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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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

Example Questions

Example Question #254 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{-4}^{8}\int_{-10}^{-8}\int_{-6}^{-4}(\frac{(23xy^{2}cos(z + 1))}{8})dxdydz\end{align*}$

Possible Answers:

12944

862.4

Correct answer:

Explanation:

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

$\begin{align*}&\int_{-4}^{8}\int_{-10}^{-8}\int_{-6}^{-4}(\frac{(23xy^{2}cos(z + 1))}{8})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-4}^{8}\int_{-10}^{-8}\int_{-6}^{-4}(\frac{(23xy^{2}cos(z + 1))}{8})dxdydz=\int_{-4}^{8}\int_{-10}^{-8}(\frac{(23x^{2}y^{2}cos(z + 1))}{16})dydz|_{-6}^{-4}\\&\int_{-4}^{8}\int_{-10}^{-8}(-\frac{(115y^{2}cos(z + 1))}{4})dydz=\int_{-4}^{8}(-\frac{(115y^{3}cos(z + 1))}{12})dz|_{-10}^{-8}\\&\int_{-4}^{8}(-\frac{(14030cos(z + 1))}{3})dz=-\frac{(14030sin(z + 1))}{3}|_{-4}^{8}=-2587\end{align*}$

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Example Question #255 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{7}^{9}\int_{-7}^{-2}\int_{6}^{8}(\frac{(4^{(\frac{z}{2})}y^{2}sin(3x))}{46})dxdydz\end{align*}$

Possible Answers:

-529.3

635.1

-21.17

105.9

Correct answer:

105.9

Explanation:

$\begin{align*}&\int_{7}^{9}\int_{-7}^{-2}\int_{6}^{8}(\frac{(4^{(\frac{z}{2})}y^{2}sin(3x))}{46})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{7}^{9}\int_{-7}^{-2}\int_{6}^{8}(\frac{(4^{(\frac{z}{2})}y^{2}sin(3x))}{46})dxdydz=\int_{7}^{9}\int_{-7}^{-2}(-\frac{(2^zy^{2}cos(3x))}{138})dydz|_{6}^{8}\\&\int_{7}^{9}\int_{-7}^{-2}(2^zy^{2}\cdot(\frac{cos(18)}{138}-\frac{ cos(24)}{138}))dydz=\int_{7}^{9}(y^{3}\cdot(\frac{(2^zcos(18))}{414}-\frac{ (2^zcos(24))}{414}))dz|_{-7}^{-2}\\&\int_{7}^{9}(\frac{(335\cdot2^z\cdot(\frac{cos(18)}{138}-\frac{ cos(24)}{138}))}{3})dz=\frac{(335\cdot2^z\cdot(cos(18) - cos(24)))}{(414ln(2))}|_{7}^{9}=105.9\end{align*}$

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Example Question #121 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{-7}^{-3}\int_{10}^{15}\int_{-8}^{-4}(\frac{(5cos(z + 1))}{(94xy)})dxdydz\end{align*}$

Possible Answers:

0.02

0.04

-0.05

Correct answer:

0.02

Explanation:

$\begin{align*}&\int_{-7}^{-3}\int_{10}^{15}\int_{-8}^{-4}(\frac{(5cos(z + 1))}{(94xy)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-7}^{-3}\int_{10}^{15}\int_{-8}^{-4}(\frac{(5cos(z + 1))}{(94xy)})dxdydz=\int_{-7}^{-3}\int_{10}^{15}(\frac{(5cos(z + 1)ln(x))}{(94y)})dydz|_{-8}^{-4}\\&\int_{-7}^{-3}\int_{10}^{15}(-\frac{(5cos(z + 1)ln(2))}{(94y)})dydz=\int_{-7}^{-3}(-\frac{(5cos(z + 1)ln(2)ln(y))}{94})dz|_{10}^{15}\\&\int_{-7}^{-3}(-\frac{(5cos(z + 1)ln(2)ln(\frac{3}{2}))}{94})dz=-\frac{(5sin(z + 1)ln(2)ln(\frac{3}{2}))}{94}|_{-7}^{-3}=0.02\end{align*}$

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Example Question #257 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{-6}^{-4}\int_{-8}^{-7}\int_{-10}^{-5}(\frac{(4zsin(4x))}{(17y)})dxdydz\end{align*}$

Possible Answers:

-0.02

0.34

-0.25

-0.08

Correct answer:

-0.08

Explanation:

$\begin{align*}&\int_{-6}^{-4}\int_{-8}^{-7}\int_{-10}^{-5}(\frac{(4zsin(4x))}{(17y)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-6}^{-4}\int_{-8}^{-7}\int_{-10}^{-5}(\frac{(4zsin(4x))}{(17y)})dxdydz=\int_{-6}^{-4}\int_{-8}^{-7}(-\frac{(zcos(4x))}{(17y)})dydz|_{-10}^{-5}\\&\int_{-6}^{-4}\int_{-8}^{-7}(-\frac{(4z\cdot(\frac{cos(20)}{4}-\frac{ cos(40)}{4}))}{(17y)})dydz=\int_{-6}^{-4}(-ln(y)\cdot(\frac{(zcos(20))}{17}-\frac{ (zcos(40))}{17}))dz|_{-8}^{-7}\\&\int_{-6}^{-4}(-\frac{(zln(\frac{7}{8})\cdot(cos(20) - cos(40)))}{17})dz=-z^{2}\cdot(\frac{(cos(20)ln(\frac{7}{8}))}{34}-\frac{ (cos(40)ln(\frac{7}{8}))}{34})|_{-6}^{-4}=-0.08\end{align*}$

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Example Question #121 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{10}^{13}\int_{-6}^{-3}\int_{-10}^{-8}(\frac{(4ysin(4x)sin(z + 2))}{39})dxdydz\end{align*}$

Possible Answers:

0.83

0.14

4.17

-2.5

Correct answer:

0.83

Explanation:

$\begin{align*}&\int_{10}^{13}\int_{-6}^{-3}\int_{-10}^{-8}(\frac{(4ysin(4x)sin(z + 2))}{39})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{10}^{13}\int_{-6}^{-3}\int_{-10}^{-8}(\frac{(4ysin(4x)sin(z + 2))}{39})dxdydz=\int_{10}^{13}\int_{-6}^{-3}(-\frac{(ycos(4x)sin(z + 2))}{39})dydz|_{-10}^{-8}\\&\int_{10}^{13}\int_{-6}^{-3}(-\frac{(4ysin(z + 2)\cdot(\frac{cos(32)}{4}-\frac{ cos(40)}{4}))}{39})dydz=\int_{10}^{13}(-y^{2}\cdot(\frac{(sin(z + 2)cos(32))}{78}-\frac{ (sin(z + 2)cos(40))}{78}))dz|_{-6}^{-3}\\&\int_{10}^{13}(\frac{(18sin(z + 2)\cdot(\frac{cos(32)}{4}-\frac{ cos(40)}{4}))}{13})dz=-cos(z + 2)\cdot(\frac{(9cos(32))}{26}-\frac{ (9cos(40))}{26})|_{10}^{13}=0.83\end{align*}$

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Example Question #122 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{-9}^{-8}\int_{6}^{9}\int_{-7}^{-3}(\frac{(19cos(y + 1))}{(5\cdot4^{(\frac{x}{2})}z^{2})})dxdydz\end{align*}$

Possible Answers:

21.95

-1.83

-43.89

-10.97

Correct answer:

-10.97

Explanation:

$\begin{align*}&\int_{-9}^{-8}\int_{6}^{9}\int_{-7}^{-3}(\frac{(19cos(y + 1))}{(5\cdot4^{(\frac{x}{2})}z^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-9}^{-8}\int_{6}^{9}\int_{-7}^{-3}(\frac{(19cos(y + 1))}{(5\cdot4^{(\frac{x}{2})}z^{2})})dxdydz=\int_{-9}^{-8}\int_{6}^{9}(-\frac{(19cos(y + 1))}{(5\cdot2^xz^{2}ln(2))})dydz|_{-7}^{-3}\\&\int_{-9}^{-8}\int_{6}^{9}(\frac{(456cos(y + 1))}{(z^{2}ln(2))})dydz=\int_{-9}^{-8}(\frac{(456sin(y + 1))}{(z^{2}ln(2))})dz|_{6}^{9}\\&\int_{-9}^{-8}(-\frac{(456\cdot(sin(7) - sin(10)))}{(z^{2}ln(2))})dz=\frac{(456\cdot(sin(7) - sin(10)))}{(zln(2))}|_{-9}^{-8}=-10.97\end{align*}$

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Example Question #731 : Calculus 3

$\begin{align*}&\text{Evaluate the triple integral}\int_{-2}^{3}\int_{-6}^{-5}\int_{7}^{10}(\frac{(10sin(3y)e^{(2z)})}{(43\cdot2^{(\frac{x}{2})})})dxdydz\end{align*}$

Possible Answers:

-1.22

3.66

-7.32

21.96

Correct answer:

3.66

Explanation:

$\begin{align*}&\int_{-2}^{3}\int_{-6}^{-5}\int_{7}^{10}(\frac{(10sin(3y)e^{(2z)})}{(43\cdot2^{(\frac{x}{2})})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-2}^{3}\int_{-6}^{-5}\int_{7}^{10}(\frac{(10sin(3y)e^{(2z)})}{(43\cdot2^{(\frac{x}{2})})})dxdydz=\int_{-2}^{3}\int_{-6}^{-5}(-\frac{(20sin(3y)e^{(2z)})}{(43\cdot2^{(\frac{x}{2})}ln(2))})dydz|_{7}^{10}\\&\int_{-2}^{3}\int_{-6}^{-5}(\frac{(5sin(3y)e^{(2z)}\cdot(2\cdot2^{(\frac{1}{2})} - 1))}{(344ln(2))})dydz=\int_{-2}^{3}(-\frac{(5cos(3y)e^{(2z)}\cdot(2\cdot2^{(\frac{1}{2})} - 1))}{(1032ln(2))})dz|_{-6}^{-5}\\&\int_{-2}^{3}(-\frac{(5e^{(2z)}\cdot(2\cdot2^{(\frac{1}{2})} - 1)\cdot(cos(15) - cos(18)))}{(1032ln(2))})dz=-\frac{(5e^{(2z)}\cdot(2\cdot2^{(\frac{1}{2})} - 1)\cdot(cos(15) - cos(18)))}{(2064ln(2))}|_{-2}^{3}=3.66\end{align*}$

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Example Question #732 : Calculus 3

$\begin{align*}&\text{Evaluate the triple integral}\int_{8}^{12}\int_{-9}^{-7}\int_{-8}^{-4}(\frac{(4\cdot4^{(\frac{z}{2})}cos(y + 1)e^{(-2x)})}{67})dxdydz\end{align*}$

Possible Answers:

$-7.455\cdot10^{9}$

$1.864\cdot10^{9}$

$9.319\cdot10^{8}$

$5.592\cdot10^{9}$

Correct answer:

$1.864\cdot10^{9}$

Explanation:

$\begin{align*}&\int_{8}^{12}\int_{-9}^{-7}\int_{-8}^{-4}(\frac{(4\cdot4^{(\frac{z}{2})}cos(y + 1)e^{(-2x)})}{67})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{8}^{12}\int_{-9}^{-7}\int_{-8}^{-4}(\frac{(4\cdot4^{(\frac{z}{2})}cos(y + 1)e^{(-2x)})}{67})dxdydz=\int_{8}^{12}\int_{-9}^{-7}(-\frac{(2\cdot2^zcos(y + 1)e^{(-2x)})}{67})dydz|_{-8}^{-4}\\&\int_{8}^{12}\int_{-9}^{-7}(\frac{(2\cdot2^zcos(y + 1)e^{(8)}\cdot(e^{(8)} - 1))}{67})dydz=\int_{8}^{12}(\frac{(2\cdot2^zsin(y + 1)e^{(8)}\cdot(e^{(8)} - 1))}{67})dz|_{-9}^{-7}\\&\int_{8}^{12}(-\frac{(2\cdot2^ze^{(8)}\cdot(sin(6) - sin(8))\cdot(e^{(8)} - 1))}{67})dz=\frac{(2\cdot2^z\cdot(e^{(8)}sin(6) - e^{(8)}sin(8) - e^{(16)}sin(6) + e^{(16)}sin(8)))}{(67ln(2))}|_{8}^{12}=1.864\cdot10^{9}\end{align*}$

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Example Question #733 : Calculus 3

$\begin{align*}&\text{Evaluate the triple integral}\int_{-9}^{-7}\int_{9}^{14}\int_{8}^{12}(\frac{(7\cdot2^{(\frac{z}{2})}cos(4x))}{2})dxdydz\end{align*}$

Possible Answers:

-0.37

4.42

-0.74

-2.21

Correct answer:

-0.74

Explanation:

$\begin{align*}&\int_{-9}^{-7}\int_{9}^{14}\int_{8}^{12}(\frac{(7\cdot2^{(\frac{z}{2})}cos(4x))}{2})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-9}^{-7}\int_{9}^{14}\int_{8}^{12}(\frac{(7\cdot2^{(\frac{z}{2})}cos(4x))}{2})dxdydz=\int_{-9}^{-7}\int_{9}^{14}(\frac{(7\cdot2^{(\frac{z}{2})}sin(4x))}{8})dydz|_{8}^{12}\\&\int_{-9}^{-7}\int_{9}^{14}(-\frac{(7\cdot2^{(\frac{z}{2})}\cdot(\frac{sin(32)}{4}-\frac{ sin(48)}{4}))}{2})dydz=\int_{-9}^{-7}(-\frac{(7\cdot2^{(\frac{z}{2})}y\cdot(\frac{sin(32)}{4}-\frac{ sin(48)}{4}))}{2})dz|_{9}^{14}\\&\int_{-9}^{-7}(-\frac{(35\cdot2^{(\frac{z}{2})}\cdot(\frac{sin(32)}{4}-\frac{ sin(48)}{4}))}{2})dz=-\frac{(35\cdot2^{(\frac{z}{2})}\cdot(sin(32) - sin(48)))}{(4ln(2))}|_{-9}^{-7}=-0.74\end{align*}$

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Example Question #734 : Calculus 3

$\begin{align*}&\text{Evaluate the triple integral}\int_{7}^{10}\int_{-9}^{-4}\int_{-5}^{-3}(\frac{(11cos(z + 2)e^{(-2x)})}{(56y)})dxdydz\end{align*}$

Possible Answers:

1634

3268

-272.3

Correct answer:

1634

Explanation:

$\begin{align*}&\int_{7}^{10}\int_{-9}^{-4}\int_{-5}^{-3}(\frac{(11cos(z + 2)e^{(-2x)})}{(56y)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{7}^{10}\int_{-9}^{-4}\int_{-5}^{-3}(\frac{(11cos(z + 2)e^{(-2x)})}{(56y)})dxdydz=\int_{7}^{10}\int_{-9}^{-4}(-\frac{(11cos(z + 2)e^{(-2x)})}{(112y)})dydz|_{-5}^{-3}\\&\int_{7}^{10}\int_{-9}^{-4}(\frac{(11cos(z + 2)e^{(6)}\cdot(e^{(4)} - 1))}{(112y)})dydz=\int_{7}^{10}(-ln(y)\cdot(\frac{(11cos(z + 2)e^{(6)})}{112}-\frac{ (11cos(z + 2)e^{(10)})}{112}))dz|_{-9}^{-4}\\&\int_{7}^{10}(\frac{(11cos(z + 2)e^{(6)}ln(\frac{4}{9})\cdot(e^{(4)} - 1))}{112})dz=\frac{(11sin(z + 2)e^{(6)}ln(\frac{4}{9})\cdot(e^{(4)} - 1))}{112}|_{7}^{10}=1634\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #254 : Triple Integrals

Example Question #255 : Triple Integrals

Example Question #121 : Triple Integration In Cartesian Coordinates

Example Question #257 : Triple Integrals

Example Question #121 : Triple Integration In Cartesian Coordinates

Example Question #122 : Triple Integration In Cartesian Coordinates

Example Question #731 : Calculus 3

Example Question #732 : Calculus 3

Example Question #733 : Calculus 3

Example Question #734 : Calculus 3

All Calculus 3 Resources

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