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Calculus 3 : Calculus 3

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Example Questions

Example Question #131 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{7}^{9}\int_{-9}^{-4}\int_{9}^{14}(\frac{(15\cdot2^ycos(z + 1))}{(2x)})dxdydz\end{align*}$

Possible Answers:

-2.22

0.11

-0.44

1.33

Correct answer:

-0.44

Explanation:

$\begin{align*}&\int_{7}^{9}\int_{-9}^{-4}\int_{9}^{14}(\frac{(15\cdot2^ycos(z + 1))}{(2x)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{7}^{9}\int_{-9}^{-4}\int_{9}^{14}(\frac{(15\cdot2^ycos(z + 1))}{(2x)})dxdydz=\int_{7}^{9}\int_{-9}^{-4}(\frac{(15\cdot2^ycos(z + 1)ln(x))}{2})dydz|_{9}^{14}\\&\int_{7}^{9}\int_{-9}^{-4}(\frac{(15\cdot2^ycos(z + 1)ln(\frac{14}{9}))}{2})dydz=\int_{7}^{9}(\frac{(15\cdot2^ycos(z + 1)ln(\frac{14}{9}))}{(2ln(2))})dz|_{-9}^{-4}\\&\int_{7}^{9}(\frac{(465cos(z + 1)ln(\frac{14}{9}))}{(1024ln(2))})dz=\frac{(465sin(z + 1)ln(\frac{14}{9}))}{(1024ln(2))}|_{7}^{9}=-0.44\end{align*}$

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Example Question #741 : Calculus 3

$\begin{align*}&\text{Evaluate the triple integral}\int_{8}^{10}\int_{-9}^{-4}\int_{7}^{11}(\frac{(5\cdot3^{(\frac{x}{4})}\cdot3^{(\frac{y}{2})}cos(z + 2))}{24})dxdydz\end{align*}$

Possible Answers:

-0.03

273.5

0.57

0.01

Correct answer:

0.01

Explanation:

$\begin{align*}&\text{Performing a triple integral, the order of integration does}\\&\text{not entirely matter.}\\&\text{For example:}\\&\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx\\&\text{Considering our integral, and keeping savvy about }\\&\text{utilizing integral rules:}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\end{align*}$ \ $\begin{align*}&\int_{8}^{10}\int_{-9}^{-4}\int_{7}^{11}(\frac{(5\cdot3^{(\frac{x}{4})}\cdot3^{(\frac{y}{2})}cos(z + 2))}{24})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{8}^{10}\int_{-9}^{-4}\int_{7}^{11}(\frac{(5\cdot3^{(\frac{x}{4})}\cdot3^{(\frac{y}{2})}cos(z + 2))}{24})dxdydz=\int_{8}^{10}\int_{-9}^{-4}(\frac{(5\cdot3^{(\frac{x}{4}+\frac{ y}{2})}cos(z + 2))}{(6ln(3))})dydz|_{7}^{11}\\&\int_{8}^{10}\int_{-9}^{-4}(\frac{(5\cdot3^{(\frac{3}{4})}\cdot3^{(\frac{y}{2})}cos(z + 2))}{ln(3)})dydz=\int_{8}^{10}(\frac{(10\cdot3^{(\frac{y}{2})}\cdot27^{(\frac{1}{4})}cos(z + 2))}{ln(3)^{2}})dz|_{-9}^{-4}\\&\int_{8}^{10}(-\frac{(5\cdot3^{(\frac{3}{4})}cos(z + 2)\cdot(2\cdot3^{(\frac{1}{2})} - 54))}{(243ln(3)^{2})})dz=-\frac{(5\cdot3^{(\frac{3}{4})}sin(z + 2)\cdot(2\cdot3^{(\frac{1}{2})} - 54))}{(243ln(3)^{2})}|_{8}^{10}=0.01\end{align*}$

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Example Question #261 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{7}^{9.5}\int_{9}^{14}\int_{-5}^{-1}(\frac{(3\cdot4^zcos(y + 1)e^{(-x)})}{68})dxdydz\end{align*}$

Possible Answers:

468800

$-1.688\cdot10^{7}$

$2.813\cdot10^{6}$

Correct answer:

$2.813\cdot10^{6}$

Explanation:

$\begin{align*}&\int_{7}^{9.5}\int_{9}^{14}\int_{-5}^{-1}(\frac{(3\cdot4^zcos(y + 1)e^{(-x)})}{68})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{7}^{9.5}\int_{9}^{14}\int_{-5}^{-1}(\frac{(3\cdot4^zcos(y + 1)e^{(-x)})}{68})dxdydz=\int_{7}^{9.5}\int_{9}^{14}(-\frac{(3\cdot2^{(2z)}cos(y + 1)e^{(-x)})}{68})dydz|_{-5}^{-1}\\&\int_{7}^{9.5}\int_{9}^{14}(-\frac{(3\cdot2^{(2z)}cos(y + 1)\cdot(e^{(1)} - e^{(5)}))}{68})dydz=\int_{7}^{9.5}(-\frac{(3\cdot2^{(2z)}sin(y + 1)\cdot(e^{(1)} - e^{(5)}))}{68})dz|_{9}^{14}\\&\int_{7}^{9.5}(-\frac{(3\cdot2^{(2z)}e^{(1)}\cdot(sin(10) - sin(15))\cdot(e^{(4)} - 1))}{68})dz=\frac{(3\cdot2^{(2z)}\cdot(e^{(1)}sin(10) - e^{(5)}sin(10) - e^{(1)}sin(15) + e^{(5)}sin(15)))}{(136ln(2))}|_{7}^{9.5}=2.813\cdot10^{6}\end{align*}$

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Example Question #741 : Calculus 3

$\begin{align*}&\text{Evaluate the triple integral}\int_{5}^{9}\int_{-3.5}^{-2.5}\int_{7}^{12}(\frac{(39x^{2})}{(32\cdot2^yz)})dxdydz\end{align*}$

Possible Answers:

-674.8

10800

2699

Correct answer:

2699

Explanation:

$\begin{align*}&\int_{5}^{9}\int_{-3.5}^{-2.5}\int_{7}^{12}(\frac{(39x^{2})}{(32\cdot2^yz)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{5}^{9}\int_{-3.5}^{-2.5}\int_{7}^{12}(\frac{(39x^{2})}{(32\cdot2^yz)})dxdydz=\int_{5}^{9}\int_{-3.5}^{-2.5}(\frac{(13x^{3})}{(32\cdot2^yz)})dydz|_{7}^{12}\\&\int_{5}^{9}\int_{-3.5}^{-2.5}(\frac{18005}{(32\cdot2^yz)})dydz=\int_{5}^{9}(-\frac{18005}{(32\cdot2^yzln(2))})dz|_{-3.5}^{-2.5}\\&\int_{5}^{9}(\frac{(18005\cdot2^{(\frac{1}{2})})}{(8zln(2))})dz=\frac{(18005\cdot2^{(\frac{1}{2})}ln(z))}{(8ln(2))}|_{5}^{9}=2699\end{align*}$

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Example Question #133 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{4}^{6.5}\int_{4.5}^{5.5}\int_{-9}^{-7}(\frac{(cos(z + 1)e^{(2y)})}{(3x^{2})})dxdydz\end{align*}$

Possible Answers:

103.9

-86.6

519.6

Correct answer:

519.6

Explanation:

$\begin{align*}&\int_{4}^{6.5}\int_{4.5}^{5.5}\int_{-9}^{-7}(\frac{(cos(z + 1)e^{(2y)})}{(3x^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{4}^{6.5}\int_{4.5}^{5.5}\int_{-9}^{-7}(\frac{(cos(z + 1)e^{(2y)})}{(3x^{2})})dxdydz=\int_{4}^{6.5}\int_{4.5}^{5.5}(-\frac{(cos(z + 1)e^{(2y)})}{(3x)})dydz|_{-9}^{-7}\\&\int_{4}^{6.5}\int_{4.5}^{5.5}(\frac{(2cos(z + 1)e^{(2y)})}{189})dydz=\int_{4}^{6.5}(\frac{(cos(z + 1)e^{(2y)})}{189})dz|_{4.5}^{5.5}\\&\int_{4}^{6.5}(\frac{(cos(z + 1)e^{(9)}\cdot(e^{(2)} - 1))}{189})dz=\frac{(sin(z + 1)e^{(9)}\cdot(e^{(2)} - 1))}{189}|_{4}^{6.5}=519.6\end{align*}$

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Example Question #742 : Calculus 3

$\begin{align*}&\text{Evaluate the triple integral}\int_{-1.5}^{3}\int_{-8}^{-5}\int_{8}^{11.5}(\frac{(z^{2}cos(x + 1)sin(3y))}{68})dxdydz\end{align*}$

Possible Answers:

-0.03

0.17

0.01

Correct answer:

-0.03

Explanation:

$\begin{align*}&\int_{-1.5}^{3}\int_{-8}^{-5}\int_{8}^{11.5}(\frac{(z^{2}cos(x + 1)sin(3y))}{68})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-1.5}^{3}\int_{-8}^{-5}\int_{8}^{11.5}(\frac{(z^{2}cos(x + 1)sin(3y))}{68})dxdydz=\int_{-1.5}^{3}\int_{-8}^{-5}(\frac{(z^{2}sin(x + 1)sin(3y))}{68})dydz|_{8}^{11.5}\\&\int_{-1.5}^{3}\int_{-8}^{-5}(-\frac{(z^{2}sin(3y)\cdot(sin(9) - sin(\frac{25}{2})))}{68})dydz=\int_{-1.5}^{3}(\frac{(z^{2}cos(3y)\cdot(sin(9) - sin(\frac{25}{2})))}{204})dz|_{-8}^{-5}\\&\int_{-1.5}^{3}(\frac{(z^{2}\cdot(cos(15) - cos(24))\cdot(sin(9) - sin(\frac{25}{2})))}{204})dz=z^{3}\cdot(\frac{(cos(15)sin(9))}{612}-\frac{ (cos(24)sin(9))}{612}-\frac{ (cos(15)sin(\frac{25}{2}))}{612}+\frac{ (cos(24)sin(\frac{25}{2}))}{612})|_{-1.5}^{3}=-0.03\end{align*}$

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Example Question #742 : Calculus 3

$\begin{align*}&\text{Evaluate the triple integral}\int_{-4.5}^{2.25}\int_{7}^{11.5}\int_{-6}^{-4.5}(\frac{(17cos(x + 1)e^{(-2z)})}{(10y)})dxdydz\end{align*}$

Possible Answers:

346.6

6238

Correct answer:

Explanation:

$\begin{align*}&\int_{-4.5}^{2.25}\int_{7}^{11.5}\int_{-6}^{-4.5}(\frac{(17cos(x + 1)e^{(-2z)})}{(10y)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-4.5}^{2.25}\int_{7}^{11.5}\int_{-6}^{-4.5}(\frac{(17cos(x + 1)e^{(-2z)})}{(10y)})dxdydz=\int_{-4.5}^{2.25}\int_{7}^{11.5}(\frac{(17sin(x + 1)e^{(-2z)})}{(10y)})dydz|_{-6}^{-4.5}\\&\int_{-4.5}^{2.25}\int_{7}^{11.5}(\frac{(17e^{(-2z)}\cdot(sin(5) - sin(\frac{7}{2})))}{(10y)})dydz=\int_{-4.5}^{2.25}(\frac{(e^{(-2z)}ln(y)\cdot(17sin(5) - 17sin(\frac{7}{2})))}{10})dz|_{7}^{11.5}\\&\int_{-4.5}^{2.25}(\frac{(17e^{(-2z)}ln(\frac{23}{14})\cdot(sin(5) - sin(\frac{7}{2})))}{10})dz=-\frac{(17e^{(-2z)}ln(\frac{23}{14})\cdot(sin(5) - sin(\frac{7}{2})))}{20}|_{-4.5}^{2.25}=-2079\end{align*}$

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Example Question #743 : Calculus 3

$\begin{align*}&\text{Evaluate the triple integral}\int_{-3.5}^{0.5}\int_{7}^{12}\int_{-8}^{-6}(\frac{(66\cdot2^zcos(3y)sin(x + 2))}{7})dxdydz\end{align*}$

Possible Answers:

35.48

-35.48

4.43

-17.74

Correct answer:

-17.74

Explanation:

$\begin{align*}&\int_{-3.5}^{0.5}\int_{7}^{12}\int_{-8}^{-6}(\frac{(66\cdot2^zcos(3y)sin(x + 2))}{7})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-3.5}^{0.5}\int_{7}^{12}\int_{-8}^{-6}(\frac{(66\cdot2^zcos(3y)sin(x + 2))}{7})dxdydz=\int_{-3.5}^{0.5}\int_{7}^{12}(-\frac{(66\cdot2^zcos(x + 2)cos(3y))}{7})dydz|_{-8}^{-6}\\&\int_{-3.5}^{0.5}\int_{7}^{12}(-\frac{(66\cdot2^zcos(3y)\cdot(cos(4) - cos(6)))}{7})dydz=\int_{-3.5}^{0.5}(-\frac{(22\cdot2^zsin(3y)\cdot(cos(4) - cos(6)))}{7})dz|_{7}^{12}\\&\int_{-3.5}^{0.5}(\frac{(22\cdot2^z\cdot(cos(4) - cos(6))\cdot(sin(21) - sin(36)))}{7})dz=-\frac{(22\cdot2^z\cdot(\frac{sin(15)}{2}-\frac{ sin(17)}{2}-\frac{ sin(25)}{2}+\frac{ sin(27)}{2}-\frac{ sin(30)}{2}+\frac{ sin(32)}{2}+\frac{ sin(40)}{2}-\frac{ sin(42)}{2}))}{(7ln(2))}|_{-3.5}^{0.5}=-17.74\end{align*}$

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Example Question #136 : Triple Integration In Cartesian Coordinates

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

Possible Answers:

-0.06

0.35

-0.69

2.07

Correct answer:

0.35

Explanation:

$\begin{align*}&\int_{-6}^{-4}\int_{-3}^{-1}\int_{6}^{8}(\frac{(sin(4z)e^{(-y)})}{(5x)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-6}^{-4}\int_{-3}^{-1}\int_{6}^{8}(\frac{(sin(4z)e^{(-y)})}{(5x)})dxdydz=\int_{-6}^{-4}\int_{-3}^{-1}(\frac{(sin(4z)e^{(-y)}ln(x))}{5})dydz|_{6}^{8}\\&\int_{-6}^{-4}\int_{-3}^{-1}(\frac{(sin(4z)e^{(-y)}ln(\frac{4}{3}))}{5})dydz=\int_{-6}^{-4}(-\frac{(sin(4z)e^{(-y)}ln(\frac{4}{3}))}{5})dz|_{-3}^{-1}\\&\int_{-6}^{-4}(-\frac{(sin(4z)ln(\frac{4}{3})\cdot(e^{(1)} - e^{(3)}))}{5})dz=\frac{(cos(4z)ln(\frac{4}{3})\cdot(e^{(1)} - e^{(3)}))}{20}|_{-6}^{-4}=0.35\end{align*}$

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Example Question #271 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{-2.5}^{-1.5}\int_{-6}^{-2.5}\int_{10}^{12.5}(\frac{(5cos(3x)e^{(-2z)})}{y^{3}})dxdydz\end{align*}$

Possible Answers:

-5.59

27.93

-22.35

1.12

Correct answer:

-5.59

Explanation:

$\begin{align*}&\int_{-2.5}^{-1.5}\int_{-6}^{-2.5}\int_{10}^{12.5}(\frac{(5cos(3x)e^{(-2z)})}{y^{3}})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-2.5}^{-1.5}\int_{-6}^{-2.5}\int_{10}^{12.5}(\frac{(5cos(3x)e^{(-2z)})}{y^{3}})dxdydz=\int_{-2.5}^{-1.5}\int_{-6}^{-2.5}(\frac{(5sin(3x)e^{(-2z)})}{(3y^{3})})dydz|_{10}^{12.5}\\&\int_{-2.5}^{-1.5}\int_{-6}^{-2.5}(-\frac{(5e^{(-2z)}\cdot(sin(30) - sin(\frac{75}{2})))}{(3y^{3})})dydz=\int_{-2.5}^{-1.5}(\frac{(5e^{(-2z)}\cdot(sin(30) - sin(\frac{75}{2})))}{(6y^{2})})dz|_{-6}^{-2.5}\\&\int_{-2.5}^{-1.5}(\frac{(119e^{(-2z)}\cdot(sin(30) - sin(\frac{75}{2})))}{1080})dz=-\frac{(119e^{(-2z)}\cdot(sin(30) - sin(\frac{75}{2})))}{2160}|_{-2.5}^{-1.5}=-5.59\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #131 : Triple Integration In Cartesian Coordinates

Example Question #741 : Calculus 3

Example Question #261 : Triple Integrals

Example Question #741 : Calculus 3

Example Question #133 : Triple Integration In Cartesian Coordinates

Example Question #742 : Calculus 3

Example Question #742 : Calculus 3

Example Question #743 : Calculus 3

Example Question #136 : Triple Integration In Cartesian Coordinates

Example Question #271 : Multiple Integration

All Calculus 3 Resources

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