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Calculus 3 : Triple Integrals

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Example Questions

Example Question #251 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{-8}^{-7}\int_{7}^{10.5}\int_{10}^{13}(\frac{(10cos(x + 2))}{(7y^{2})})dxdydz\end{align*}$

Possible Answers:

0.08

-0.32

-0.02

0.03

Correct answer:

0.08

Explanation:

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

$\begin{align*}&\int_{-8}^{-7}\int_{7}^{10.5}\int_{10}^{13}(\frac{(10cos(x + 2))}{(7y^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-8}^{-7}\int_{7}^{10.5}\int_{10}^{13}(\frac{(10cos(x + 2))}{(7y^{2})})dxdydz=\int_{-8}^{-7}\int_{7}^{10.5}(\frac{(10sin(x + 2))}{(7y^{2})})dydz|_{10}^{13}\\&\int_{-8}^{-7}\int_{7}^{10.5}(-\frac{(10\cdot (sin(12) - sin(15)))}{(7y^{2})})dydz=\int_{-8}^{-7}(\frac{(\frac{(10sin(12))}{7}-\frac{ (10sin(15))}{7})}{y})dz|_{7}^{10.5}\\&\int_{-8}^{-7}(\frac{(10sin(15))}{147}-\frac{ (10sin(12))}{147})dz=-z\cdot (\frac{(10sin(12))}{147}-\frac{ (10sin(15))}{147})|_{-8}^{-7}=0.08\end{align*}$

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Example Question #252 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{-3}^{2}\int_{11}^{14}\int_{6}^{10.5}(\frac{(67xe^{(2z)})}{(5y)})dxdydz\end{align*}$

Possible Answers:

3275

13100

Correct answer:

3275

Explanation:

$\begin{align*}&\int_{-3}^{2}\int_{11}^{14}\int_{6}^{10.5}(\frac{(67xe^{(2z)})}{(5y)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-3}^{2}\int_{11}^{14}\int_{6}^{10.5}(\frac{(67xe^{(2z)})}{(5y)})dxdydz=\int_{-3}^{2}\int_{11}^{14}(\frac{(67x^{2}e^{(2z)})}{(10y)})dydz|_{6}^{10.5}\\&\int_{-3}^{2}\int_{11}^{14}(\frac{(19899e^{(2z)})}{(40y)})dydz=\int_{-3}^{2}(\frac{(19899e^{(2z)}ln(y))}{40})dz|_{11}^{14}\\&\int_{-3}^{2}(\frac{(19899e^{(2z)}ln(\frac{14}{11}))}{40})dz=\frac{(19899e^{(2z)}ln(\frac{14}{11}))}{80}|_{-3}^{2}=3275\end{align*}$

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Example Question #253 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{-4}^{-2.5}\int_{4}^{7}\int_{9}^{14}(\frac{(2e^{(-y)}e^{(-z)})}{(5x)})dxdydz\end{align*}$

Possible Answers:

0.13

0.03

-0.03

-0.65

Correct answer:

0.13

Explanation:

$\begin{align*}&\int_{-4}^{-2.5}\int_{4}^{7}\int_{9}^{14}(\frac{(2e^{(-y)}e^{(-z)})}{(5x)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-4}^{-2.5}\int_{4}^{7}\int_{9}^{14}(\frac{(2e^{(-y)}e^{(-z)})}{(5x)})dxdydz=\int_{-4}^{-2.5}\int_{4}^{7}(\frac{(2e^{(- y - z)}ln(x))}{5})dydz|_{9}^{14}\\&\int_{-4}^{-2.5}\int_{4}^{7}(\frac{(2e^{(- y - z)}ln(\frac{14}{9}))}{5})dydz=\int_{-4}^{-2.5}(-\frac{(2e^{(- y - z)}ln(\frac{14}{9}))}{5})dz|_{4}^{7}\\&\int_{-4}^{-2.5}(\frac{(2e^{(- z - 7)}ln(\frac{14}{9})\cdot (e^{(3)} - 1))}{5})dz=-\frac{(2e^{(- z - 7)}ln(\frac{14}{9})\cdot (e^{(3)} - 1))}{5}|_{-4}^{-2.5}=0.13\end{align*}$

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Example Question #261 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{8}^{10}\int_{9}^{13.5}\int_{-3.5}^{-1.5}(\frac{(17\cdot 3^{(\frac{x}{3})}sin(y + 2))}{10})dxdydz\end{align*}$

Possible Answers:

-8.21

13.68

-0.68

2.74

Correct answer:

2.74

Explanation:

$\begin{align*}&\int_{8}^{10}\int_{9}^{13.5}\int_{-3.5}^{-1.5}(\frac{(17\cdot 3^{(\frac{x}{3})}sin(y + 2))}{10})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{8}^{10}\int_{9}^{13.5}\int_{-3.5}^{-1.5}(\frac{(17\cdot 3^{(\frac{x}{3})}sin(y + 2))}{10})dxdydz=\int_{8}^{10}\int_{9}^{13.5}(\frac{(51\cdot 3^{(\frac{x}{3})}sin(y + 2))}{(10ln(3))})dydz|_{-3.5}^{-1.5}\\&\int_{8}^{10}\int_{9}^{13.5}(\frac{(17sin(y + 2)\cdot (3\cdot 3^{(\frac{1}{2})} - 3^{(\frac{5}{6})}))}{(30ln(3))})dydz=\int_{8}^{10}(-\frac{(17cos(y + 2)\cdot (3\cdot 3^{(\frac{1}{2})} - 3^{(\frac{5}{6})}))}{(30ln(3))})dz|_{9}^{13.5}\\&\int_{8}^{10}(-\frac{(17\cdot 3^{(\frac{1}{2})}\cdot (cos(11) - cos(\frac{31}{2}))\cdot (3^{(\frac{1}{3})} - 3))}{(30ln(3))})dz=-\frac{(17\cdot 3^{(\frac{1}{2})}z\cdot (cos(11) - cos(\frac{31}{2}))\cdot (3^{(\frac{1}{3})} - 3))}{(30ln(3))}|_{8}^{10}=2.74\end{align*}$

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Example Question #871 : Calculus 3

$\begin{align*}&\text{Evaluate the triple integral}\int_{8}^{11.5}\int_{-10}^{-5}\int_{-9}^{-4}(\frac{(3cos(x + 1)cos(z + 2))}{(23y)})dxdydz\end{align*}$

Possible Answers:

0.1

-0.1

-0.05

0.02

Correct answer:

-0.1

Explanation:

$\begin{align*}&\int_{8}^{11.5}\int_{-10}^{-5}\int_{-9}^{-4}(\frac{(3cos(x + 1)cos(z + 2))}{(23y)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{8}^{11.5}\int_{-10}^{-5}\int_{-9}^{-4}(\frac{(3cos(x + 1)cos(z + 2))}{(23y)})dxdydz=\int_{8}^{11.5}\int_{-10}^{-5}(\frac{(3cos(z + 2)sin(x + 1))}{(23y)})dydz|_{-9}^{-4}\\&\int_{8}^{11.5}\int_{-10}^{-5}(-\frac{(3cos(z + 2)\cdot (sin(3) - sin(8)))}{(23y)})dydz=\int_{8}^{11.5}(-ln(y)\cdot (\frac{(3cos(z + 2)sin(3))}{23}-\frac{ (3cos(z + 2)sin(8))}{23}))dz|_{-10}^{-5}\\&\int_{8}^{11.5}(\frac{(3cos(z + 2)ln(2)\cdot (sin(3) - sin(8)))}{23})dz=\frac{(3sin(z + 2)ln(2)\cdot (sin(3) - sin(8)))}{23}|_{8}^{11.5}=-0.1\end{align*}$

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Example Question #872 : Calculus 3

$\begin{align*}&\text{Evaluate the triple integral}\int_{-8}^{-4.5}\int_{-6}^{-4}\int_{-3}^{-1}(\frac{(61sin(z + 1))}{(33\cdot 3^xy)})dxdydz\end{align*}$

Possible Answers:

-6.92

138.4

9.23

-27.68

Correct answer:

-27.68

Explanation:

$\begin{align*}&\int_{-8}^{-4.5}\int_{-6}^{-4}\int_{-3}^{-1}(\frac{(61sin(z + 1))}{(33\cdot 3^xy)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-8}^{-4.5}\int_{-6}^{-4}\int_{-3}^{-1}(\frac{(61sin(z + 1))}{(33\cdot 3^xy)})dxdydz=\int_{-8}^{-4.5}\int_{-6}^{-4}(-\frac{(61sin(z + 1))}{(33\cdot 3^xyln(3))})dydz|_{-3}^{-1}\\&\int_{-8}^{-4.5}\int_{-6}^{-4}(\frac{(488sin(z + 1))}{(11yln(3))})dydz=\int_{-8}^{-4.5}(\frac{(488sin(z + 1)ln(y))}{(11ln(3))})dz|_{-6}^{-4}\\&\int_{-8}^{-4.5}(\frac{(488sin(z + 1)ln(\frac{2}{3}))}{(11ln(3))})dz=-\frac{(488cos(z + 1)ln(\frac{2}{3}))}{(11ln(3))}|_{-8}^{-4.5}=-27.68\end{align*}$

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Example Question #261 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{-4.5}^{-2}\int_{-3}^{-1.5}\int_{3.5}^{6.5}(\frac{(49\cdot 2^{(\frac{x}{4})}e^{(2y)}e^{(-z)})}{2})dxdydz\end{align*}$

Possible Answers:

345.5

-115.2

1728

Correct answer:

345.5

Explanation:

$\begin{align*}&\int_{-4.5}^{-2}\int_{-3}^{-1.5}\int_{3.5}^{6.5}(\frac{(49\cdot 2^{(\frac{x}{4})}e^{(2y)}e^{(-z)})}{2})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-4.5}^{-2}\int_{-3}^{-1.5}\int_{3.5}^{6.5}(\frac{(49\cdot 2^{(\frac{x}{4})}e^{(2y)}e^{(-z)})}{2})dxdydz=\int_{-4.5}^{-2}\int_{-3}^{-1.5}(\frac{(98\cdot 2^{(\frac{x}{4})}e^{(2y - z)})}{ln(2)})dydz|_{3.5}^{6.5}\\&\int_{-4.5}^{-2}\int_{-3}^{-1.5}(-\frac{(98\cdot 2^{(\frac{5}{8})}e^{(2y - z)}\cdot (2^{(\frac{1}{4})} - 2))}{ln(2)})dydz=\int_{-4.5}^{-2}(-\frac{(49\cdot 2^{(\frac{5}{8})}e^{(2y - z)}\cdot (2^{(\frac{1}{4})} - 2))}{ln(2)})dz|_{-3}^{-1.5}\\&\int_{-4.5}^{-2}(-\frac{(49\cdot 2^{(\frac{5}{8})}e^{(- z - 6)}\cdot (2^{(\frac{1}{4})} - 2)\cdot (e^{(3)} - 1))}{ln(2)})dz=\frac{(49\cdot 2^{(\frac{5}{8})}e^{(- z - 6)}\cdot (2^{(\frac{1}{4})} - 2)\cdot (e^{(3)} - 1))}{ln(2)}|_{-4.5}^{-2}=345.5\end{align*}$

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Example Question #265 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{6}^{11}\int_{10}^{11.5}\int_{9}^{13.5}(\frac{(25ysin(x + 2))}{(2z^{3})})dxdydz\end{align*}$

Possible Answers:

-0.48

7.73

-3.87

1.93

Correct answer:

1.93

Explanation:

$\begin{align*}&\int_{6}^{11}\int_{10}^{11.5}\int_{9}^{13.5}(\frac{(25ysin(x + 2))}{(2z^{3})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{6}^{11}\int_{10}^{11.5}\int_{9}^{13.5}(\frac{(25ysin(x + 2))}{(2z^{3})})dxdydz=\int_{6}^{11}\int_{10}^{11.5}(-\frac{(25ycos(x + 2))}{(2z^{3})})dydz|_{9}^{13.5}\\&\int_{6}^{11}\int_{10}^{11.5}(\frac{(25y\cdot (cos(11) - cos(\frac{31}{2})))}{(2z^{3})})dydz=\int_{6}^{11}(\frac{(y^{2}\cdot (25cos(11) - 25cos(\frac{31}{2})))}{(4z^{3})})dz|_{10}^{11.5}\\&\int_{6}^{11}(\frac{(3225\cdot (cos(11) - cos(\frac{31}{2})))}{(16z^{3})})dz=-\frac{(\frac{(3225cos(11))}{32}-\frac{ (3225cos(\frac{31}{2}))}{32})}{z^{2}}|_{6}^{11}=1.93\end{align*}$

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Example Question #266 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{7}^{11.5}\int_{-3}^{-2}\int_{-8}^{-6}(\frac{(49cos(3z)e^{(-y)})}{(12x)})dxdydz\end{align*}$

Possible Answers:

3.87

-0.77

7.75

-19.37

Correct answer:

3.87

Explanation:

$\begin{align*}&\int_{7}^{11.5}\int_{-3}^{-2}\int_{-8}^{-6}(\frac{(49cos(3z)e^{(-y)})}{(12x)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{7}^{11.5}\int_{-3}^{-2}\int_{-8}^{-6}(\frac{(49cos(3z)e^{(-y)})}{(12x)})dxdydz=\int_{7}^{11.5}\int_{-3}^{-2}(\frac{(49cos(3z)e^{(-y)}ln(x))}{12})dydz|_{-8}^{-6}\\&\int_{7}^{11.5}\int_{-3}^{-2}(\frac{(49cos(3z)e^{(-y)}ln(\frac{3}{4}))}{12})dydz=\int_{7}^{11.5}(-\frac{(49cos(3z)e^{(-y)}ln(\frac{3}{4}))}{12})dz|_{-3}^{-2}\\&\int_{7}^{11.5}(-\frac{(49cos(3z)ln(\frac{3}{4})\cdot (e^{(2)} - e^{(3)}))}{12})dz=-\frac{(49sin(3z)ln(\frac{3}{4})\cdot (e^{(2)} - e^{(3)}))}{36}|_{7}^{11.5}=3.87\end{align*}$

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Example Question #267 : Triple Integration In Cartesian Coordinates

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

Possible Answers:

47.84

-239.2

-47.84

478.4

Correct answer:

-239.2

Explanation:

$\begin{align*}&\int_{-6}^{-3}\int_{4}^{5.5}\int_{-3}^{-1}(\frac{(7z)}{(2^x\cdot 2^{(\frac{y}{2})})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-6}^{-3}\int_{4}^{5.5}\int_{-3}^{-1}(\frac{(7z)}{(2^x\cdot 2^{(\frac{y}{2})})})dxdydz=\int_{-6}^{-3}\int_{4}^{5.5}(-\frac{(\frac{7}{2^{(x +\frac{ y}{2})}z})}{ln(2)})dydz|_{-3}^{-1}\\&\int_{-6}^{-3}\int_{4}^{5.5}(\frac{(42z)}{(2^{(\frac{y}{2})}ln(2))})dydz=\int_{-6}^{-3}(-\frac{(84z)}{(2^{(\frac{y}{2})}ln(2)^{2})})dz|_{4}^{5.5}\\&\int_{-6}^{-3}(-\frac{(21z\cdot (2^{(\frac{1}{4})} - 2))}{(2ln(2)^{2})})dz=-\frac{(z^{2}\cdot (21\cdot 2^{(\frac{1}{4})} - 42))}{(4ln(2)^{2})}|_{-6}^{-3}=-239.2\end{align*}$

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Calculus 3 : Triple Integrals

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #251 : Triple Integration In Cartesian Coordinates

Example Question #252 : Triple Integration In Cartesian Coordinates

Example Question #253 : Triple Integration In Cartesian Coordinates

Example Question #261 : Triple Integration In Cartesian Coordinates

Example Question #871 : Calculus 3

Example Question #872 : Calculus 3

Example Question #261 : Triple Integration In Cartesian Coordinates

Example Question #265 : Triple Integration In Cartesian Coordinates

Example Question #266 : Triple Integration In Cartesian Coordinates

Example Question #267 : Triple Integration In Cartesian Coordinates

All Calculus 3 Resources

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