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Calculus 3 : Calculus 3

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

Example Questions

Example Question #244 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{4}^{8}\int_{7}^{9}\int_{8}^{13}(\frac{(33\cdot3^{(\frac{z}{2})}sin(x + 2)e^{(\frac{y}{2})})}{4})dxdydz\end{align*}$

Possible Answers:

2442

39088

Correct answer:

Explanation:

$\begin{align*}\\&\int_{4}^{8}\int_{7}^{9}\int_{8}^{13}(\frac{(33\cdot3^{(\frac{z}{2})}sin(x + 2)e^{(\frac{y}{2})})}{4})dxdydz=\int_{4}^{8}\int_{7}^{9}(-\frac{(33\cdot3^{(\frac{z}{2})}cos(x + 2)e^{(\frac{y}{2})})}{4})dydz|_{8}^{13}\\&\int_{4}^{8}\int_{7}^{9}(\frac{(33\cdot3^{(\frac{z}{2})}e^{(\frac{y}{2})}\cdot(cos(10) - cos(15)))}{4})dydz=\int_{4}^{8}(\frac{(33\cdot3^{(\frac{z}{2})}e^{(\frac{y}{2})}\cdot(cos(10) - cos(15)))}{2})dz|_{7}^{9}\\&\int_{4}^{8}(\frac{(33\cdot3^{(\frac{z}{2})}e^{(\frac{7}{2})}\cdot(cos(10) - cos(15))\cdot(e^{(1)} - 1))}{2})dz=-\frac{(33\cdot3^{(\frac{z}{2})}\cdot(cos(10)e^{(\frac{7}{2})} - cos(10)e^{(\frac{9}{2})} - cos(15)e^{(\frac{7}{2})} + cos(15)e^{(\frac{9}{2})}))}{ln(3)}|_{4}^{8}=-9769\end{align*}$

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Example Question #245 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{-7}^{-2}\int_{8}^{13}\int_{-8}^{-6}(\frac{(2cos(x + 1)cos(y + 2))}{(33z)})dxdydz\end{align*}$

Possible Answers:

-0.15

-0.29

0.29

0.07

Correct answer:

-0.15

Explanation:

$\begin{align*}&\text{Performing a triple integral, the order of integration does}\\&\text{not entirely matter.}\\&\text{For example:}\\&\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx\\&\text{Considering our integral:}\\&\int_{-7}^{-2}\int_{8}^{13}\int_{-8}^{-6}(\frac{(2cos(x + 1)cos(y + 2))}{(33z)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-7}^{-2}\int_{8}^{13}\int_{-8}^{-6}(\frac{(2cos(x + 1)cos(y + 2))}{(33z)})dxdydz=\int_{-7}^{-2}\int_{8}^{13}(\frac{(2cos(y + 2)sin(x + 1))}{(33z)})dydz|_{-8}^{-6}\\&\int_{-7}^{-2}\int_{8}^{13}(-\frac{(2cos(y + 2)\cdot(sin(5) - sin(7)))}{(33z)})dydz=\int_{-7}^{-2}(-\frac{(2sin(y + 2)\cdot(sin(5) - sin(7)))}{(33z)})dz|_{8}^{13}\\&\int_{-7}^{-2}(\frac{(2\cdot(sin(5) - sin(7))\cdot(sin(10) - sin(15)))}{(33z)})dz=ln(z)\cdot(\frac{(2sin(5)sin(10))}{33}-\frac{ (2sin(7)sin(10))}{33}-\frac{ (2sin(5)sin(15))}{33}+\frac{ (2sin(7)sin(15))}{33})|_{-7}^{-2}=-0.15\end{align*}$

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Example Question #246 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{-8}^{-7}\int_{8}^{10}\int_{7}^{12}(\frac{(25sin(3z))}{x})dxdydz\end{align*}$

Possible Answers:

-34.92

-1.46

8.73

17.46

Correct answer:

8.73

Explanation:

$\begin{align*}&\text{Performing a triple integral, the order of integration does}\\&\text{not entirely matter.}\\&\text{For example:}\\&\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx\\&\text{Considering our integral:}\\&\int_{-8}^{-7}\int_{8}^{10}\int_{7}^{12}(\frac{(25sin(3z))}{x})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-8}^{-7}\int_{8}^{10}\int_{7}^{12}(\frac{(25sin(3z))}{x})dxdydz=\int_{-8}^{-7}\int_{8}^{10}(25sin(3z)ln(x))dydz|_{7}^{12}\\&\int_{-8}^{-7}\int_{8}^{10}(25sin(3z)ln(\frac{12}{7}))dydz=\int_{-8}^{-7}(25ysin(3z)ln(\frac{12}{7}))dz|_{8}^{10}\\&\int_{-8}^{-7}(50sin(3z)ln(\frac{12}{7}))dz=-\frac{(50cos(3z)ln(\frac{12}{7}))}{3}|_{-8}^{-7}=8.73\end{align*}$

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Example Question #247 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{-9}^{-6}\int_{7}^{10}\int_{6}^{9}(\frac{(5ysin(z + 1))}{(3x)})dxdydz\end{align*}$

Possible Answers:

36.98

1.48

-36.98

-7.39

Correct answer:

-7.39

Explanation:

$\begin{align*}&\text{Performing a triple integral, the order of integration does}\\&\text{not entirely matter.}\\&\text{For example:}\\&\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx\\&\text{Considering our integral:}\\&\int_{-9}^{-6}\int_{7}^{10}\int_{6}^{9}(\frac{(5ysin(z + 1))}{(3x)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-9}^{-6}\int_{7}^{10}\int_{6}^{9}(\frac{(5ysin(z + 1))}{(3x)})dxdydz=\int_{-9}^{-6}\int_{7}^{10}(\frac{(5ysin(z + 1)ln(x))}{3})dydz|_{6}^{9}\\&\int_{-9}^{-6}\int_{7}^{10}(\frac{(5ysin(z + 1)ln(\frac{3}{2}))}{3})dydz=\int_{-9}^{-6}(\frac{(5y^{2}sin(z + 1)ln(\frac{3}{2}))}{6})dz|_{7}^{10}\\&\int_{-9}^{-6}(\frac{(85sin(z + 1)ln(\frac{3}{2}))}{2})dz=-\frac{(85cos(z + 1)ln(\frac{3}{2}))}{2}|_{-9}^{-6}=-7.39\end{align*}$

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Example Question #248 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{-9}^{-6}\int_{6}^{9}\int_{10}^{15}(\frac{(11e^{(-\frac{x}{2})})}{(8z)})dxdydz\end{align*}$

Possible Answers:

-0.1

-0.02

0.04

Correct answer:

-0.02

Explanation:

$\begin{align*}&\int_{-9}^{-6}\int_{6}^{9}\int_{10}^{15}(\frac{(11e^{(-\frac{x}{2})})}{(8z)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-9}^{-6}\int_{6}^{9}\int_{10}^{15}(\frac{(11e^{(-\frac{x}{2})})}{(8z)})dxdydz=\int_{-9}^{-6}\int_{6}^{9}(-\frac{(11e^{(-\frac{x}{2})})}{(4z)})dydz|_{10}^{15}\\&\int_{-9}^{-6}\int_{6}^{9}(\frac{(11e^{(-\frac{15}{2})}\cdot(e^{(\frac{5}{2})} - 1))}{(4z)})dydz=\int_{-9}^{-6}(\frac{(11ye^{(-\frac{15}{2})}\cdot(e^{(\frac{5}{2})} - 1))}{(4z)})dz|_{6}^{9}\\&\int_{-9}^{-6}(\frac{(33e^{(-\frac{15}{2})}\cdot(e^{(\frac{5}{2})} - 1))}{(4z)})dz=\frac{(33e^{(-\frac{15}{2})}ln(z)\cdot(e^{(\frac{5}{2})} - 1))}{4}|_{-9}^{-6}=-0.02\end{align*}$

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Example Question #249 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{1}^{3}\int_{-8}^{4}\int_{-8}^{-6}(\frac{(5sin(y + 1))}{(33z)})dxdydz\end{align*}$

Possible Answers:

-0.08

0.16

0.47

-0.31

Correct answer:

0.16

Explanation:

$\begin{align*}&\int_{1}^{3}\int_{-8}^{4}\int_{-8}^{-6}(\frac{(5sin(y + 1))}{(33z)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{1}^{3}\int_{-8}^{4}\int_{-8}^{-6}(\frac{(5sin(y + 1))}{(33z)})dxdydz=\int_{1}^{3}\int_{-8}^{4}(\frac{(5xsin(y + 1))}{(33z)})dydz|_{-8}^{-6}\\&\int_{1}^{3}\int_{-8}^{4}(\frac{(10sin(y + 1))}{(33z)})dydz=\int_{1}^{3}(-\frac{(10cos(y + 1))}{(33z)})dz|_{-8}^{4}\\&\int_{1}^{3}(-\frac{(10\cdot(cos(5) - cos(7)))}{(33z)})dz=-ln(z)\cdot(\frac{(10cos(5))}{33}-\frac{ (10cos(7))}{33})|_{1}^{3}=0.16\end{align*}$

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Example Question #250 : Multiple Integration

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

Possible Answers:

$-1.428\cdot10^{7}$

$5.714\cdot10^{7}$

$-2.857\cdot10^{7}$

$2.857\cdot10^{6}$

Correct answer:

$-1.428\cdot10^{7}$

Explanation:

$\begin{align*}&\int_{9}^{12}\int_{6}^{9}\int_{-8}^{-4}(\frac{(5cos(x + 2)e^{(\frac{y}{2})}e^{(z)})}{9})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{9}^{12}\int_{6}^{9}\int_{-8}^{-4}(\frac{(5cos(x + 2)e^{(\frac{y}{2})}e^{(z)})}{9})dxdydz=\int_{9}^{12}\int_{6}^{9}(\frac{(5sin(x + 2)e^{(\frac{y}{2}+ z)})}{9})dydz|_{-8}^{-4}\\&\int_{9}^{12}\int_{6}^{9}(-e^{(\frac{y}{2}+ z)}\cdot(\frac{(5sin(2))}{9}-\frac{ (5sin(6))}{9}))dydz=\int_{9}^{12}(-\frac{(10e^{(\frac{y}{2}+ z)}\cdot(sin(2) - sin(6)))}{9})dz|_{6}^{9}\\&\int_{9}^{12}(-\frac{(10e^{(3)}e^{(z)}\cdot(sin(2) - sin(6))\cdot(e^{(\frac{3}{2})} - 1))}{9})dz=-\frac{(10e^{(z + 3)}\cdot(sin(2) - sin(6))\cdot(e^{(\frac{3}{2})} - 1))}{9}|_{9}^{12}=-1.428\cdot10^{7}\end{align*}$

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Example Question #111 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{6}^{7}\int_{10}^{15}\int_{-9}^{-7}(\frac{1}{(9\cdot3^xyz^{2})})dxdydz\end{align*}$

Possible Answers:

34.17

5.69

17.08

-34.17

Correct answer:

17.08

Explanation:

$\begin{align*}&\int_{6}^{7}\int_{10}^{15}\int_{-9}^{-7}(\frac{1}{(9\cdot3^xyz^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{6}^{7}\int_{10}^{15}\int_{-9}^{-7}(\frac{1}{(9\cdot3^xyz^{2})})dxdydz=\int_{6}^{7}\int_{10}^{15}(-\frac{1}{(9\cdot3^xyz^{2}ln(3))})dydz|_{-9}^{-7}\\&\int_{6}^{7}\int_{10}^{15}(\frac{1944}{(yz^{2}ln(3))})dydz=\int_{6}^{7}(\frac{(1944ln(y))}{(z^{2}ln(3))})dz|_{10}^{15}\\&\int_{6}^{7}(\frac{(1944ln(\frac{3}{2}))}{(z^{2}ln(3))})dz=-\frac{(1944ln(\frac{3}{2}))}{(zln(3))}|_{6}^{7}=17.08\end{align*}$

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Example Question #112 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{-9}^{-8}\int_{-7}^{-4}\int_{7}^{11}(\frac{(7cos(4z)sin(x + 2)e^{(-\frac{y}{2})})}{51})dxdydz\end{align*}$

Possible Answers:

4.96

-19.82

24.77

-0.99

Correct answer:

4.96

Explanation:

$\begin{align*}&\int_{-9}^{-8}\int_{-7}^{-4}\int_{7}^{11}(\frac{(7cos(4z)sin(x + 2)e^{(-\frac{y}{2})})}{51})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-9}^{-8}\int_{-7}^{-4}\int_{7}^{11}(\frac{(7cos(4z)sin(x + 2)e^{(-\frac{y}{2})})}{51})dxdydz=\int_{-9}^{-8}\int_{-7}^{-4}(-\frac{(7cos(x + 2)cos(4z)e^{(-\frac{y}{2})})}{51})dydz|_{7}^{11}\\&\int_{-9}^{-8}\int_{-7}^{-4}(\frac{(7cos(4z)e^{(-\frac{y}{2})}\cdot(cos(9) - cos(13)))}{51})dydz=\int_{-9}^{-8}(-\frac{(14cos(4z)e^{(-\frac{y}{2})}\cdot(cos(9) - cos(13)))}{51})dz|_{-7}^{-4}\\&\int_{-9}^{-8}(\frac{(14cos(4z)e^{(2)}\cdot(cos(9) - cos(13))\cdot(e^{(\frac{3}{2})} - 1))}{51})dz=\frac{(7sin(4z)e^{(2)}\cdot(cos(9) - cos(13))\cdot(e^{(\frac{3}{2})} - 1))}{102}|_{-9}^{-8}=4.96\end{align*}$

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Example Question #113 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{2}^{6}\int_{-10}^{-7}\int_{7}^{12}(\frac{(7e^{(\frac{y}{2})})}{(3\cdot2^zx)})dxdydz\end{align*}$

Possible Answers:

0.02

0.08

-0.04

-0.01

Correct answer:

0.02

Explanation:

$\begin{align*}&\int_{2}^{6}\int_{-10}^{-7}\int_{7}^{12}(\frac{(7e^{(\frac{y}{2})})}{(3\cdot2^zx)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{2}^{6}\int_{-10}^{-7}\int_{7}^{12}(\frac{(7e^{(\frac{y}{2})})}{(3\cdot2^zx)})dxdydz=\int_{2}^{6}\int_{-10}^{-7}(\frac{(7e^{(\frac{y}{2})}ln(x))}{(3\cdot2^z)})dydz|_{7}^{12}\\&\int_{2}^{6}\int_{-10}^{-7}(\frac{(7e^{(\frac{y}{2})}ln(\frac{12}{7}))}{(3\cdot2^z)})dydz=\int_{2}^{6}(\frac{(14e^{(\frac{y}{2})}ln(\frac{12}{7}))}{(3\cdot2^z)})dz|_{-10}^{-7}\\&\int_{2}^{6}(\frac{(14e^{(-5)}ln(\frac{12}{7})\cdot(e^{(\frac{3}{2})} - 1))}{(3\cdot2^z)})dz=-\frac{(14e^{(-5)}ln(\frac{12}{7})\cdot(e^{(\frac{3}{2})} - 1))}{(3\cdot2^zln(2))}|_{2}^{6}=0.02\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #244 : Multiple Integration

Example Question #245 : Multiple Integration

Example Question #246 : Multiple Integration

Example Question #247 : Multiple Integration

Example Question #248 : Multiple Integration

Example Question #249 : Multiple Integration

Example Question #250 : Multiple Integration

Example Question #111 : Triple Integration In Cartesian Coordinates

Example Question #112 : Triple Integration In Cartesian Coordinates

Example Question #113 : Triple Integration In Cartesian Coordinates

All Calculus 3 Resources

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