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Calculus 3 : Triple Integrals

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Example Questions

Example Question #441 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{-9}^{-5}\int_{3}^{7}\int_{-4.5}^{-1.5}(\frac{(5z^{2}e^{(2y)})}{(47\cdot 2^x)})dxdydz\end{align*}$

Possible Answers:

$-1.103\cdot10^{9}$

$3.677\cdot10^{8}$

$2.206\cdot10^{9}$

$-1.839\cdot10^{8}$

Correct answer:

$3.677\cdot10^{8}$

Explanation:

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

$\begin{align*}&\int_{-9}^{-5}\int_{3}^{7}\int_{-4.5}^{-1.5}(\frac{(5z^{2}e^{(2y)})}{(47\cdot 2^x)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-9}^{-5}\int_{3}^{7}\int_{-4.5}^{-1.5}(\frac{(5z^{2}e^{(2y)})}{(47\cdot 2^x)})dxdydz=\int_{-9}^{-5}\int_{3}^{7}(-\frac{(5z^{2}e^{(2y)})}{(47\cdot 2^xln(2))})dydz|_{-4.5}^{-1.5}\\&\int_{-9}^{-5}\int_{3}^{7}(\frac{(70\cdot 2^{(\frac{1}{2})}z^{2}e^{(2y)})}{(47ln(2))})dydz=\int_{-9}^{-5}(\frac{(35\cdot 2^{(\frac{1}{2})}z^{2}e^{(2y)})}{(47ln(2))})dz|_{3}^{7}\\&\int_{-9}^{-5}(\frac{(35\cdot 2^{(\frac{1}{2})}z^{2}e^{(6)}\cdot (e^{(8)} - 1))}{(47ln(2))})dz=-\frac{(z^{3}\cdot (35\cdot 2^{(\frac{1}{2})}e^{(6)} - 35\cdot 2^{(\frac{1}{2})}e^{(14)}))}{(141ln(2))}|_{-9}^{-5}=3.677\cdot10^{8}\end{align*}$

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Example Question #441 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{-3}^{2}\int_{8}^{11.5}\int_{10}^{11}(\frac{(3ye^{(-z)})}{(43x)})dxdydz\end{align*}$

Possible Answers:

-13.58

0.755

4.53

-1.51

Correct answer:

4.53

Explanation:

$\begin{align*}&\int_{-3}^{2}\int_{8}^{11.5}\int_{10}^{11}(\frac{(3ye^{(-z)})}{(43x)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-3}^{2}\int_{8}^{11.5}\int_{10}^{11}(\frac{(3ye^{(-z)})}{(43x)})dxdydz=\int_{-3}^{2}\int_{8}^{11.5}(\frac{(3ye^{(-z)}ln(x))}{43})dydz|_{10}^{11}\\&\int_{-3}^{2}\int_{8}^{11.5}(\frac{(3ye^{(-z)}ln(\frac{11}{10}))}{43})dydz=\int_{-3}^{2}(\frac{(3y^{2}e^{(-z)}ln(\frac{11}{10}))}{86})dz|_{8}^{11.5}\\&\int_{-3}^{2}(\frac{(819e^{(-z)}ln(\frac{11}{10}))}{344})dz=-\frac{(819e^{(-z)}ln(\frac{11}{10}))}{344}|_{-3}^{2}=4.53\end{align*}$

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Example Question #911 : Calculus 3

$\begin{align*}&\text{Evaluate the triple integral}\int_{3}^{7.5}\int_{10}^{15}\int_{8}^{11.5}(\frac{(5sin(x + 1))}{(33yz)})dxdydz\end{align*}$

Possible Answers:

-0.215

0.0215

-0.107

0.322

Correct answer:

-0.107

Explanation:

$\begin{align*}&\int_{3}^{7.5}\int_{10}^{15}\int_{8}^{11.5}(\frac{(5sin(x + 1))}{(33yz)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{3}^{7.5}\int_{10}^{15}\int_{8}^{11.5}(\frac{(5sin(x + 1))}{(33yz)})dxdydz=\int_{3}^{7.5}\int_{10}^{15}(-\frac{(5cos(x + 1))}{(33yz)})dydz|_{8}^{11.5}\\&\int_{3}^{7.5}\int_{10}^{15}(\frac{(5\cdot (cos(9) - cos(\frac{25}{2})))}{(33yz)})dydz=\int_{3}^{7.5}(\frac{(ln(y)\cdot (5cos(9) - 5cos(\frac{25}{2})))}{(33z)})dz|_{10}^{15}\\&\int_{3}^{7.5}(\frac{(5ln(\frac{3}{2})\cdot (cos(9) - cos(\frac{25}{2})))}{(33z)})dz=ln(z)\cdot (\frac{(5cos(9)ln(\frac{3}{2}))}{33}-\frac{ (5cos(\frac{25}{2})ln(\frac{3}{2}))}{33})|_{3}^{7.5}=-0.107\end{align*}$

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Example Question #444 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{-6}^{-3}\int_{10}^{15}\int_{7}^{8}(\frac{(9sin(4z))}{(320xy^{3})})dxdydz\end{align*}$

Possible Answers:

$6.57\cdot10^{-6}$

$3.65\cdot10^{-7 }$

$-2.19\cdot10^{-6}$

$-1.09\cdot10^{-6 }$

Correct answer:

$-1.09\cdot10^{-6 }$

Explanation:

$\begin{align*}&\int_{-6}^{-3}\int_{10}^{15}\int_{7}^{8}(\frac{(9sin(4z))}{(320xy^{3})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-6}^{-3}\int_{10}^{15}\int_{7}^{8}(\frac{(9sin(4z))}{(320xy^{3})})dxdydz=\int_{-6}^{-3}\int_{10}^{15}(\frac{(9sin(4z)ln(x))}{(320y^{3})})dydz|_{7}^{8}\\&\int_{-6}^{-3}\int_{10}^{15}(\frac{(9sin(4z)ln(\frac{8}{7}))}{(320y^{3})})dydz=\int_{-6}^{-3}(-\frac{(9sin(4z)ln(\frac{8}{7}))}{(640y^{2})})dz|_{10}^{15}\\&\int_{-6}^{-3}(\frac{(sin(4z)ln(\frac{8}{7}))}{12800})dz=-\frac{(cos(4z)ln(\frac{8}{7}))}{51200}|_{-6}^{-3}=-1.09\cdot10^{-6}\end{align*}$

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Example Question #441 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{8}^{9.5}\int_{9}^{10.5}\int_{4}^{8}(\frac{(3cos(4y)sin(z + 2))}{(8x)})dxdydz\end{align*}$

Possible Answers:

0.00129

-0.00647

0.0388

-0.00216

Correct answer:

-0.00647

Explanation:

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

$\begin{align*}&\int_{8}^{9.5}\int_{9}^{10.5}\int_{4}^{8}(\frac{(3cos(4y)sin(z + 2))}{(8x)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{8}^{9.5}\int_{9}^{10.5}\int_{4}^{8}(\frac{(3cos(4y)sin(z + 2))}{(8x)})dxdydz=\int_{8}^{9.5}\int_{9}^{10.5}(\frac{(3cos(4y)sin(z + 2)ln(x))}{8})dydz|_{4}^{8}\\&\int_{8}^{9.5}\int_{9}^{10.5}(\frac{(3cos(4y)sin(z + 2)ln(2))}{8})dydz=\int_{8}^{9.5}(\frac{(3sin(4y)sin(z + 2)ln(2))}{32})dz|_{9}^{10.5}\\&\int_{8}^{9.5}(-\frac{(3sin(z + 2)ln(2)\cdot (sin(36) - sin(42)))}{32})dz=\frac{(3cos(z + 2)ln(2)\cdot (sin(36) - sin(42)))}{32}|_{8}^{9.5}=-0.00647\end{align*}$

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Example Question #446 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{-8}^{-3.5}\int_{8}^{13}\int_{-9}^{-8}(\frac{(19sin(4z))}{(8xy)})dxdydz\end{align*}$

Possible Answers:

-0.0237

-0.0474

0.00789

0.071

Correct answer:

-0.0237

Explanation:

$\begin{align*}&\int_{-8}^{-3.5}\int_{8}^{13}\int_{-9}^{-8}(\frac{(19sin(4z))}{(8xy)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-8}^{-3.5}\int_{8}^{13}\int_{-9}^{-8}(\frac{(19sin(4z))}{(8xy)})dxdydz=\int_{-8}^{-3.5}\int_{8}^{13}(\frac{(19sin(4z)ln(x))}{(8y)})dydz|_{-9}^{-8}\\&\int_{-8}^{-3.5}\int_{8}^{13}(\frac{(19sin(4z)ln(\frac{8}{9}))}{(8y)})dydz=\int_{-8}^{-3.5}(\frac{(19sin(4z)ln(\frac{8}{9})ln(y))}{8})dz|_{8}^{13}\\&\int_{-8}^{-3.5}(\frac{(19sin(4z)ln(\frac{8}{9})ln(\frac{13}{8}))}{8})dz=-\frac{(19cos(4z)ln(\frac{8}{9})ln(\frac{13}{8}))}{32}|_{-8}^{-3.5}=-0.0237\end{align*}$

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Example Question #447 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{-6}^{-2}\int_{10}^{11.5}\int_{5}^{6.5}(\frac{(61e^{(-x)})}{(36yz^{2})})dxdydz\end{align*}$

Possible Answers:

-0.000826

-0.000138

$8.26\cdot10^{-5}$

0.000413

Correct answer:

0.000413

Explanation:

$\begin{align*}&\int_{-6}^{-2}\int_{10}^{11.5}\int_{5}^{6.5}(\frac{(61e^{(-x)})}{(36yz^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-6}^{-2}\int_{10}^{11.5}\int_{5}^{6.5}(\frac{(61e^{(-x)})}{(36yz^{2})})dxdydz=\int_{-6}^{-2}\int_{10}^{11.5}(-\frac{(61e^{(-x)})}{(36yz^{2})})dydz|_{5}^{6.5}\\&\int_{-6}^{-2}\int_{10}^{11.5}(\frac{(61\cdot (e^{(-5)} - e^{(-\frac{13}{2})}))}{(36yz^{2})})dydz=\int_{-6}^{-2}(\frac{(61e^{(-\frac{13}{2})}ln(y)\cdot (e^{(\frac{3}{2})} - 1))}{(36z^{2})})dz|_{10}^{11.5}\\&\int_{-6}^{-2}(\frac{(61e^{(-\frac{13}{2})}ln(\frac{23}{20})\cdot (e^{(\frac{3}{2})} - 1))}{(36z^{2})})dz=-\frac{(61e^{(-\frac{13}{2})}ln(\frac{23}{20})\cdot (e^{(\frac{3}{2})} - 1))}{(36z)}|_{-6}^{-2}=0.000413\end{align*}$

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Example Question #442 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{-6}^{-3}\int_{-4.5}^{-1}\int_{6}^{8}(\frac{(2e^{(y)})}{(x^{2}z^{2})})dxdydz\end{align*}$

Possible Answers:

$-9.910\cdot10^{-4}$

$4.955\cdot10^{-3}$

$1.239\cdot10^{-3}$

$-2.973\cdot10^{-2}$

Correct answer:

$4.955\cdot10^{-3}$

Explanation:

$\begin{align*}&\int_{-6}^{-3}\int_{-4.5}^{-1}\int_{6}^{8}(\frac{(2e^{(y)})}{(x^{2}z^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-6}^{-3}\int_{-4.5}^{-1}\int_{6}^{8}(\frac{(2e^{(y)})}{(x^{2}z^{2})})dxdydz=\int_{-6}^{-3}\int_{-4.5}^{-1}(-\frac{(2e^{(y)})}{(xz^{2})})dydz|_{6}^{8}\\&\int_{-6}^{-3}\int_{-4.5}^{-1}(\frac{e^{(y)}}{(12z^{2})})dydz=\int_{-6}^{-3}(\frac{e^{(y)}}{(12z^{2})})dz|_{-4.5}^{-1}\\&\int_{-6}^{-3}(\frac{(e^{(-1)} - e^{(-\frac{9}{2})})}{(12z^{2})})dz=-\frac{(\frac{e^{(-1)}}{12}-\frac{ e^{(-\frac{9}{2})}}{12})}{z}|_{-6}^{-3}=4.955\cdot10^{-3}\end{align*}$

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Example Question #926 : Calculus 3

$\begin{align*}&\text{Evaluate the triple integral}\int_{-10}^{-9}\int_{10}^{12.5}\int_{-9}^{-5.5}(\frac{(19cos(z + 1))}{(16x^{3}y)})dxdydz\end{align*}$

Possible Answers:

$1.584\cdot10^{-3}$

$-3.168\cdot10^{-4}$

$5.280\cdot10^{-4}$

$-9.504\cdot10^{-3}$

Correct answer:

$1.584\cdot10^{-3}$

Explanation:

$\begin{align*}&\int_{-10}^{-9}\int_{10}^{12.5}\int_{-9}^{-5.5}(\frac{(19cos(z + 1))}{(16x^{3}y)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-10}^{-9}\int_{10}^{12.5}\int_{-9}^{-5.5}(\frac{(19cos(z + 1))}{(16x^{3}y)})dxdydz=\int_{-10}^{-9}\int_{10}^{12.5}(-\frac{(19cos(z + 1))}{(32x^{2}y)})dydz|_{-9}^{-5.5}\\&\int_{-10}^{-9}\int_{10}^{12.5}(-\frac{(3857cos(z + 1))}{(313632y)})dydz=\int_{-10}^{-9}(-\frac{(3857cos(z + 1)ln(y))}{313632})dz|_{10}^{12.5}\\&\int_{-10}^{-9}(-\frac{(3857cos(z + 1)ln(\frac{5}{4}))}{313632})dz=-\frac{(3857sin(z + 1)ln(\frac{5}{4}))}{313632}|_{-10}^{-9}=1.584\cdot10^{-3}\end{align*}$

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Example Question #449 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{7}^{10}\int_{9}^{12}\int_{-4}^{-0.5}(\frac{(sin(y + 1)e^{(x)})}{z^{2}})dxdydz\end{align*}$

Possible Answers:

$7.338\cdot10^{-3}$

$-8.806\cdot10^{-2}$

$2.642\cdot10^{-1}$

$-4.403\cdot10^{-2}$

Correct answer:

$-4.403\cdot10^{-2}$

Explanation:

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

$\begin{align*}&\int_{7}^{10}\int_{9}^{12}\int_{-4}^{-0.5}(\frac{(sin(y + 1)e^{(x)})}{z^{2}})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{7}^{10}\int_{9}^{12}\int_{-4}^{-0.5}(\frac{(sin(y + 1)e^{(x)})}{z^{2}})dxdydz=\int_{7}^{10}\int_{9}^{12}(\frac{(sin(y + 1)e^{(x)})}{z^{2}})dydz|_{-4}^{-0.5}\\&\int_{7}^{10}\int_{9}^{12}(\frac{(sin(y + 1)e^{(-4)}\cdot (e^{(\frac{7}{2})} - 1))}{z^{2}})dydz=\int_{7}^{10}(-\frac{(cos(y + 1)e^{(-4)}\cdot (e^{(\frac{7}{2})} - 1))}{z^{2}})dz|_{9}^{12}\\&\int_{7}^{10}(\frac{(e^{(-4)}\cdot (cos(10) - cos(13))\cdot (e^{(\frac{7}{2})} - 1))}{z^{2}})dz=-\frac{(e^{(-4)}\cdot (cos(10) - cos(13))\cdot (e^{(\frac{7}{2})} - 1))}{z}|_{7}^{10}=-4.403\cdot10^{-2}\end{align*}$

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Calculus 3 : Triple Integrals

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #441 : Multiple Integration

Example Question #441 : Multiple Integration

Example Question #911 : Calculus 3

Example Question #444 : Multiple Integration

Example Question #441 : Multiple Integration

Example Question #446 : Multiple Integration

Example Question #447 : Multiple Integration

Example Question #442 : Multiple Integration

Example Question #926 : Calculus 3

Example Question #449 : Multiple Integration

All Calculus 3 Resources

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