}{y})dxdydz\end{align*}$

Possible Answers:

0.33

-3.94

1.97

-0.33

Correct answer:

1.97

Explanation:

$\begin{align*}&\int_{-3}^{-2}\int_{8}^{10}\int_{3}^{7.5}(\frac{(3^{(\frac{x}{3})}\cdot 3^{(\frac{z}{2})})}{y})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-3}^{-2}\int_{8}^{10}\int_{3}^{7.5}(\frac{(3^{(\frac{x}{3})}\cdot 3^{(\frac{z}{2})})}{y})dxdydz=\int_{-3}^{-2}\int_{8}^{10}(\frac{(3\cdot 3^{(\frac{x}{3}+\frac{ z}{2})})}{(yln(3))})dydz|_{3}^{7.5}\\&\int_{-3}^{-2}\int_{8}^{10}(\frac{(9\cdot 3^{(\frac{z}{2})}\cdot (3\cdot 3^{(\frac{1}{2})} - 1))}{(yln(3))})dydz=\int_{-3}^{-2}(-\frac{(ln(y)\cdot (9\cdot 3^{(\frac{z}{2})} - 27\cdot 3^{(\frac{1}{2})}\cdot 3^{(\frac{z}{2})}))}{ln(3)})dz|_{8}^{10}\\&\int_{-3}^{-2}(\frac{(9\cdot 3^{(\frac{z}{2})}ln(\frac{5}{4})\cdot (3\cdot 3^{(\frac{1}{2})} - 1))}{ln(3)})dz=-\frac{(18\cdot 3^{(\frac{z}{2})}\cdot (ln(\frac{5}{4}) - 3\cdot 3^{(\frac{1}{2})}ln(\frac{5}{4})))}{ln(3)^{2}}|_{-3}^{-2}=1.97\end{align*}$

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Example Question #421 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{-8}^{-4.5}\int_{8}^{9}\int_{-4}^{-2}(5y^{2}cos(z + 1)e^{(2x)})dxdydz\end{align*}$

Possible Answers:

16.38

-0.82

3.28

-13.11

Correct answer:

3.28

Explanation:

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

$\begin{align*}&\int_{-8}^{-4.5}\int_{8}^{9}\int_{-4}^{-2}(5y^{2}cos(z + 1)e^{(2x)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-8}^{-4.5}\int_{8}^{9}\int_{-4}^{-2}(5y^{2}cos(z + 1)e^{(2x)})dxdydz=\int_{-8}^{-4.5}\int_{8}^{9}(\frac{(5y^{2}cos(z + 1)e^{(2x)})}{2})dydz|_{-4}^{-2}\\&\int_{-8}^{-4.5}\int_{8}^{9}(\frac{(5y^{2}cos(z + 1)e^{(-8)}\cdot (e^{(4)} - 1))}{2})dydz=\int_{-8}^{-4.5}(\frac{(5y^{3}cos(z + 1)e^{(-8)}\cdot (e^{(4)} - 1))}{6})dz|_{8}^{9}\\&\int_{-8}^{-4.5}(\frac{(1085cos(z + 1)e^{(-8)}\cdot (e^{(4)} - 1))}{6})dz=\frac{(1085sin(z + 1)e^{(-8)}\cdot (e^{(4)} - 1))}{6}|_{-8}^{-4.5}=3.28\end{align*}$

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Example Question #429 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{-10}^{-5}\int_{7}^{11}\int_{-9}^{-6.5}(\frac{(8cos(x + 2)sin(y + 1))}{(207z)})dxdydz\end{align*}$

Possible Answers:

0.17

-0.01

-0.04

0.04

Correct answer:

0.04

Explanation:

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

$\begin{align*}&\int_{-10}^{-5}\int_{7}^{11}\int_{-9}^{-6.5}(\frac{(8cos(x + 2)sin(y + 1))}{(207z)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-10}^{-5}\int_{7}^{11}\int_{-9}^{-6.5}(\frac{(8cos(x + 2)sin(y + 1))}{(207z)})dxdydz=\int_{-10}^{-5}\int_{7}^{11}(\frac{(8sin(x + 2)sin(y + 1))}{(207z)})dydz|_{-9}^{-6.5}\\&\int_{-10}^{-5}\int_{7}^{11}(\frac{(8sin(y + 1)\cdot (sin(7) - sin(\frac{9}{2})))}{(207z)})dydz=\int_{-10}^{-5}(-\frac{(8cos(y + 1)\cdot (sin(7) - sin(\frac{9}{2})))}{(207z)})dz|_{7}^{11}\\&\int_{-10}^{-5}(\frac{(8\cdot (cos(8) - cos(12))\cdot (sin(7) - sin(\frac{9}{2})))}{(207z)})dz=ln(z)\cdot (\frac{(8cos(8)sin(7))}{207}-\frac{ (8cos(8)sin(\frac{9}{2}))}{207}-\frac{ (8cos(12)sin(7))}{207}+\frac{ (8cos(12)sin(\frac{9}{2}))}{207})|_{-10}^{-5}=0.04\end{align*}$

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Example Question #430 : Triple Integrals

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

Possible Answers:

0.46

-3.64

-1.82

7.29

Correct answer:

-1.82

Explanation:

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

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

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Calculus 3 : Multiple Integration

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #281 : Triple Integration In Cartesian Coordinates

Example Question #281 : Triple Integration In Cartesian Coordinates

Example Question #423 : Triple Integrals

Example Question #424 : Triple Integrals

Example Question #425 : Triple Integrals

Example Question #426 : Triple Integrals

Example Question #427 : Triple Integrals

Example Question #421 : Triple Integrals

Example Question #429 : Triple Integrals

Example Question #430 : Triple Integrals

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