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Calculus 3 : Triple Integrals

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

Example Questions

Example Question #137 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{-3.5}^{0.5}\int_{7}^{12}\int_{-8}^{-6}(\frac{(66\cdot2^zcos(3y)sin(x + 2))}{7})dxdydz\end{align*}$

Possible Answers:

-35.48

35.48

4.43

-17.74

Correct answer:

-17.74

Explanation:

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

$\begin{align*}&\int_{-3.5}^{0.5}\int_{7}^{12}\int_{-8}^{-6}(\frac{(66\cdot2^zcos(3y)sin(x + 2))}{7})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-3.5}^{0.5}\int_{7}^{12}\int_{-8}^{-6}(\frac{(66\cdot2^zcos(3y)sin(x + 2))}{7})dxdydz=\int_{-3.5}^{0.5}\int_{7}^{12}(-\frac{(66\cdot2^zcos(x + 2)cos(3y))}{7})dydz|_{-8}^{-6}\\&\int_{-3.5}^{0.5}\int_{7}^{12}(-\frac{(66\cdot2^zcos(3y)\cdot(cos(4) - cos(6)))}{7})dydz=\int_{-3.5}^{0.5}(-\frac{(22\cdot2^zsin(3y)\cdot(cos(4) - cos(6)))}{7})dz|_{7}^{12}\\&\int_{-3.5}^{0.5}(\frac{(22\cdot2^z\cdot(cos(4) - cos(6))\cdot(sin(21) - sin(36)))}{7})dz=-\frac{(22\cdot2^z\cdot(\frac{sin(15)}{2}-\frac{ sin(17)}{2}-\frac{ sin(25)}{2}+\frac{ sin(27)}{2}-\frac{ sin(30)}{2}+\frac{ sin(32)}{2}+\frac{ sin(40)}{2}-\frac{ sin(42)}{2}))}{(7ln(2))}|_{-3.5}^{0.5}=-17.74\end{align*}$

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Example Question #138 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{-6}^{-4}\int_{-3}^{-1}\int_{6}^{8}(\frac{(sin(4z)e^{(-y)})}{(5x)})dxdydz\end{align*}$

Possible Answers:

-0.69

0.35

2.07

-0.06

Correct answer:

0.35

Explanation:

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

$\begin{align*}&\int_{-6}^{-4}\int_{-3}^{-1}\int_{6}^{8}(\frac{(sin(4z)e^{(-y)})}{(5x)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-6}^{-4}\int_{-3}^{-1}\int_{6}^{8}(\frac{(sin(4z)e^{(-y)})}{(5x)})dxdydz=\int_{-6}^{-4}\int_{-3}^{-1}(\frac{(sin(4z)e^{(-y)}ln(x))}{5})dydz|_{6}^{8}\\&\int_{-6}^{-4}\int_{-3}^{-1}(\frac{(sin(4z)e^{(-y)}ln(\frac{4}{3}))}{5})dydz=\int_{-6}^{-4}(-\frac{(sin(4z)e^{(-y)}ln(\frac{4}{3}))}{5})dz|_{-3}^{-1}\\&\int_{-6}^{-4}(-\frac{(sin(4z)ln(\frac{4}{3})\cdot(e^{(1)} - e^{(3)}))}{5})dz=\frac{(cos(4z)ln(\frac{4}{3})\cdot(e^{(1)} - e^{(3)}))}{20}|_{-6}^{-4}=0.35\end{align*}$

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Example Question #139 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{-2.5}^{-1.5}\int_{-6}^{-2.5}\int_{10}^{12.5}(\frac{(5cos(3x)e^{(-2z)})}{y^{3}})dxdydz\end{align*}$

Possible Answers:

-22.35

-5.59

27.93

1.12

Correct answer:

-5.59

Explanation:

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

$\begin{align*}&\int_{-2.5}^{-1.5}\int_{-6}^{-2.5}\int_{10}^{12.5}(\frac{(5cos(3x)e^{(-2z)})}{y^{3}})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-2.5}^{-1.5}\int_{-6}^{-2.5}\int_{10}^{12.5}(\frac{(5cos(3x)e^{(-2z)})}{y^{3}})dxdydz=\int_{-2.5}^{-1.5}\int_{-6}^{-2.5}(\frac{(5sin(3x)e^{(-2z)})}{(3y^{3})})dydz|_{10}^{12.5}\\&\int_{-2.5}^{-1.5}\int_{-6}^{-2.5}(-\frac{(5e^{(-2z)}\cdot(sin(30) - sin(\frac{75}{2})))}{(3y^{3})})dydz=\int_{-2.5}^{-1.5}(\frac{(5e^{(-2z)}\cdot(sin(30) - sin(\frac{75}{2})))}{(6y^{2})})dz|_{-6}^{-2.5}\\&\int_{-2.5}^{-1.5}(\frac{(119e^{(-2z)}\cdot(sin(30) - sin(\frac{75}{2})))}{1080})dz=-\frac{(119e^{(-2z)}\cdot(sin(30) - sin(\frac{75}{2})))}{2160}|_{-2.5}^{-1.5}=-5.59\end{align*}$

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Example Question #271 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{-7}^{-5.5}\int_{6}^{8}\int_{5}^{9}(\frac{(2^{(\frac{x}{3})}sin(z + 1))}{(6y)})dxdydz\end{align*}$

Possible Answers:

0.2

-0.29

-1.17

1.17

Correct answer:

1.17

Explanation:

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

$\begin{align*}&\int_{-7}^{-5.5}\int_{6}^{8}\int_{5}^{9}(\frac{(2^{(\frac{x}{3})}sin(z + 1))}{(6y)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-7}^{-5.5}\int_{6}^{8}\int_{5}^{9}(\frac{(2^{(\frac{x}{3})}sin(z + 1))}{(6y)})dxdydz=\int_{-7}^{-5.5}\int_{6}^{8}(\frac{(2^{(\frac{x}{3})}sin(z + 1))}{(2yln(2))})dydz|_{5}^{9}\\&\int_{-7}^{-5.5}\int_{6}^{8}(-\frac{(sin(z + 1)\cdot(6\cdot2^{(\frac{2}{3})} - 24))}{(6yln(2))})dydz=\int_{-7}^{-5.5}(\frac{(ln(y)\cdot(4sin(z + 1) - 4^{(\frac{1}{3})}sin(z + 1)))}{ln(2)})dz|_{6}^{8}\\&\int_{-7}^{-5.5}(\frac{(4sin(z + 1)ln(\frac{4}{3}) - 2^{(\frac{2}{3})}sin(z + 1)ln(\frac{4}{3}))}{ln(2)})dz=\frac{(2ln(\frac{4}{3})\cdot(2^{(\frac{2}{3})} - 4))}{(ln(2)\cdot(tan(\frac{z}{2}+\frac{ 1}{2})^{2} + 1))}|_{-7}^{-5.5}=1.17\end{align*}$

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Example Question #141 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{-6}^{-4}\int_{-3}^{1.5}\int_{6}^{8.5}(\frac{(53sin(z + 1))}{(4\cdot3^{(\frac{y}{2})}x)})dxdydz\end{align*}$

Possible Answers:

-203.6

-16.97

101.8

50.91

Correct answer:

50.91

Explanation:

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

$\begin{align*}&\int_{-6}^{-4}\int_{-3}^{1.5}\int_{6}^{8.5}(\frac{(53sin(z + 1))}{(4\cdot3^{(\frac{y}{2})}x)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-6}^{-4}\int_{-3}^{1.5}\int_{6}^{8.5}(\frac{(53sin(z + 1))}{(4\cdot3^{(\frac{y}{2})}x)})dxdydz=\int_{-6}^{-4}\int_{-3}^{1.5}(\frac{(53sin(z + 1)ln(x))}{(4\cdot3^{(\frac{y}{2})})})dydz|_{6}^{8.5}\\&\int_{-6}^{-4}\int_{-3}^{1.5}(\frac{(53sin(z + 1)ln(\frac{17}{12}))}{(4\cdot3^{(\frac{y}{2})})})dydz=\int_{-6}^{-4}(-\frac{(53sin(z + 1)ln(\frac{17}{12}))}{(2\cdot3^{(\frac{y}{2})}ln(3))})dz|_{-3}^{1.5}\\&\int_{-6}^{-4}(\frac{(53sin(z + 1)ln(\frac{17}{12})\cdot(18\cdot3^{(\frac{1}{2})} - 2\cdot3^{(\frac{1}{4})}))}{(12ln(3))})dz=-\frac{(53cos(z + 1)ln(\frac{17}{12})\cdot(18\cdot3^{(\frac{1}{2})} - 2\cdot3^{(\frac{1}{4})}))}{(12ln(3))}|_{-6}^{-4}=50.91\end{align*}$

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Example Question #142 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{7}^{10}\int_{-3}^{-1}\int_{10}^{14.5}(\frac{(51\cdot3^ysin(3x)sin(z + 1))}{10})dxdydz\end{align*}$

Possible Answers:

-0.1

0.05

-0.02

0.1

Correct answer:

0.05

Explanation:

$\begin{align*}&\int_{7}^{10}\int_{-3}^{-1}\int_{10}^{14.5}(\frac{(51\cdot3^ysin(3x)sin(z + 1))}{10})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{7}^{10}\int_{-3}^{-1}\int_{10}^{14.5}(\frac{(51\cdot3^ysin(3x)sin(z + 1))}{10})dxdydz=\int_{7}^{10}\int_{-3}^{-1}(-\frac{(17\cdot3^ycos(3x)sin(z + 1))}{10})dydz|_{10}^{14.5}\\&\int_{7}^{10}\int_{-3}^{-1}(\frac{(51\cdot3^ysin(z + 1)\cdot(\frac{cos(30)}{3}-\frac{ cos(\frac{87}{2})}{3}))}{10})dydz=\int_{7}^{10}(\frac{(17\cdot3^ysin(z + 1)cos(30) - 17\cdot3^ysin(z + 1)cos(\frac{87}{2}))}{(10ln(3))})dz|_{-3}^{-1}\\&\int_{7}^{10}(\frac{(68sin(z + 1)cos(30) - 68sin(z + 1)cos(\frac{87}{2}))}{(135ln(3))})dz=-\frac{(136cos(30)cos(\frac{z}{2}+\frac{ 1}{2})^{2} - 136cos(\frac{87}{2})cos(\frac{z}{2}+\frac{ 1}{2})^{2})}{(135ln(3))}|_{7}^{10}=0.05\end{align*}$

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Example Question #143 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{-6}^{-1}\int_{-5}^{-2}\int_{4}^{5.5}(\frac{(17sin(z + 2))}{(2\cdot2^yx)})dxdydz\end{align*}$

Possible Answers:

-21.76

65.28

-130.6

783.3

Correct answer:

-130.6

Explanation:

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

$\begin{align*}&\int_{-6}^{-1}\int_{-5}^{-2}\int_{4}^{5.5}(\frac{(17sin(z + 2))}{(2\cdot2^yx)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-6}^{-1}\int_{-5}^{-2}\int_{4}^{5.5}(\frac{(17sin(z + 2))}{(2\cdot2^yx)})dxdydz=\int_{-6}^{-1}\int_{-5}^{-2}(\frac{(17\cdot(\frac{1}{2})^ysin(z + 2)ln(x))}{2})dydz|_{4}^{5.5}\\&\int_{-6}^{-1}\int_{-5}^{-2}(\frac{(17sin(z + 2)ln(\frac{11}{8}))}{(2\cdot2^y)})dydz=\int_{-6}^{-1}(-\frac{(17sin(z + 2)ln(\frac{11}{8}))}{(2\cdot2^yln(2))})dz|_{-5}^{-2}\\&\int_{-6}^{-1}(\frac{(238sin(z + 2)ln(\frac{11}{8}))}{ln(2)})dz=-\frac{(238cos(z + 2)ln(\frac{11}{8}))}{ln(2)}|_{-6}^{-1}=-130.6\end{align*}$

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Example Question #144 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{9}^{11.5}\int_{-7}^{-6}\int_{8}^{11}(\frac{(5z^{2})}{(41xy^{2})})dxdydz\end{align*}$

Possible Answers:

-1.22

-0.06

0.98

0.24

Correct answer:

0.24

Explanation:

$\begin{align*}&\int_{9}^{11.5}\int_{-7}^{-6}\int_{8}^{11}(\frac{(5z^{2})}{(41xy^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{9}^{11.5}\int_{-7}^{-6}\int_{8}^{11}(\frac{(5z^{2})}{(41xy^{2})})dxdydz=\int_{9}^{11.5}\int_{-7}^{-6}(\frac{(5z^{2}ln(x))}{(41y^{2})})dydz|_{8}^{11}\\&\int_{9}^{11.5}\int_{-7}^{-6}(\frac{(5z^{2}ln(\frac{11}{8}))}{(41y^{2})})dydz=\int_{9}^{11.5}(-\frac{(5z^{2}ln(\frac{11}{8}))}{(41y)})dz|_{-7}^{-6}\\&\int_{9}^{11.5}(\frac{(5z^{2}ln(\frac{11}{8}))}{1722})dz=\frac{(5z^{3}ln(\frac{11}{8}))}{5166}|_{9}^{11.5}=0.24\end{align*}$

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Example Question #145 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{-9}^{-5}\int_{3.5}^{8}\int_{6}^{9}(\frac{(2\cdot2^{(\frac{y}{2})}sin(3z))}{13})dxdydz\end{align*}$

Possible Answers:

-1.31

-13.11

15.74

2.62

Correct answer:

2.62

Explanation:

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

$\begin{align*}&\int_{-9}^{-5}\int_{3.5}^{8}\int_{6}^{9}(\frac{(2\cdot2^{(\frac{y}{2})}sin(3z))}{13})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-9}^{-5}\int_{3.5}^{8}\int_{6}^{9}(\frac{(2\cdot2^{(\frac{y}{2})}sin(3z))}{13})dxdydz=\int_{-9}^{-5}\int_{3.5}^{8}(\frac{(2\cdot2^{(\frac{y}{2})}xsin(3z))}{13})dydz|_{6}^{9}\\&\int_{-9}^{-5}\int_{3.5}^{8}(\frac{(6\cdot2^{(\frac{y}{2})}sin(3z))}{13})dydz=\int_{-9}^{-5}(\frac{(12\cdot2^{(\frac{y}{2})}sin(3z))}{(13ln(2))})dz|_{3.5}^{8}\\&\int_{-9}^{-5}(-\frac{(6sin(3z)\cdot(4\cdot2^{(\frac{3}{4})} - 32))}{(13ln(2))})dz=\frac{(2cos(3z)\cdot(4\cdot2^{(\frac{3}{4})} - 32))}{(13ln(2))}|_{-9}^{-5}=2.62\end{align*}$

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Example Question #146 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{9}^{12}\int_{-8}^{-5}\int_{6}^{11}(\frac{(3x^{2}sin(4y))}{(10z)})dxdydz\end{align*}$

Possible Answers:

-1.71

0.85

-10.25

3.42

Correct answer:

3.42

Explanation:

$\begin{align*}&\int_{9}^{12}\int_{-8}^{-5}\int_{6}^{11}(\frac{(3x^{2}sin(4y))}{(10z)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{9}^{12}\int_{-8}^{-5}\int_{6}^{11}(\frac{(3x^{2}sin(4y))}{(10z)})dxdydz=\int_{9}^{12}\int_{-8}^{-5}(\frac{(x^{3}sin(4y))}{(10z)})dydz|_{6}^{11}\\&\int_{9}^{12}\int_{-8}^{-5}(\frac{(223sin(4y))}{(2z)})dydz=\int_{9}^{12}(-\frac{(223cos(4y))}{(8z)})dz|_{-8}^{-5}\\&\int_{9}^{12}(-\frac{(223\cdot(\frac{cos(20)}{4}-\frac{ cos(32)}{4}))}{(2z)})dz=-ln(z)\cdot(\frac{(223cos(20))}{8}-\frac{ (223cos(32))}{8})|_{9}^{12}=3.42\end{align*}$

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Calculus 3 : Triple Integrals

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #137 : Triple Integration In Cartesian Coordinates

Example Question #138 : Triple Integration In Cartesian Coordinates

Example Question #139 : Triple Integration In Cartesian Coordinates

Example Question #271 : Triple Integrals

Example Question #141 : Triple Integration In Cartesian Coordinates

Example Question #142 : Triple Integration In Cartesian Coordinates

Example Question #143 : Triple Integration In Cartesian Coordinates

Example Question #144 : Triple Integration In Cartesian Coordinates

Example Question #145 : Triple Integration In Cartesian Coordinates

Example Question #146 : Triple Integration In Cartesian Coordinates

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