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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 #384 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{6}^{7.5}\int_{4.5}^{8.5}\int_{-5}^{-0.5}(\frac{(11z^{2}e^{(-x)}e^{(-y)})}{38})dxdydz\end{align*}$

Possible Answers:

31.8

5.3

-7.95

-190.8

Correct answer:

31.8

Explanation:

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

$\begin{align*}&\int_{6}^{7.5}\int_{4.5}^{8.5}\int_{-5}^{-0.5}(\frac{(11z^{2}e^{(-x)}e^{(-y)})}{38})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{6}^{7.5}\int_{4.5}^{8.5}\int_{-5}^{-0.5}(\frac{(11z^{2}e^{(-x)}e^{(-y)})}{38})dxdydz=\int_{6}^{7.5}\int_{4.5}^{8.5}(-\frac{(11z^{2}e^{(- x - y)})}{38})dydz|_{-5}^{-0.5}\\&\int_{6}^{7.5}\int_{4.5}^{8.5}(-\frac{(11z^{2}e^{(-y)}\cdot (e^{(\frac{1}{2})} - e^{(5)}))}{38})dydz=\int_{6}^{7.5}(\frac{(11z^{2}e^{(-y)}\cdot (e^{(\frac{1}{2})} - e^{(5)}))}{38})dz|_{4.5}^{8.5}\\&\int_{6}^{7.5}(-\frac{(11z^{2}\cdot (e^{(-8)} - e^{(-\frac{7}{2})})\cdot (e^{(4)} - 1))}{38})dz=z^{3}\cdot (\frac{(11e^{(\frac{1}{2})})}{114}-\frac{ (11e^{(-4)})}{114}+\frac{ (11e^{(-8)})}{114}-\frac{ (11e^{(-\frac{7}{2})})}{114})|_{6}^{7.5}=31.8\end{align*}$

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Example Question #385 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{-6}^{-5}\int_{-8}^{-4.5}\int_{-4}^{-2}(\frac{(4cos(3y)e^{(-2x)})}{z^{2}})dxdydz\end{align*}$

Possible Answers:

-333.5

555.8

-111.2

55.58

Correct answer:

-111.2

Explanation:

$\begin{align*}&\int_{-6}^{-5}\int_{-8}^{-4.5}\int_{-4}^{-2}(\frac{(4cos(3y)e^{(-2x)})}{z^{2}})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-6}^{-5}\int_{-8}^{-4.5}\int_{-4}^{-2}(\frac{(4cos(3y)e^{(-2x)})}{z^{2}})dxdydz=\int_{-6}^{-5}\int_{-8}^{-4.5}(-\frac{(2cos(3y)e^{(-2x)})}{z^{2}})dydz|_{-4}^{-2}\\&\int_{-6}^{-5}\int_{-8}^{-4.5}(\frac{(2cos(3y)e^{(4)}\cdot (e^{(4)} - 1))}{z^{2}})dydz=\int_{-6}^{-5}(\frac{(2sin(3y)e^{(4)}\cdot (e^{(4)} - 1))}{(3z^{2})})dz|_{-8}^{-4.5}\\&\int_{-6}^{-5}(\frac{(2e^{(4)}\cdot (sin(24) - sin(\frac{27}{2}))\cdot (e^{(4)} - 1))}{(3z^{2})})dz=-\frac{(2e^{(4)}\cdot (sin(24) - sin(\frac{27}{2}))\cdot (e^{(4)} - 1))}{(3z)}|_{-6}^{-5}=-111.2\end{align*}$

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Example Question #386 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{-6}^{-1.5}\int_{-8}^{-6}\int_{4}^{8}(\frac{(65\cdot 3^{(\frac{x}{3})}sin(z + 2))}{(8y^{2})})dxdydz\end{align*}$

Possible Answers:

-20.38

-122.3

122.3

3.4

Correct answer:

-20.38

Explanation:

$\begin{align*}&\int_{-6}^{-1.5}\int_{-8}^{-6}\int_{4}^{8}(\frac{(65\cdot 3^{(\frac{x}{3})}sin(z + 2))}{(8y^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-6}^{-1.5}\int_{-8}^{-6}\int_{4}^{8}(\frac{(65\cdot 3^{(\frac{x}{3})}sin(z + 2))}{(8y^{2})})dxdydz=\int_{-6}^{-1.5}\int_{-8}^{-6}(\frac{(195\cdot 3^{(\frac{x}{3})}sin(z + 2))}{(8y^{2}ln(3))})dydz|_{4}^{8}\\&\int_{-6}^{-1.5}\int_{-8}^{-6}(-\frac{(65sin(z + 2)\cdot (9\cdot 3^{(\frac{1}{3})} - 27\cdot 3^{(\frac{2}{3})}))}{(8y^{2}ln(3))})dydz=\int_{-6}^{-1.5}(\frac{(585\cdot 3^{(\frac{1}{3})}sin(z + 2) - 1755\cdot 9^{(\frac{1}{3})}sin(z + 2))}{(8yln(3))})dz|_{-8}^{-6}\\&\int_{-6}^{-1.5}(-\frac{(195\cdot 3^{(\frac{1}{3})}sin(z + 2) - 585\cdot 3^{(\frac{2}{3})}sin(z + 2))}{(64ln(3))})dz=\frac{(195\cdot (3^{(\frac{1}{3})} - 3\cdot 9^{(\frac{1}{3})}))}{(32ln(3)\cdot (tan(\frac{z}{2}+ 1)^{2} + 1))}|_{-6}^{-1.5}=-20.38\end{align*}$

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Example Question #387 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{8}^{13}\int_{10}^{12}\int_{-4}^{-2.5}(\frac{(z^{2}cos(3y))}{(7\cdot 3^x)})dxdydz\end{align*}$

Possible Answers:

5.97

-5.97

1.19

-1.19

Correct answer:

-5.97

Explanation:

$\begin{align*}&\int_{8}^{13}\int_{10}^{12}\int_{-4}^{-2.5}(\frac{(z^{2}cos(3y))}{(7\cdot 3^x)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{8}^{13}\int_{10}^{12}\int_{-4}^{-2.5}(\frac{(z^{2}cos(3y))}{(7\cdot 3^x)})dxdydz=\int_{8}^{13}\int_{10}^{12}(-\frac{(z^{2}cos(3y))}{(7\cdot 3^xln(3))})dydz|_{-4}^{-2.5}\\&\int_{8}^{13}\int_{10}^{12}(-\frac{(9z^{2}cos(3y)\cdot (3^{(\frac{1}{2})} - 9))}{(7ln(3))})dydz=\int_{8}^{13}(-\frac{(3z^{2}sin(3y)\cdot (3^{(\frac{1}{2})} - 9))}{(7ln(3))})dz|_{10}^{12}\\&\int_{8}^{13}(\frac{(3z^{2}\cdot (sin(30) - sin(36))\cdot (3^{(\frac{1}{2})} - 9))}{(7ln(3))})dz=-\frac{(z^{3}\cdot (27sin(30) - 27sin(36) - 3\cdot 3^{(\frac{1}{2})}sin(30) + 3\cdot 3^{(\frac{1}{2})}sin(36)))}{(21ln(3))}|_{8}^{13}=-5.97\end{align*}$

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Example Question #388 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{10}^{15}\int_{-10}^{-7}\int_{8}^{9}(\frac{(23cos(z + 2)sin(3y))}{(18x)})dxdydz\end{align*}$

Possible Answers:

0.09

-0.11

-0.01

-0.06

Correct answer:

-0.01

Explanation:

$\begin{align*}&\int_{10}^{15}\int_{-10}^{-7}\int_{8}^{9}(\frac{(23cos(z + 2)sin(3y))}{(18x)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{10}^{15}\int_{-10}^{-7}\int_{8}^{9}(\frac{(23cos(z + 2)sin(3y))}{(18x)})dxdydz=\int_{10}^{15}\int_{-10}^{-7}(\frac{(23cos(z + 2)sin(3y)ln(x))}{18})dydz|_{8}^{9}\\&\int_{10}^{15}\int_{-10}^{-7}(\frac{(23cos(z + 2)sin(3y)ln(\frac{9}{8}))}{18})dydz=\int_{10}^{15}(-\frac{(23cos(3y)cos(z + 2)ln(\frac{9}{8}))}{54})dz|_{-10}^{-7}\\&\int_{10}^{15}(-\frac{(23cos(z + 2)ln(\frac{9}{8})\cdot (\frac{cos(21)}{3}-\frac{ cos(30)}{3}))}{18})dz=-\frac{(23sin(z + 2)ln(\frac{9}{8})\cdot (\frac{cos(21)}{3}-\frac{ cos(30)}{3}))}{18}|_{10}^{15}=-0.01\end{align*}$

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Example Question #389 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{-3}^{-0.5}\int_{-4.5}^{-1.5}\int_{5}^{10}(\frac{(21\cdot 2^{(\frac{x}{4})}\cdot 3^{(\frac{y}{2})}\cdot 3^{(\frac{z}{2})})}{2})dxdydz\end{align*}$

Possible Answers:

-26.47

132.3

264.7

Correct answer:

132.3

Explanation:

$\begin{align*}&\int_{-3}^{-0.5}\int_{-4.5}^{-1.5}\int_{5}^{10}(\frac{(21\cdot 2^{(\frac{x}{4})}\cdot 3^{(\frac{y}{2})}\cdot 3^{(\frac{z}{2})})}{2})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-3}^{-0.5}\int_{-4.5}^{-1.5}\int_{5}^{10}(\frac{(21\cdot 2^{(\frac{x}{4})}\cdot 3^{(\frac{y}{2})}\cdot 3^{(\frac{z}{2})})}{2})dxdydz=\int_{-3}^{-0.5}\int_{-4.5}^{-1.5}(\frac{(42\cdot 2^{(\frac{x}{4})}\cdot 3^{(\frac{y}{2}+\frac{ z}{2})})}{ln(2)})dydz|_{5}^{10}\end{align*}$

$\begin{align*}\\&\int_{-3}^{-0.5}\int_{-4.5}^{-1.5}(\frac{(84\cdot 2^{(\frac{1}{4})}\cdot 3^{(\frac{y}{2})}\cdot 3^{(\frac{z}{2})}\cdot (2\cdot 2^{(\frac{1}{4})} - 1))}{ln(2)})dydz=\int_{-3}^{-0.5}(\frac{(2\cdot (168\cdot 2^{(\frac{1}{2})}\cdot 3^{(\frac{y}{2})}\cdot 3^{(\frac{z}{2})} - 84\cdot 2^{(\frac{1}{4})}\cdot 3^{(\frac{y}{2})}\cdot 3^{(\frac{z}{2})}))}{(ln(2)ln(3))})dz|_{-4.5}^{-1.5}\\&\int_{-3}^{-0.5}(-\frac{(56\cdot 2^{(\frac{1}{4})}\cdot 3^{(\frac{1}{4})}\cdot 3^{(\frac{z}{2})}\cdot (2\cdot 2^{(\frac{1}{4})} - 1)\cdot (3^{(\frac{1}{2})} - 9))}{(9ln(2)ln(3))})dz=\frac{(2\cdot 3^{(\frac{z}{2})}\cdot (1008\cdot 2^{(\frac{1}{2})}\cdot 3^{(\frac{1}{4})} - 112\cdot 2^{(\frac{1}{2})}\cdot 27^{(\frac{1}{4})} - 504\cdot 6^{(\frac{1}{4})} + 56\cdot 54^{(\frac{1}{4})}))}{(9ln(2)ln(3)^{2})}|_{-3}^{-0.5}=132.3\end{align*}$

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Example Question #390 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{-6}^{-4}\int_{-10}^{-7.5}\int_{4}^{8.5}(\frac{(cos(3y)cos(z + 1))}{(2\cdot 2^{(\frac{x}{2})})})dxdydz\end{align*}$

Possible Answers:

0.01

-0.31

-0.01

0.05

Correct answer:

0.05

Explanation:

$\begin{align*}&\int_{-6}^{-4}\int_{-10}^{-7.5}\int_{4}^{8.5}(\frac{(cos(3y)cos(z + 1))}{(2\cdot 2^{(\frac{x}{2})})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-6}^{-4}\int_{-10}^{-7.5}\int_{4}^{8.5}(\frac{(cos(3y)cos(z + 1))}{(2\cdot 2^{(\frac{x}{2})})})dxdydz=\int_{-6}^{-4}\int_{-10}^{-7.5}(-\frac{(cos(3y)cos(z + 1))}{(2^{(\frac{x}{2})}ln(2))})dydz|_{4}^{8.5}\\&\int_{-6}^{-4}\int_{-10}^{-7.5}(-\frac{(cos(3y)cos(z + 1)\cdot (2^{(\frac{3}{4})} - 8))}{(32ln(2))})dydz=\int_{-6}^{-4}(-\frac{(cos(z + 1)sin(3y)\cdot (2^{(\frac{3}{4})} - 8))}{(96ln(2))})dz|_{-10}^{-7.5}\\&\int_{-6}^{-4}(-\frac{(cos(z + 1)\cdot (sin(30) - sin(\frac{45}{2}))\cdot (2^{(\frac{3}{4})} - 8))}{(96ln(2))})dz=-\frac{(sin(z + 1)\cdot (sin(30) - sin(\frac{45}{2}))\cdot (2^{(\frac{3}{4})} - 8))}{(96ln(2))}|_{-6}^{-4}=0.05\end{align*}$

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Example Question #391 : Triple Integrals

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

Possible Answers:

0.08

-0.02

0.03

-0.32

Correct answer:

0.08

Explanation:

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

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Example Question #391 : Triple Integrals

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

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 #393 : Triple Integrals

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

Possible Answers:

-0.03

0.13

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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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #384 : Triple Integrals

Example Question #385 : Triple Integrals

Example Question #386 : Triple Integrals

Example Question #387 : Triple Integrals

Example Question #388 : Triple Integrals

Example Question #389 : Triple Integrals

Example Question #390 : Triple Integrals

Example Question #391 : Triple Integrals

Example Question #391 : Triple Integrals

Example Question #393 : Triple Integrals

All Calculus 3 Resources

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