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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 #344 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{7}^{9}\int_{9}^{10}\int_{-3.5}^{-1}(\frac{(65z^{2}e^{(2x)})}{8})dxdydz\end{align*}$

Possible Answers:

-70.26

351.3

70.26

-35.13

Correct answer:

70.26

Explanation:

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

$\begin{align*}&\int_{7}^{9}\int_{9}^{10}\int_{-3.5}^{-1}(\frac{(65z^{2}e^{(2x)})}{8})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{7}^{9}\int_{9}^{10}\int_{-3.5}^{-1}(\frac{(65z^{2}e^{(2x)})}{8})dxdydz=\int_{7}^{9}\int_{9}^{10}(\frac{(65z^{2}e^{(2x)})}{16})dydz|_{-3.5}^{-1}\\&\int_{7}^{9}\int_{9}^{10}(\frac{(65z^{2}e^{(-7)}\cdot (e^{(5)} - 1))}{16})dydz=\int_{7}^{9}(\frac{(65yz^{2}e^{(-7)}\cdot (e^{(5)} - 1))}{16})dz|_{9}^{10}\\&\int_{7}^{9}(\frac{(65z^{2}e^{(-7)}\cdot (e^{(5)} - 1))}{16})dz=\frac{(65z^{3}e^{(-7)}\cdot (e^{(5)} - 1))}{48}|_{7}^{9}=70.26\end{align*}$

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Example Question #345 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{7}^{8}\int_{5}^{6.5}\int_{4.5}^{9}(\frac{(31cos(3z))}{(2\cdot 3^{(\frac{x}{2})}y)})dxdydz\end{align*}$

Possible Answers:

-0.33

-1.33

0.11

0.33

Correct answer:

-0.33

Explanation:

$\begin{align*}&\int_{7}^{8}\int_{5}^{6.5}\int_{4.5}^{9}(\frac{(31cos(3z))}{(2\cdot 3^{(\frac{x}{2})}y)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{7}^{8}\int_{5}^{6.5}\int_{4.5}^{9}(\frac{(31cos(3z))}{(2\cdot 3^{(\frac{x}{2})}y)})dxdydz=\int_{7}^{8}\int_{5}^{6.5}(-\frac{(31cos(3z))}{(3^{(\frac{x}{2})}yln(3))})dydz|_{4.5}^{9}\\&\int_{7}^{8}\int_{5}^{6.5}(-\frac{(31cos(3z)\cdot (2\cdot 3^{(\frac{1}{2})} - 18\cdot 3^{(\frac{3}{4})}))}{(486yln(3))})dydz=\int_{7}^{8}(-\frac{(ln(y)\cdot (31\cdot 3^{(\frac{1}{2})}cos(3z) - 279\cdot 27^{(\frac{1}{4})}cos(3z)))}{(243ln(3))})dz|_{5}^{6.5}\\&\int_{7}^{8}(-\frac{(31\cdot 3^{(\frac{1}{2})}cos(3z)ln(\frac{13}{10}) - 279\cdot 3^{(\frac{3}{4})}cos(3z)ln(\frac{13}{10}))}{(243ln(3))})dz=-\frac{(31\cdot 3^{(\frac{1}{2})}sin(3z)ln(\frac{13}{10}) - 279\cdot 27^{(\frac{1}{4})}sin(3z)ln(\frac{13}{10}))}{(729ln(3))}|_{7}^{8}=-0.33\end{align*}$

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Example Question #346 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{10}^{13.5}\int_{5}^{7}\int_{-7}^{-2}(\frac{(8\cdot 3^ysin(3x))}{(31z)})dxdydz\end{align*}$

Possible Answers:

137.8

-22.96

22.96

-68.88

Correct answer:

-68.88

Explanation:

$\begin{align*}&\int_{10}^{13.5}\int_{5}^{7}\int_{-7}^{-2}(\frac{(8\cdot 3^ysin(3x))}{(31z)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{10}^{13.5}\int_{5}^{7}\int_{-7}^{-2}(\frac{(8\cdot 3^ysin(3x))}{(31z)})dxdydz=\int_{10}^{13.5}\int_{5}^{7}(-\frac{(8\cdot 3^ycos(3x))}{(93z)})dydz|_{-7}^{-2}\\&\int_{10}^{13.5}\int_{5}^{7}(-\frac{(8\cdot 3^y\cdot (\frac{cos(6)}{3}-\frac{ cos(21)}{3}))}{(31z)})dydz=\int_{10}^{13.5}(-\frac{(8\cdot 3^y\cdot (cos(6) - cos(21)))}{(93zln(3))})dz|_{5}^{7}\\&\int_{10}^{13.5}(-\frac{(5184\cdot (cos(6) - cos(21)))}{(31zln(3))})dz=-\frac{(ln(z)\cdot (5184cos(6) - 5184cos(21)))}{(31ln(3))}|_{10}^{13.5}=-68.88\end{align*}$

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Example Question #347 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{-7}^{-5.5}\int_{-9}^{-7}\int_{9}^{11.5}(\frac{(ysin(x + 1)sin(z + 1))}{2})dxdydz\end{align*}$

Possible Answers:

-17.21

-2.87

51.62

17.21

Correct answer:

17.21

Explanation:

$\begin{align*}&\int_{-7}^{-5.5}\int_{-9}^{-7}\int_{9}^{11.5}(\frac{(ysin(x + 1)sin(z + 1))}{2})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-7}^{-5.5}\int_{-9}^{-7}\int_{9}^{11.5}(\frac{(ysin(x + 1)sin(z + 1))}{2})dxdydz=\int_{-7}^{-5.5}\int_{-9}^{-7}(-\frac{(ycos(x + 1)sin(z + 1))}{2})dydz|_{9}^{11.5}\\&\int_{-7}^{-5.5}\int_{-9}^{-7}(\frac{(ysin(z + 1)\cdot (cos(10) - cos(\frac{25}{2})))}{2})dydz=\int_{-7}^{-5.5}(y^{2}\cdot (\frac{(sin(z + 1)cos(10))}{4}-\frac{ (sin(z + 1)cos(\frac{25}{2}))}{4}))dz|_{-9}^{-7}\\&\int_{-7}^{-5.5}(-8sin(z + 1)\cdot (cos(10) - cos(\frac{25}{2})))dz=8cos(z + 1)\cdot (cos(10) - cos(\frac{25}{2}))|_{-7}^{-5.5}=17.21\end{align*}$

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Example Question #348 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{-10}^{-5.5}\int_{8}^{11.5}\int_{9}^{12.5}(\frac{(37sin(x + 2)sin(4z))}{y^{2}})dxdydz\end{align*}$

Possible Answers:

-0.21

0.17

-0.01

0.04

Correct answer:

0.04

Explanation:

$\begin{align*}&\int_{-10}^{-5.5}\int_{8}^{11.5}\int_{9}^{12.5}(\frac{(37sin(x + 2)sin(4z))}{y^{2}})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-10}^{-5.5}\int_{8}^{11.5}\int_{9}^{12.5}(\frac{(37sin(x + 2)sin(4z))}{y^{2}})dxdydz=\int_{-10}^{-5.5}\int_{8}^{11.5}(-\frac{(37cos(x + 2)sin(4z))}{y^{2}})dydz|_{9}^{12.5}\\&\int_{-10}^{-5.5}\int_{8}^{11.5}(\frac{(37sin(4z)\cdot (cos(11) - cos(\frac{29}{2})))}{y^{2}})dydz=\int_{-10}^{-5.5}(-\frac{(37sin(4z)cos(11) - 37sin(4z)cos(\frac{29}{2}))}{y})dz|_{8}^{11.5}\\&\int_{-10}^{-5.5}(\frac{(259sin(4z)\cdot (cos(11) - cos(\frac{29}{2})))}{184})dz=-cos(4z)\cdot (\frac{(259cos(11))}{736}-\frac{ (259cos(\frac{29}{2}))}{736})|_{-10}^{-5.5}=0.04\end{align*}$

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Example Question #349 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{-8}^{-3}\int_{-4}^{1}\int_{-9}^{-5.5}(\frac{(53\cdot 2^y)}{(10x^{3}z)})dxdydz\end{align*}$

Possible Answers:

-0.45

0.15

-0.03

0.08

Correct answer:

0.15

Explanation:

$\begin{align*}&\int_{-8}^{-3}\int_{-4}^{1}\int_{-9}^{-5.5}(\frac{(53\cdot 2^y)}{(10x^{3}z)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-8}^{-3}\int_{-4}^{1}\int_{-9}^{-5.5}(\frac{(53\cdot 2^y)}{(10x^{3}z)})dxdydz=\int_{-8}^{-3}\int_{-4}^{1}(-\frac{(53\cdot 2^y)}{(20x^{2}z)})dydz|_{-9}^{-5.5}\\&\int_{-8}^{-3}\int_{-4}^{1}(-\frac{(10759\cdot 2^y)}{(196020z)})dydz=\int_{-8}^{-3}(-\frac{(10759\cdot 2^y)}{(196020zln(2))})dz|_{-4}^{1}\\&\int_{-8}^{-3}(-\frac{333529}{(3136320zln(2))})dz=-\frac{(333529ln(z))}{(3136320ln(2))}|_{-8}^{-3}=0.15\end{align*}$

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Example Question #350 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{10}^{12.5}\int_{7}^{8.5}\int_{-5}^{-0.5}(\frac{(2z^{2}cos(3y)e^{(x)})}{21})dxdydz\end{align*}$

Possible Answers:

-2.89

5.78

-0.72

0.96

Correct answer:

-2.89

Explanation:

$\begin{align*}&\int_{10}^{12.5}\int_{7}^{8.5}\int_{-5}^{-0.5}(\frac{(2z^{2}cos(3y)e^{(x)})}{21})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{10}^{12.5}\int_{7}^{8.5}\int_{-5}^{-0.5}(\frac{(2z^{2}cos(3y)e^{(x)})}{21})dxdydz=\int_{10}^{12.5}\int_{7}^{8.5}(\frac{(2z^{2}cos(3y)e^{(x)})}{21})dydz|_{-5}^{-0.5}\\&\int_{10}^{12.5}\int_{7}^{8.5}(\frac{(2z^{2}cos(3y)\cdot (e^{(-\frac{1}{2})} - e^{(-5)}))}{21})dydz=\int_{10}^{12.5}(\frac{(2z^{2}sin(3y)\cdot (e^{(-\frac{1}{2})} - e^{(-5)}))}{63})dz|_{7}^{8.5}\\&\int_{10}^{12.5}(-\frac{(2z^{2}e^{(-5)}\cdot (sin(21) - sin(\frac{51}{2}))\cdot (e^{(\frac{9}{2})} - 1))}{63})dz=-z^{3}\cdot (\frac{(2e^{(-\frac{1}{2})}sin(21))}{189}-\frac{ (2e^{(-5)}sin(21))}{189}-\frac{ (2e^{(-\frac{1}{2})}sin(\frac{51}{2}))}{189}+\frac{ (2e^{(-5)}sin(\frac{51}{2}))}{189})|_{10}^{12.5}=-2.89\end{align*}$

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Example Question #351 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{-9}^{-8}\int_{8}^{11.5}\int_{-6}^{-3}(\frac{(72cos(x + 2)sin(4z))}{(11y^{2})})dxdydz\end{align*}$

Possible Answers:

0.03

-0.02

-0.29

0.1

Correct answer:

0.1

Explanation:

$\begin{align*}&\int_{-9}^{-8}\int_{8}^{11.5}\int_{-6}^{-3}(\frac{(72cos(x + 2)sin(4z))}{(11y^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-9}^{-8}\int_{8}^{11.5}\int_{-6}^{-3}(\frac{(72cos(x + 2)sin(4z))}{(11y^{2})})dxdydz=\int_{-9}^{-8}\int_{8}^{11.5}(\frac{(72sin(x + 2)sin(4z))}{(11y^{2})})dydz|_{-6}^{-3}\\&\int_{-9}^{-8}\int_{8}^{11.5}(-\frac{(72sin(4z)\cdot (sin(1) - sin(4)))}{(11y^{2})})dydz=\int_{-9}^{-8}(\frac{(\frac{(72sin(4z)sin(1))}{11}-\frac{ (72sin(4z)sin(4))}{11})}{y})dz|_{8}^{11.5}\\&\int_{-9}^{-8}(-\frac{(63sin(4z)\cdot (sin(1) - sin(4)))}{253})dz=cos(4z)\cdot (\frac{(63sin(1))}{1012}-\frac{ (63sin(4))}{1012})|_{-9}^{-8}=0.1\end{align*}$

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Example Question #352 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{-3.5}^{1.75}\int_{-6}^{-4}\int_{6}^{9.5}(10cos(4y)sin(x + 1)e^{(2z)})dxdydz\end{align*}$

Possible Answers:

10.48

-31.43

-188.6

125.7

Correct answer:

-31.43

Explanation:

$\begin{align*}&\int_{-3.5}^{1.75}\int_{-6}^{-4}\int_{6}^{9.5}(10cos(4y)sin(x + 1)e^{(2z)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-3.5}^{1.75}\int_{-6}^{-4}\int_{6}^{9.5}(10cos(4y)sin(x + 1)e^{(2z)})dxdydz=\int_{-3.5}^{1.75}\int_{-6}^{-4}(-10cos(x + 1)cos(4y)e^{(2z)})dydz|_{6}^{9.5}\\&\int_{-3.5}^{1.75}\int_{-6}^{-4}(10cos(4y)e^{(2z)}\cdot (cos(7) - cos(\frac{21}{2})))dydz=\int_{-3.5}^{1.75}(\frac{(5sin(4y)e^{(2z)}\cdot (cos(7) - cos(\frac{21}{2})))}{2})dz|_{-6}^{-4}\\&\int_{-3.5}^{1.75}(-\frac{(5e^{(2z)}\cdot (cos(7) - cos(\frac{21}{2}))\cdot (sin(16) - sin(24)))}{2})dz=-\frac{(5e^{(2z)}\cdot (cos(7) - cos(\frac{21}{2}))\cdot (sin(16) - sin(24)))}{4}|_{-3.5}^{1.75}=-31.43\end{align*}$

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Example Question #353 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{3}^{6.5}\int_{-8}^{-7}\int_{-4}^{1}(\frac{(y^{2}e^{(-x)}e^{(z)})}{14})dxdydz\end{align*}$

Possible Answers:

140800

422300

Correct answer:

140800

Explanation:

$\begin{align*}&\int_{3}^{6.5}\int_{-8}^{-7}\int_{-4}^{1}(\frac{(y^{2}e^{(-x)}e^{(z)})}{14})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{3}^{6.5}\int_{-8}^{-7}\int_{-4}^{1}(\frac{(y^{2}e^{(-x)}e^{(z)})}{14})dxdydz=\int_{3}^{6.5}\int_{-8}^{-7}(-\frac{(y^{2}e^{(z - x)})}{14})dydz|_{-4}^{1}\\&\int_{3}^{6.5}\int_{-8}^{-7}(\frac{(y^{2}e^{(-1)}e^{(z)}\cdot (e^{(5)} - 1))}{14})dydz=\int_{3}^{6.5}(\frac{(y^{3}e^{(-1)}e^{(z)}\cdot (e^{(5)} - 1))}{42})dz|_{-8}^{-7}\\&\int_{3}^{6.5}(\frac{(169e^{(-1)}e^{(z)}\cdot (e^{(5)} - 1))}{42})dz=e^{(z - 1)}\cdot (\frac{(169e^{(5)})}{42}-\frac{ 169}{42})|_{3}^{6.5}=140800\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #344 : Triple Integrals

Example Question #345 : Triple Integrals

Example Question #346 : Triple Integrals

Example Question #347 : Triple Integrals

Example Question #348 : Triple Integrals

Example Question #349 : Triple Integrals

Example Question #350 : Triple Integrals

Example Question #351 : Triple Integrals

Example Question #352 : Triple Integrals

Example Question #353 : Triple Integrals

All Calculus 3 Resources

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