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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 #361 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{-8}^{-4}\int_{-7}^{-2.5}\int_{-4}^{-3}(\frac{(3cos(y + 2)cos(z + 1)e^{(-x)})}{47})dxdydz\end{align*}$

Possible Answers:

-3.27

-1.64

3.27

0.33

Correct answer:

-1.64

Explanation:

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

$\begin{align*}&\int_{-8}^{-4}\int_{-7}^{-2.5}\int_{-4}^{-3}(\frac{(3cos(y + 2)cos(z + 1)e^{(-x)})}{47})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-8}^{-4}\int_{-7}^{-2.5}\int_{-4}^{-3}(\frac{(3cos(y + 2)cos(z + 1)e^{(-x)})}{47})dxdydz=\int_{-8}^{-4}\int_{-7}^{-2.5}(-\frac{(3cos(y + 2)cos(z + 1)e^{(-x)})}{47})dydz|_{-4}^{-3}\\&\int_{-8}^{-4}\int_{-7}^{-2.5}(-\frac{(3cos(y + 2)cos(z + 1)\cdot (e^{(3)} - e^{(4)}))}{47})dydz=\int_{-8}^{-4}(-\frac{(3cos(z + 1)sin(y + 2)\cdot (e^{(3)} - e^{(4)}))}{47})dz|_{-7}^{-2.5}\\&\int_{-8}^{-4}(-\frac{(3cos(z + 1)e^{(3)}\cdot (sin(\frac{1}{2}) - sin(5))\cdot (e^{(1)} - 1))}{47})dz=-\frac{(3sin(z + 1)e^{(3)}\cdot (sin(\frac{1}{2}) - sin(5))\cdot (e^{(1)} - 1))}{47}|_{-8}^{-4}=-1.64\end{align*}$

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Example Question #362 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{8}^{12}\int_{3}^{6}\int_{-6}^{-1.5}(\frac{(7z^{2}cos(3x)e^{(2y)})}{68})dxdydz\end{align*}$

Possible Answers:

$-1.279\cdot10^{6}$

511500

255800

Correct answer:

255800

Explanation:

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

$\begin{align*}&\int_{8}^{12}\int_{3}^{6}\int_{-6}^{-1.5}(\frac{(7z^{2}cos(3x)e^{(2y)})}{68})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{8}^{12}\int_{3}^{6}\int_{-6}^{-1.5}(\frac{(7z^{2}cos(3x)e^{(2y)})}{68})dxdydz=\int_{8}^{12}\int_{3}^{6}(\frac{(7z^{2}sin(3x)e^{(2y)})}{204})dydz|_{-6}^{-1.5}\\&\int_{8}^{12}\int_{3}^{6}(-\frac{(7z^{2}e^{(2y)}\cdot (\frac{sin(\frac{9}{2})}{3}-\frac{ sin(18)}{3}))}{68})dydz=\int_{8}^{12}(-\frac{(7z^{2}e^{(2y)}\cdot (\frac{sin(\frac{9}{2})}{3}-\frac{ sin(18)}{3}))}{136})dz|_{3}^{6}\\&\int_{8}^{12}(-\frac{(7z^{2}e^{(6)}\cdot (sin(\frac{9}{2}) - sin(18))\cdot (e^{(6)} - 1))}{408})dz=z^{3}\cdot (\frac{(7e^{(6)}sin(\frac{9}{2}))}{1224}-\frac{ (7e^{(12)}sin(\frac{9}{2}))}{1224}-\frac{ (7e^{(6)}sin(18))}{1224}+\frac{ (7e^{(12)}sin(18))}{1224})|_{8}^{12}=255800\end{align*}$

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Example Question #363 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{-7}^{-2.5}\int_{-5}^{-1}\int_{-8}^{-6.5}(\frac{12}{(31\cdot 3^yz)})dxdydz\end{align*}$

Possible Answers:

-130.6

32.65

653

-522.4

Correct answer:

-130.6

Explanation:

$\begin{align*}&\int_{-7}^{-2.5}\int_{-5}^{-1}\int_{-8}^{-6.5}(\frac{12}{(31\cdot 3^yz)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-7}^{-2.5}\int_{-5}^{-1}\int_{-8}^{-6.5}(\frac{12}{(31\cdot 3^yz)})dxdydz=\int_{-7}^{-2.5}\int_{-5}^{-1}(\frac{(12x)}{(31\cdot 3^yz)})dydz|_{-8}^{-6.5}\\&\int_{-7}^{-2.5}\int_{-5}^{-1}(\frac{18}{(31\cdot 3^yz)})dydz=\int_{-7}^{-2.5}(-\frac{18}{(31\cdot 3^yzln(3))})dz|_{-5}^{-1}\\&\int_{-7}^{-2.5}(\frac{4320}{(31zln(3))})dz=\frac{(4320ln(z))}{(31ln(3))}|_{-7}^{-2.5}=-130.6\end{align*}$

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Example Question #364 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{3.5}^{6}\int_{-7}^{-2.5}\int_{8}^{10}(\frac{(23e^{(z)})}{(3x^{2}y^{2})})dxdydz\end{align*}$

Possible Answers:

18.25

-9.13

-18.25

4.56

Correct answer:

18.25

Explanation:

$\begin{align*}&\int_{3.5}^{6}\int_{-7}^{-2.5}\int_{8}^{10}(\frac{(23e^{(z)})}{(3x^{2}y^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{3.5}^{6}\int_{-7}^{-2.5}\int_{8}^{10}(\frac{(23e^{(z)})}{(3x^{2}y^{2})})dxdydz=\int_{3.5}^{6}\int_{-7}^{-2.5}(-\frac{(23e^{(z)})}{(3xy^{2})})dydz|_{8}^{10}\\&\int_{3.5}^{6}\int_{-7}^{-2.5}(\frac{(23e^{(z)})}{(120y^{2})})dydz=\int_{3.5}^{6}(-\frac{(23e^{(z)})}{(120y)})dz|_{-7}^{-2.5}\\&\int_{3.5}^{6}(\frac{(69e^{(z)})}{1400})dz=\frac{(69e^{(z)})}{1400}|_{3.5}^{6}=18.25\end{align*}$

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Example Question #361 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{-5}^{-1.5}\int_{-5}^{-4}\int_{8}^{12.5}(\frac{(19sin(4x)e^{(-z)}e^{(y)})}{4})dxdydz\end{align*}$

Possible Answers:

0.09

-0.06

-0.26

0.26

Correct answer:

-0.26

Explanation:

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

$\begin{align*}&\int_{-5}^{-1.5}\int_{-5}^{-4}\int_{8}^{12.5}(\frac{(19sin(4x)e^{(-z)}e^{(y)})}{4})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-5}^{-1.5}\int_{-5}^{-4}\int_{8}^{12.5}(\frac{(19sin(4x)e^{(-z)}e^{(y)})}{4})dxdydz=\int_{-5}^{-1.5}\int_{-5}^{-4}(-\frac{(19cos(4x)e^{(y - z)})}{16})dydz|_{8}^{12.5}\\&\int_{-5}^{-1.5}\int_{-5}^{-4}(e^{(y - z)}\cdot (\frac{(19cos(32))}{16}-\frac{ (19cos(50))}{16}))dydz=\int_{-5}^{-1.5}(\frac{(19e^{(y - z)}\cdot (cos(32) - cos(50)))}{16})dz|_{-5}^{-4}\\&\int_{-5}^{-1.5}(\frac{(19e^{(- z - 5)}\cdot (cos(32) - cos(50))\cdot (e^{(1)} - 1))}{16})dz=-\frac{(19e^{(- z - 5)}\cdot (cos(32) - cos(50))\cdot (e^{(1)} - 1))}{16}|_{-5}^{-1.5}=-0.26\end{align*}$

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Example Question #365 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{-5}^{-3.5}\int_{10}^{11.5}\int_{3.5}^{8.5}(\frac{(3^{(\frac{x}{3})}\cdot 3^zsin(y + 2))}{8})dxdydz\end{align*}$

Possible Answers:

0.03

-0.01

0.15

-0.15

Correct answer:

0.03

Explanation:

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

$\begin{align*}&\int_{-5}^{-3.5}\int_{10}^{11.5}\int_{3.5}^{8.5}(\frac{(3^{(\frac{x}{3})}\cdot 3^zsin(y + 2))}{8})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-5}^{-3.5}\int_{10}^{11.5}\int_{3.5}^{8.5}(\frac{(3^{(\frac{x}{3})}\cdot 3^zsin(y + 2))}{8})dxdydz=\int_{-5}^{-3.5}\int_{10}^{11.5}(\frac{(3\cdot 3^{(\frac{x}{3}+ z)}sin(y + 2))}{(8ln(3))})dydz|_{3.5}^{8.5}\end{align*}$

$\begin{align*}\\&\int_{-5}^{-3.5}\int_{10}^{11.5}(-\frac{(3^zsin(y + 2)\cdot (9\cdot 3^{(\frac{1}{6})} - 27\cdot 3^{(\frac{5}{6})}))}{(8ln(3))})dydz=\int_{-5}^{-3.5}(\frac{(3^zcos(y + 2)\cdot (9\cdot 3^{(\frac{1}{6})} - 27\cdot 3^{(\frac{5}{6})}))}{(8ln(3))})dz|_{10}^{11.5}\\&\int_{-5}^{-3.5}(\frac{(9\cdot 3^{(\frac{1}{6})}\cdot 3^z\cdot (3\cdot 3^{(\frac{2}{3})} - 1)\cdot (cos(12) - cos(\frac{27}{2})))}{(8ln(3))})dz=-\frac{(9\cdot 3^z\cdot (3^{(\frac{1}{6})}cos(12) - 3^{(\frac{1}{6})}cos(\frac{27}{2}) - 3\cdot 243^{(\frac{1}{6})}cos(12) + 3\cdot 243^{(\frac{1}{6})}cos(\frac{27}{2})))}{(8ln(3)^{2})}|_{-5}^{-3.5}=0.03\end{align*}$

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Example Question #366 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{-7}^{-3}\int_{8}^{12}\int_{10}^{13.5}(\frac{(10cos(4y)sin(x + 2))}{(9z^{3})})dxdydz\end{align*}$

Possible Answers:

0.09

-0.15

0.03

-0.01

Correct answer:

0.03

Explanation:

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

$\begin{align*}&\int_{-7}^{-3}\int_{8}^{12}\int_{10}^{13.5}(\frac{(10cos(4y)sin(x + 2))}{(9z^{3})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-7}^{-3}\int_{8}^{12}\int_{10}^{13.5}(\frac{(10cos(4y)sin(x + 2))}{(9z^{3})})dxdydz=\int_{-7}^{-3}\int_{8}^{12}(-\frac{(10cos(x + 2)cos(4y))}{(9z^{3})})dydz|_{10}^{13.5}\\&\int_{-7}^{-3}\int_{8}^{12}(\frac{(10cos(4y)\cdot (cos(12) - cos(\frac{31}{2})))}{(9z^{3})})dydz=\int_{-7}^{-3}(\frac{(5sin(4y)\cdot (cos(12) - cos(\frac{31}{2})))}{(18z^{3})})dz|_{8}^{12}\\&\int_{-7}^{-3}(-\frac{(5\cdot (cos(12) - cos(\frac{31}{2}))\cdot (sin(32) - sin(48)))}{(18z^{3})})dz=\frac{(\frac{(5cos(12)sin(32))}{36}-\frac{ (5cos(12)sin(48))}{36}-\frac{ (5cos(\frac{31}{2})sin(32))}{36}+\frac{ (5cos(\frac{31}{2})sin(48))}{36})}{z^{2}}|_{-7}^{-3}=0.03\end{align*}$

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Example Question #371 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{-7}^{-5.5}\int_{-10}^{-7.5}\int_{7}^{11.5}(\frac{(27cos(4y)sin(z + 1))}{(8x^{2})})dxdydz\end{align*}$

Possible Answers:

-0.02

0.1

-0.29

0.48

Correct answer:

0.1

Explanation:

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

$\begin{align*}&\int_{-7}^{-5.5}\int_{-10}^{-7.5}\int_{7}^{11.5}(\frac{(27cos(4y)sin(z + 1))}{(8x^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-7}^{-5.5}\int_{-10}^{-7.5}\int_{7}^{11.5}(\frac{(27cos(4y)sin(z + 1))}{(8x^{2})})dxdydz=\int_{-7}^{-5.5}\int_{-10}^{-7.5}(-\frac{(27cos(4y)sin(z + 1))}{(8x)})dydz|_{7}^{11.5}\\&\int_{-7}^{-5.5}\int_{-10}^{-7.5}(\frac{(243cos(4y)sin(z + 1))}{1288})dydz=\int_{-7}^{-5.5}(\frac{(243sin(4y)sin(z + 1))}{5152})dz|_{-10}^{-7.5}\\&\int_{-7}^{-5.5}(-\frac{(243sin(z + 1)\cdot (\frac{sin(30)}{4}-\frac{ sin(40)}{4}))}{1288})dz=cos(z + 1)\cdot (\frac{(243sin(30))}{5152}-\frac{ (243sin(40))}{5152})|_{-7}^{-5.5}=0.1\end{align*}$

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Example Question #372 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{7}^{12}\int_{-4}^{-2.5}\int_{9}^{10}(\frac{(8\cdot 2^y)}{(19z)})dxdydz\end{align*}$

Possible Answers:

0.01

-0.01

0.04

-0.11

Correct answer:

0.04

Explanation:

$\begin{align*}&\int_{7}^{12}\int_{-4}^{-2.5}\int_{9}^{10}(\frac{(8\cdot 2^y)}{(19z)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{7}^{12}\int_{-4}^{-2.5}\int_{9}^{10}(\frac{(8\cdot 2^y)}{(19z)})dxdydz=\int_{7}^{12}\int_{-4}^{-2.5}(\frac{(8\cdot 2^yx)}{(19z)})dydz|_{9}^{10}\\&\int_{7}^{12}\int_{-4}^{-2.5}(\frac{(8\cdot 2^y)}{(19z)})dydz=\int_{7}^{12}(\frac{(8\cdot 2^y)}{(19zln(2))})dz|_{-4}^{-2.5}\\&\int_{7}^{12}(\frac{(2\cdot 2^{(\frac{1}{2})} - 1)}{(38zln(2))})dz=\frac{(ln(z)\cdot (2\cdot 2^{(\frac{1}{2})} - 1))}{(38ln(2))}|_{7}^{12}=0.04\end{align*}$

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Example Question #371 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{-3.5}^{-1.5}\int_{-3.5}^{-0.5}\int_{-9}^{-8}(\frac{(4\cdot 3^ysin(x + 1)e^{(-z)})}{3})dxdydz\end{align*}$

Possible Answers:

-17.38

-86.89

8.69

17.38

Correct answer:

-17.38

Explanation:

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

$\begin{align*}&\int_{-3.5}^{-1.5}\int_{-3.5}^{-0.5}\int_{-9}^{-8}(\frac{(4\cdot 3^ysin(x + 1)e^{(-z)})}{3})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-3.5}^{-1.5}\int_{-3.5}^{-0.5}\int_{-9}^{-8}(\frac{(4\cdot 3^ysin(x + 1)e^{(-z)})}{3})dxdydz=\int_{-3.5}^{-1.5}\int_{-3.5}^{-0.5}(-\frac{(4\cdot 3^ycos(x + 1)e^{(-z)})}{3})dydz|_{-9}^{-8}\\&\int_{-3.5}^{-1.5}\int_{-3.5}^{-0.5}(-\frac{(4\cdot 3^ye^{(-z)}\cdot (cos(7) - cos(8)))}{3})dydz=\int_{-3.5}^{-1.5}(-\frac{(4\cdot 3^ye^{(-z)}\cdot (cos(7) - cos(8)))}{(3ln(3))})dz|_{-3.5}^{-0.5}\\&\int_{-3.5}^{-1.5}(-\frac{(104\cdot 3^{(\frac{1}{2})}cos(7)e^{(-z)} - 104\cdot 3^{(\frac{1}{2})}cos(8)e^{(-z)})}{(243ln(3))})dz=\frac{(e^{(-z)}\cdot (104\cdot 3^{(\frac{1}{2})}cos(7) - 104\cdot 3^{(\frac{1}{2})}cos(8)))}{(243ln(3))}|_{-3.5}^{-1.5}=-17.38\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #361 : Triple Integrals

Example Question #362 : Triple Integrals

Example Question #363 : Triple Integrals

Example Question #364 : Triple Integrals

Example Question #361 : Triple Integrals

Example Question #365 : Triple Integrals

Example Question #366 : Triple Integrals

Example Question #371 : Multiple Integration

Example Question #372 : Multiple Integration

Example Question #371 : Triple Integrals

All Calculus 3 Resources

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