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Calculus 3 : Calculus 3

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Example Questions

Example Question #161 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{-6}^{-3.5}\int_{3}^{6}\int_{-8}^{-6}(\frac{(9cos(x + 2)e^{(2y)})}{(28z)})dxdydz\end{align*}$

Possible Answers:

33577

1678

Correct answer:

Explanation:

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

$\begin{align*}&\int_{-6}^{-3.5}\int_{3}^{6}\int_{-8}^{-6}(\frac{(9cos(x + 2)e^{(2y)})}{(28z)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-6}^{-3.5}\int_{3}^{6}\int_{-8}^{-6}(\frac{(9cos(x + 2)e^{(2y)})}{(28z)})dxdydz=\int_{-6}^{-3.5}\int_{3}^{6}(\frac{(9sin(x + 2)e^{(2y)})}{(28z)})dydz|_{-8}^{-6}\\&\int_{-6}^{-3.5}\int_{3}^{6}(-\frac{(9e^{(2y)}\cdot(sin(4) - sin(6)))}{(28z)})dydz=\int_{-6}^{-3.5}(-\frac{(9e^{(2y)}\cdot(sin(4) - sin(6)))}{(56z)})dz|_{3}^{6}\\&\int_{-6}^{-3.5}(-\frac{(9e^{(6)}\cdot(sin(4) - sin(6))\cdot(e^{(6)} - 1))}{(56z)})dz=ln(z)\cdot(\frac{(9e^{(6)}sin(4))}{56}-\frac{ (9e^{(6)}sin(6))}{56}-\frac{ (9e^{(12)}sin(4))}{56}+\frac{ (9e^{(12)}sin(6))}{56})|_{-6}^{-3.5}=-6714\end{align*}$

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Example Question #771 : Calculus 3

$\begin{align*}&\text{Evaluate the triple integral}\int_{-8}^{-6.5}\int_{-9}^{-5.5}\int_{10}^{13.5}(\frac{(67x^{2})}{(7y^{2})})dxdydz\end{align*}$

Possible Answers:

-98.83

494.2

1977

-494.2

Correct answer:

494.2

Explanation:

$\begin{align*}&\int_{-8}^{-6.5}\int_{-9}^{-5.5}\int_{10}^{13.5}(\frac{(67x^{2})}{(7y^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-8}^{-6.5}\int_{-9}^{-5.5}\int_{10}^{13.5}(\frac{(67x^{2})}{(7y^{2})})dxdydz=\int_{-8}^{-6.5}\int_{-9}^{-5.5}(\frac{(67x^{3})}{(21y^{2})})dydz|_{10}^{13.5}\\&\int_{-8}^{-6.5}\int_{-9}^{-5.5}(\frac{111823}{(24y^{2})})dydz=\int_{-8}^{-6.5}(-\frac{111823}{(24y)})dz|_{-9}^{-5.5}\\&\int_{-8}^{-6.5}(\frac{782761}{2376})dz=\frac{(782761z)}{2376}|_{-8}^{-6.5}=494.2\end{align*}$

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Example Question #163 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{4}^{7}\int_{8}^{12}\int_{9}^{10}(\frac{(17cos(x + 2))}{(4yz)})dxdydz\end{align*}$

Possible Answers:

0.45

0.22

-0.22

-1.79

Correct answer:

0.45

Explanation:

$\begin{align*}&\int_{4}^{7}\int_{8}^{12}\int_{9}^{10}(\frac{(17cos(x + 2))}{(4yz)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{4}^{7}\int_{8}^{12}\int_{9}^{10}(\frac{(17cos(x + 2))}{(4yz)})dxdydz=\int_{4}^{7}\int_{8}^{12}(\frac{(17sin(x + 2))}{(4yz)})dydz|_{9}^{10}\\&\int_{4}^{7}\int_{8}^{12}(-\frac{(17\cdot(sin(11) - sin(12)))}{(4yz)})dydz=\int_{4}^{7}(-\frac{(17ln(y)\cdot(sin(11) - sin(12)))}{(4z)})dz|_{8}^{12}\\&\int_{4}^{7}(-\frac{(17ln(\frac{3}{2})\cdot(sin(11) - sin(12)))}{(4z)})dz=-ln(z)\cdot(\frac{(17ln(\frac{3}{2})sin(11))}{4}-\frac{ (17ln(\frac{3}{2})sin(12))}{4})|_{4}^{7}=0.45\end{align*}$

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Example Question #164 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{-4.5}^{-2.5}\int_{9}^{13}\int_{-5}^{-0.5}(\frac{(4e^{(2x)}e^{(-2z)})}{(33y^{2})})dxdydz\end{align*}$

Possible Answers:

-12.13

1.52

-0.61

3.03

Correct answer:

3.03

Explanation:

$\begin{align*}&\int_{-4.5}^{-2.5}\int_{9}^{13}\int_{-5}^{-0.5}(\frac{(4e^{(2x)}e^{(-2z)})}{(33y^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-4.5}^{-2.5}\int_{9}^{13}\int_{-5}^{-0.5}(\frac{(4e^{(2x)}e^{(-2z)})}{(33y^{2})})dxdydz=\int_{-4.5}^{-2.5}\int_{9}^{13}(\frac{(2e^{(2x - 2z)})}{(33y^{2})})dydz|_{-5}^{-0.5}\\&\int_{-4.5}^{-2.5}\int_{9}^{13}(\frac{(2e^{(-10)}e^{(-2z)}\cdot(e^{(9)} - 1))}{(33y^{2})})dydz=\int_{-4.5}^{-2.5}(-\frac{(2e^{(- 2z - 10)}\cdot(e^{(9)} - 1))}{(33y)})dz|_{9}^{13}\\&\int_{-4.5}^{-2.5}(\frac{(8e^{(-10)}e^{(-2z)}\cdot(e^{(9)} - 1))}{3861})dz=-\frac{(4e^{(-10)}e^{(-2z)}\cdot(e^{(9)} - 1))}{3861}|_{-4.5}^{-2.5}=3.03\end{align*}$

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Example Question #165 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{5}^{8}\int_{5}^{6}\int_{6}^{8.5}(\frac{(9\cdot2^{(\frac{z}{2})}cos(4x))}{(32\cdot2^y)})dxdydz\end{align*}$

Possible Answers:

-0.02

0.14

-0.41

0.07

Correct answer:

0.07

Explanation:

$\begin{align*}&\int_{5}^{8}\int_{5}^{6}\int_{6}^{8.5}(\frac{(9\cdot2^{(\frac{z}{2})}cos(4x))}{(32\cdot2^y)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{5}^{8}\int_{5}^{6}\int_{6}^{8.5}(\frac{(9\cdot2^{(\frac{z}{2})}cos(4x))}{(32\cdot2^y)})dxdydz=\int_{5}^{8}\int_{5}^{6}(\frac{(9\cdot2^{(\frac{z}{2}- y)}sin(4x))}{128})dydz|_{6}^{8.5}\\&\int_{5}^{8}\int_{5}^{6}(-2^{(\frac{z}{2}- y)}\cdot(\frac{(9sin(24))}{128}-\frac{ (9sin(34))}{128}))dydz=\int_{5}^{8}(\frac{(9\cdot2^{(\frac{z}{2}- y)}\cdot(sin(24) - sin(34)))}{(128ln(2))})dz|_{5}^{6}\\&\int_{5}^{8}(-\frac{(9\cdot(2^{(\frac{1}{2})})^z\cdot(sin(24) - sin(34)))}{(8192ln(2))})dz=-\frac{(9\cdot2^{(\frac{z}{2})}\cdot(sin(24) - sin(34)))}{(4096ln(2)^{2})}|_{5}^{8}=0.07\end{align*}$

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Example Question #166 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{-9}^{-4.5}\int_{8}^{12.5}\int_{4}^{5.5}(\frac{(61cos(3y)sin(z + 1))}{(2\cdot2^{(\frac{x}{2})})})dxdydz\end{align*}$

Possible Answers:

1.66

-0.55

-3.33

0.55

Correct answer:

1.66

Explanation:

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

$\begin{align*}&\int_{-9}^{-4.5}\int_{8}^{12.5}\int_{4}^{5.5}(\frac{(61cos(3y)sin(z + 1))}{(2\cdot2^{(\frac{x}{2})})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-9}^{-4.5}\int_{8}^{12.5}\int_{4}^{5.5}(\frac{(61cos(3y)sin(z + 1))}{(2\cdot2^{(\frac{x}{2})})})dxdydz=\int_{-9}^{-4.5}\int_{8}^{12.5}(-\frac{(61cos(3y)sin(z + 1))}{(2^{(\frac{x}{2})}ln(2))})dydz|_{4}^{5.5}\\&\int_{-9}^{-4.5}\int_{8}^{12.5}(-\frac{(61cos(3y)sin(z + 1)\cdot(2^{(\frac{1}{4})} - 2))}{(8ln(2))})dydz=\int_{-9}^{-4.5}(-\frac{(61sin(3y)sin(z + 1)\cdot(2^{(\frac{1}{4})} - 2))}{(24ln(2))})dz|_{8}^{12.5}\\&\int_{-9}^{-4.5}(\frac{(61sin(z + 1)\cdot(sin(24) - sin(\frac{75}{2}))\cdot(2^{(\frac{1}{4})} - 2))}{(24ln(2))})dz=-\frac{(61cos(z + 1)\cdot(sin(24) - sin(\frac{75}{2}))\cdot(2^{(\frac{1}{4})} - 2))}{(24ln(2))}|_{-9}^{-4.5}=1.66\end{align*}$

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Example Question #167 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{9}^{10.5}\int_{8}^{11}\int_{6}^{8.5}(5sin(x + 2)sin(y + 2)sin(z + 1))dxdydz\end{align*}$

Possible Answers:

-19.06

7.62

-0.76

3.81

Correct answer:

3.81

Explanation:

$\begin{align*}&\int_{9}^{10.5}\int_{8}^{11}\int_{6}^{8.5}(5sin(x + 2)sin(y + 2)sin(z + 1))dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{9}^{10.5}\int_{8}^{11}\int_{6}^{8.5}(5sin(x + 2)sin(y + 2)sin(z + 1))dxdydz=\int_{9}^{10.5}\int_{8}^{11}(-5cos(x + 2)sin(y + 2)sin(z + 1))dydz|_{6}^{8.5}\\&\int_{9}^{10.5}\int_{8}^{11}(5sin(y + 2)sin(z + 1)\cdot(cos(8) - cos(\frac{21}{2})))dydz=\int_{9}^{10.5}(-5cos(y + 2)sin(z + 1)\cdot(cos(8) - cos(\frac{21}{2})))dz|_{8}^{11}\\&\int_{9}^{10.5}(5sin(z + 1)\cdot(cos(10) - cos(13))\cdot(cos(8) - cos(\frac{21}{2})))dz=-5cos(z + 1)\cdot(cos(10) - cos(13))\cdot(cos(8) - cos(\frac{21}{2}))|_{9}^{10.5}=3.81\end{align*}$

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Example Question #771 : Calculus 3

$\begin{align*}&\text{Evaluate the triple integral}\int_{-6}^{-5}\int_{-10}^{-6}\int_{9}^{14}(\frac{(9x^{2})}{(160yz^{2})})dxdydz\end{align*}$

Possible Answers:

0.64

0.13

-0.13

-0.64

Correct answer:

-0.64

Explanation:

$\begin{align*}&\int_{-6}^{-5}\int_{-10}^{-6}\int_{9}^{14}(\frac{(9x^{2})}{(160yz^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-6}^{-5}\int_{-10}^{-6}\int_{9}^{14}(\frac{(9x^{2})}{(160yz^{2})})dxdydz=\int_{-6}^{-5}\int_{-10}^{-6}(\frac{(3x^{3})}{(160yz^{2})})dydz|_{9}^{14}\\&\int_{-6}^{-5}\int_{-10}^{-6}(\frac{1209}{(32yz^{2})})dydz=\int_{-6}^{-5}(\frac{(1209ln(y))}{(32z^{2})})dz|_{-10}^{-6}\\&\int_{-6}^{-5}(\frac{(1209ln(\frac{3}{5}))}{(32z^{2})})dz=-\frac{(1209ln(\frac{3}{5}))}{(32z)}|_{-6}^{-5}=-0.64\end{align*}$

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Example Question #772 : Calculus 3

$\begin{align*}&\text{Evaluate the triple integral}\int_{10}^{15}\int_{7}^{9.5}\int_{-7}^{-2.5}(\frac{(11cos(x + 2)sin(y + 1))}{(76z)})dxdydz\end{align*}$

Possible Answers:

-0.01

-0.03

0.08

0.13

Correct answer:

-0.03

Explanation:

$\begin{align*}&\int_{10}^{15}\int_{7}^{9.5}\int_{-7}^{-2.5}(\frac{(11cos(x + 2)sin(y + 1))}{(76z)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{10}^{15}\int_{7}^{9.5}\int_{-7}^{-2.5}(\frac{(11cos(x + 2)sin(y + 1))}{(76z)})dxdydz=\int_{10}^{15}\int_{7}^{9.5}(\frac{(11sin(x + 2)sin(y + 1))}{(76z)})dydz|_{-7}^{-2.5}\\&\int_{10}^{15}\int_{7}^{9.5}(-\frac{(11sin(y + 1)\cdot(sin(\frac{1}{2}) - sin(5)))}{(76z)})dydz=\int_{10}^{15}(\frac{(11cos(y + 1)\cdot(sin(\frac{1}{2}) - sin(5)))}{(76z)})dz|_{7}^{9.5}\\&\int_{10}^{15}(-\frac{(11\cdot(cos(8) - cos(\frac{21}{2}))\cdot(sin(\frac{1}{2}) - sin(5)))}{(76z)})dz=-ln(z)\cdot(\frac{(11cos(8)sin(\frac{1}{2}))}{76}-\frac{ (11cos(8)sin(5))}{76}-\frac{ (11cos(\frac{21}{2})sin(\frac{1}{2}))}{76}+\frac{ (11cos(\frac{21}{2})sin(5))}{76})|_{10}^{15}=-0.03\end{align*}$

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Example Question #773 : Calculus 3

$\begin{align*}&\text{Evaluate the triple integral}\int_{4}^{5.5}\int_{-9}^{-5.5}\int_{10}^{13}(\frac{(3\cdot3^z)}{(65x^{2}y^{2})})dxdydz\end{align*}$

Possible Answers:

0.02

-0.07

-0.01

0.01

Correct answer:

0.02

Explanation:

$\begin{align*}&\int_{4}^{5.5}\int_{-9}^{-5.5}\int_{10}^{13}(\frac{(3\cdot3^z)}{(65x^{2}y^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{4}^{5.5}\int_{-9}^{-5.5}\int_{10}^{13}(\frac{(3\cdot3^z)}{(65x^{2}y^{2})})dxdydz=\int_{4}^{5.5}\int_{-9}^{-5.5}(-\frac{(3\cdot3^z)}{(65xy^{2})})dydz|_{10}^{13}\\&\int_{4}^{5.5}\int_{-9}^{-5.5}(\frac{(9\cdot3^z)}{(8450y^{2})})dydz=\int_{4}^{5.5}(-\frac{(9\cdot3^z)}{(8450y)})dz|_{-9}^{-5.5}\\&\int_{4}^{5.5}(\frac{(7\cdot3^z)}{92950})dz=\frac{(7\cdot3^z)}{(92950ln(3))}|_{4}^{5.5}=0.02\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #161 : Triple Integration In Cartesian Coordinates

Example Question #771 : Calculus 3

Example Question #163 : Triple Integration In Cartesian Coordinates

Example Question #164 : Triple Integration In Cartesian Coordinates

Example Question #165 : Triple Integration In Cartesian Coordinates

Example Question #166 : Triple Integration In Cartesian Coordinates

Example Question #167 : Triple Integration In Cartesian Coordinates

Example Question #771 : Calculus 3

Example Question #772 : Calculus 3

Example Question #773 : Calculus 3

All Calculus 3 Resources

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