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Calculus 3 : Triple Integration in Cartesian Coordinates

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Example Question #311 : Triple Integration In Cartesian Coordinates

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

Possible Answers:

$6.57\cdot10^{-6}$

$-1.09\cdot10^{-6 }$

$-2.19\cdot10^{-6}$

$3.65\cdot10^{-7 }$

Correct answer:

$-1.09\cdot10^{-6 }$

Explanation:

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

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

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Example Question #311 : Triple Integration In Cartesian Coordinates

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

Possible Answers:

-0.00647

-0.00216

0.0388

0.00129

Correct answer:

-0.00647

Explanation:

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

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Example Question #311 : Triple Integration In Cartesian Coordinates

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

Possible Answers:

-0.0474

0.071

0.00789

-0.0237

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 #311 : Triple Integration In Cartesian Coordinates

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

Possible Answers:

$8.26\cdot10^{-5}$

0.000413

-0.000826

-0.000138

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 #445 : Triple Integrals

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

Possible Answers:

$-9.910\cdot10^{-4}$

$1.239\cdot10^{-3}$

$4.955\cdot10^{-3}$

$-2.973\cdot10^{-2}$

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 #446 : Triple Integrals

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

Possible Answers:

$-9.504\cdot10^{-3}$

$1.584\cdot10^{-3}$

$-3.168\cdot10^{-4}$

$5.280\cdot10^{-4}$

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 #922 : Calculus 3

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

Possible Answers:

$-8.806\cdot10^{-2}$

$7.338\cdot10^{-3}$

$2.642\cdot10^{-1}$

$-4.403\cdot10^{-2}$

Correct answer:

$-4.403\cdot10^{-2}$

Explanation:

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

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Example Question #311 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{-3.5}^{0.5}\int_{-4.5}^{2.25}\int_{-3}^{1.5}(\frac{(2^y\cdot 3^{(\frac{z}{2})})}{2^x})dxdydz\end{align*}$

Possible Answers:

-39.93

79.86

159.73

-319.46

Correct answer:

159.73

Explanation:

$\begin{align*}&\int_{-3.5}^{0.5}\int_{-4.5}^{2.25}\int_{-3}^{1.5}(\frac{(2^y\cdot 3^{(\frac{z}{2})})}{2^x})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-3.5}^{0.5}\int_{-4.5}^{2.25}\int_{-3}^{1.5}(\frac{(2^y\cdot 3^{(\frac{z}{2})})}{2^x})dxdydz=\int_{-3.5}^{0.5}\int_{-4.5}^{2.25}(-\frac{(2^{(y - x)}\cdot 3^{(\frac{z}{2})})}{ln(2)})dydz|_{-3}^{1.5}\\&\int_{-3.5}^{0.5}\int_{-4.5}^{2.25}(-\frac{(2^y\cdot 3^{(\frac{z}{2})}\cdot (2^{(\frac{1}{2})} - 32))}{(4ln(2))})dydz=\int_{-3.5}^{0.5}(\frac{(2^y\cdot (8\cdot 3^{(\frac{z}{2})} -\frac{ (2^{(\frac{1}{2})}\cdot 3^{(\frac{z}{2})})}{4}))}{ln(2)^{2}})dz|_{-4.5}^{2.25}\\&\int_{-3.5}^{0.5}(-\frac{(3^{(\frac{z}{2})}\cdot (\frac{2^{(\frac{1}{2})}}{4}- 32\cdot 2^{(\frac{1}{4})} + 2^{(\frac{3}{4})} -\frac{ 1}{64}))}{ln(2)^{2}})dz=-\frac{(2\cdot 3^{(\frac{z}{2})}\cdot (\frac{2^{(\frac{1}{2})}}{4}- 32\cdot 2^{(\frac{1}{4})} + 2^{(\frac{3}{4})} -\frac{ 1}{64}))}{(ln(2)^{2}ln(3))}|_{-3.5}^{0.5}=159.73\end{align*}$

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Example Question #452 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{6}^{9.5}\int_{-3.5}^{1.5}\int_{10}^{14}(\frac{(17\cdot 2^{(\frac{y}{2})})}{(3xz^{2})})dxdydz\end{align*}$

Possible Answers:

$-1.17\cdot10^{-1}$

$4.68\cdot10^{-1}$

-2.81

1.4 0

Correct answer:

$4.68\cdot10^{-1}$

Explanation:

$\begin{align*}&\int_{6}^{9.5}\int_{-3.5}^{1.5}\int_{10}^{14}(\frac{(17\cdot 2^{(\frac{y}{2})})}{(3xz^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{6}^{9.5}\int_{-3.5}^{1.5}\int_{10}^{14}(\frac{(17\cdot 2^{(\frac{y}{2})})}{(3xz^{2})})dxdydz=\int_{6}^{9.5}\int_{-3.5}^{1.5}(\frac{(17\cdot 2^{(\frac{y}{2})}ln(x))}{(3z^{2})})dydz|_{10}^{14}\\&\int_{6}^{9.5}\int_{-3.5}^{1.5}(\frac{(17\cdot 2^{(\frac{y}{2})}ln(\frac{7}{5}))}{(3z^{2})})dydz=\int_{6}^{9.5}(\frac{(34\cdot 2^{(\frac{y}{2})}ln(\frac{7}{5}))}{(3z^{2}ln(2))})dz|_{-3.5}^{1.5}\\&\int_{6}^{9.5}(-\frac{(17ln(\frac{7}{5})\cdot (2^{(\frac{1}{4})} - 4\cdot 2^{(\frac{3}{4})}))}{(6z^{2}ln(2))})dz=\frac{(17\cdot 2^{(\frac{1}{4})}ln(\frac{7}{5}) - 68\cdot 8^{(\frac{1}{4})}ln(\frac{7}{5}))}{(6zln(2))}|_{6}^{9.5}=4.68\cdot10^{-1}\end{align*}$

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Example Question #453 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{3.5}^{8.5}\int_{-8}^{-3}\int_{-10}^{-7.5}(\frac{(3^zcos(4x)cos(4y))}{7})dxdydz\end{align*}$

Possible Answers:

-43.36

57.81

-346.87

173.44

Correct answer:

173.44

Explanation:

$\begin{align*}&\int_{3.5}^{8.5}\int_{-8}^{-3}\int_{-10}^{-7.5}(\frac{(3^zcos(4x)cos(4y))}{7})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{3.5}^{8.5}\int_{-8}^{-3}\int_{-10}^{-7.5}(\frac{(3^zcos(4x)cos(4y))}{7})dxdydz=\int_{3.5}^{8.5}\int_{-8}^{-3}(\frac{(3^zcos(4y)sin(4x))}{28})dydz|_{-10}^{-7.5}\\&\int_{3.5}^{8.5}\int_{-8}^{-3}(-\frac{(3^zcos(4y)\cdot (\frac{sin(30)}{4}-\frac{ sin(40)}{4}))}{7})dydz=\int_{3.5}^{8.5}(-\frac{(3^zsin(4y)\cdot (sin(30) - sin(40)))}{112})dz|_{-8}^{-3}\\&\int_{3.5}^{8.5}(\frac{(3^z\cdot (sin(12) - sin(32))\cdot (sin(30) - sin(40)))}{112})dz=-\frac{(3^z\cdot (\frac{cos(2)}{2}-\frac{ cos(8)}{2}-\frac{ cos(18)}{2}+\frac{ cos(28)}{2}+\frac{ cos(42)}{2}-\frac{ cos(52)}{2}-\frac{ cos(62)}{2}+\frac{ cos(72)}{2}))}{(112ln(3))}|_{3.5}^{8.5}=173.44\end{align*}$

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Calculus 3 : Triple Integration in Cartesian Coordinates

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #311 : Triple Integration In Cartesian Coordinates

Example Question #311 : Triple Integration In Cartesian Coordinates

Example Question #311 : Triple Integration In Cartesian Coordinates

Example Question #311 : Triple Integration In Cartesian Coordinates

Example Question #445 : Triple Integrals

Example Question #446 : Triple Integrals

Example Question #922 : Calculus 3

Example Question #311 : Triple Integration In Cartesian Coordinates

Example Question #452 : Triple Integrals

Example Question #453 : Triple Integrals

All Calculus 3 Resources

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