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Calculus 3 : Triple Integration in Cartesian Coordinates

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Example Questions

Example Question #321 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{3}^{5.5}\int_{3}^{7.5}\int_{-4.5}^{-1}(8e^{(-2y)}e^{(-2z)}e^{(x)})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{3}^{5.5}\int_{3}^{7.5}\int_{-4.5}^{-1}(8e^{(-2y)}e^{(-2z)}e^{(x)})dxdydz\end{align*}$

Possible Answers:

$1.45\cdot10^{-6}$ $\displaystyle 1.45\cdot10^{-6}$

$-1.45\cdot10^{-6}$ $\displaystyle -1.45\cdot10^{-6}$

$-1.31\cdot10^{-5}$ $\displaystyle -1.31\cdot10^{-5}$

$4.35\cdot10^{-6}$ $\displaystyle 4.35\cdot10^{-6}$

Correct answer:

$4.35\cdot10^{-6}$ $\displaystyle 4.35\cdot10^{-6}$

Explanation:

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

$\begin{align*}&\int_{3}^{5.5}\int_{3}^{7.5}\int_{-4.5}^{-1}(8e^{(-2y)}e^{(-2z)}e^{(x)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{3}^{5.5}\int_{3}^{7.5}\int_{-4.5}^{-1}(8e^{(-2y)}e^{(-2z)}e^{(x)})dxdydz=\int_{3}^{5.5}\int_{3}^{7.5}(8e^{(-2y)}e^{(-2z)}e^{(x)})dydz|_{-4.5}^{-1}\\&\int_{3}^{5.5}\int_{3}^{7.5}(e^{(- 2y - 2z)}\cdot (8e^{(-1)} - 8e^{(-\frac{9}{2})}))dydz=\int_{3}^{5.5}(-4e^{(-\frac{9}{2})}e^{(- 2y - 2z)}\cdot (e^{(\frac{7}{2})} - 1))dz|_{3}^{7.5}\\&\int_{3}^{5.5}(4e^{(-2z)}\cdot (e^{(-16)} - e^{(-\frac{39}{2})})\cdot (e^{(9)} - 1))dz=-2e^{(-2z)}\cdot (e^{(-16)} - e^{(-\frac{39}{2})})\cdot (e^{(9)} - 1)|_{3}^{5.5}=4.35\cdot10^{-6}\end{align*}$ $\displaystyle \begin{align*}&\int_{3}^{5.5}\int_{3}^{7.5}\int_{-4.5}^{-1}(8e^{(-2y)}e^{(-2z)}e^{(x)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{3}^{5.5}\int_{3}^{7.5}\int_{-4.5}^{-1}(8e^{(-2y)}e^{(-2z)}e^{(x)})dxdydz=\int_{3}^{5.5}\int_{3}^{7.5}(8e^{(-2y)}e^{(-2z)}e^{(x)})dydz|_{-4.5}^{-1}\\&\int_{3}^{5.5}\int_{3}^{7.5}(e^{(- 2y - 2z)}\cdot (8e^{(-1)} - 8e^{(-\frac{9}{2})}))dydz=\int_{3}^{5.5}(-4e^{(-\frac{9}{2})}e^{(- 2y - 2z)}\cdot (e^{(\frac{7}{2})} - 1))dz|_{3}^{7.5}\\&\int_{3}^{5.5}(4e^{(-2z)}\cdot (e^{(-16)} - e^{(-\frac{39}{2})})\cdot (e^{(9)} - 1))dz=-2e^{(-2z)}\cdot (e^{(-16)} - e^{(-\frac{39}{2})})\cdot (e^{(9)} - 1)|_{3}^{5.5}=4.35\cdot10^{-6}\end{align*}$

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Example Question #322 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{7}^{8.5}\int_{-4.5}^{-2}\int_{3}^{5.5}(\frac{(53\cdot 2^ycos(3z))}{(4\cdot 2^{(\frac{x}{2})})})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{7}^{8.5}\int_{-4.5}^{-2}\int_{3}^{5.5}(\frac{(53\cdot 2^ycos(3z))}{(4\cdot 2^{(\frac{x}{2})})})dxdydz\end{align*}$

Possible Answers:

$-9.26\cdot10^{-2}$ $\displaystyle -9.26\cdot10^{-2}$

2.22 $\displaystyle 2.22$

$9.26\cdot10^{-2}$ $\displaystyle 9.26\cdot10^{-2}$

-0.37 $\displaystyle -0.37$

Correct answer:

-0.37 $\displaystyle -0.37$

Explanation:

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

$\begin{align*}&\int_{7}^{8.5}\int_{-4.5}^{-2}\int_{3}^{5.5}(\frac{(53\cdot 2^ycos(3z))}{(4\cdot 2^{(\frac{x}{2})})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{7}^{8.5}\int_{-4.5}^{-2}\int_{3}^{5.5}(\frac{(53\cdot 2^ycos(3z))}{(4\cdot 2^{(\frac{x}{2})})})dxdydz=\int_{7}^{8.5}\int_{-4.5}^{-2}(-\frac{(106\cdot 2^{(y - 2)}cos(3z))}{(2^{(\frac{x}{2})}ln(2))})dydz|_{3}^{5.5}\\&\int_{7}^{8.5}\int_{-4.5}^{-2}(\frac{(53\cdot 2^ycos(3z)\cdot (2\cdot 2^{(\frac{1}{2})} - 2^{(\frac{1}{4})}))}{(16ln(2))})dydz=\int_{7}^{8.5}(\frac{(53\cdot 2^y\cdot (2\cdot 2^{(\frac{1}{2})}cos(3z) - 2^{(\frac{1}{4})}cos(3z)))}{(16ln(2)^{2})})dz|_{-4.5}^{-2}\\&\int_{7}^{8.5}(\frac{(53cos(3z)\cdot (16\cdot 2^{(\frac{1}{2})} - 8\cdot 2^{(\frac{1}{4})} + 2^{(\frac{3}{4})} - 4))}{(512ln(2)^{2})})dz=\frac{(53sin(3z)\cdot (16\cdot 2^{(\frac{1}{2})} - 8\cdot 2^{(\frac{1}{4})} + 2^{(\frac{3}{4})} - 4))}{(1536ln(2)^{2})}|_{7}^{8.5}=-0.37\end{align*}$ $\displaystyle \begin{align*}&\int_{7}^{8.5}\int_{-4.5}^{-2}\int_{3}^{5.5}(\frac{(53\cdot 2^ycos(3z))}{(4\cdot 2^{(\frac{x}{2})})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{7}^{8.5}\int_{-4.5}^{-2}\int_{3}^{5.5}(\frac{(53\cdot 2^ycos(3z))}{(4\cdot 2^{(\frac{x}{2})})})dxdydz=\int_{7}^{8.5}\int_{-4.5}^{-2}(-\frac{(106\cdot 2^{(y - 2)}cos(3z))}{(2^{(\frac{x}{2})}ln(2))})dydz|_{3}^{5.5}\\&\int_{7}^{8.5}\int_{-4.5}^{-2}(\frac{(53\cdot 2^ycos(3z)\cdot (2\cdot 2^{(\frac{1}{2})} - 2^{(\frac{1}{4})}))}{(16ln(2))})dydz=\int_{7}^{8.5}(\frac{(53\cdot 2^y\cdot (2\cdot 2^{(\frac{1}{2})}cos(3z) - 2^{(\frac{1}{4})}cos(3z)))}{(16ln(2)^{2})})dz|_{-4.5}^{-2}\\&\int_{7}^{8.5}(\frac{(53cos(3z)\cdot (16\cdot 2^{(\frac{1}{2})} - 8\cdot 2^{(\frac{1}{4})} + 2^{(\frac{3}{4})} - 4))}{(512ln(2)^{2})})dz=\frac{(53sin(3z)\cdot (16\cdot 2^{(\frac{1}{2})} - 8\cdot 2^{(\frac{1}{4})} + 2^{(\frac{3}{4})} - 4))}{(1536ln(2)^{2})}|_{7}^{8.5}=-0.37\end{align*}$

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Example Question #323 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{6}^{10}\int_{8}^{9.5}\int_{-8}^{-4.5}(\frac{(5cos(3z)sin(x + 1))}{(37y)})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{6}^{10}\int_{8}^{9.5}\int_{-8}^{-4.5}(\frac{(5cos(3z)sin(x + 1))}{(37y)})dxdydz\end{align*}$

Possible Answers:

$5.17\cdot10^{-4}$ $\displaystyle 5.17\cdot10^{-4}$

$6.20\cdot10^{-3}$ $\displaystyle 6.20\cdot10^{-3}$

$-1.86\cdot10^{-2}$ $\displaystyle -1.86\cdot10^{-2}$

$-3.10\cdot10^{-3}$ $\displaystyle -3.10\cdot10^{-3}$

Correct answer:

$-3.10\cdot10^{-3}$ $\displaystyle -3.10\cdot10^{-3}$

Explanation:

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

$\begin{align*}&\int_{6}^{10}\int_{8}^{9.5}\int_{-8}^{-4.5}(\frac{(5cos(3z)sin(x + 1))}{(37y)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{6}^{10}\int_{8}^{9.5}\int_{-8}^{-4.5}(\frac{(5cos(3z)sin(x + 1))}{(37y)})dxdydz=\int_{6}^{10}\int_{8}^{9.5}(-\frac{(5cos(x + 1)cos(3z))}{(37y)})dydz|_{-8}^{-4.5}\\&\int_{6}^{10}\int_{8}^{9.5}(\frac{(5cos(3z)\cdot (cos(7) - cos(\frac{7}{2})))}{(37y)})dydz=\int_{6}^{10}(\frac{(5cos(3z)ln(y)\cdot (cos(7) - cos(\frac{7}{2})))}{37})dz|_{8}^{9.5}\\&\int_{6}^{10}(\frac{(5cos(3z)ln(\frac{19}{16})\cdot (cos(7) - cos(\frac{7}{2})))}{37})dz=\frac{(5sin(3z)ln(\frac{19}{16})\cdot (cos(7) - cos(\frac{7}{2})))}{111}|_{6}^{10}=-3.10\cdot10^{-3}\end{align*}$ $\displaystyle \begin{align*}&\int_{6}^{10}\int_{8}^{9.5}\int_{-8}^{-4.5}(\frac{(5cos(3z)sin(x + 1))}{(37y)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{6}^{10}\int_{8}^{9.5}\int_{-8}^{-4.5}(\frac{(5cos(3z)sin(x + 1))}{(37y)})dxdydz=\int_{6}^{10}\int_{8}^{9.5}(-\frac{(5cos(x + 1)cos(3z))}{(37y)})dydz|_{-8}^{-4.5}\\&\int_{6}^{10}\int_{8}^{9.5}(\frac{(5cos(3z)\cdot (cos(7) - cos(\frac{7}{2})))}{(37y)})dydz=\int_{6}^{10}(\frac{(5cos(3z)ln(y)\cdot (cos(7) - cos(\frac{7}{2})))}{37})dz|_{8}^{9.5}\\&\int_{6}^{10}(\frac{(5cos(3z)ln(\frac{19}{16})\cdot (cos(7) - cos(\frac{7}{2})))}{37})dz=\frac{(5sin(3z)ln(\frac{19}{16})\cdot (cos(7) - cos(\frac{7}{2})))}{111}|_{6}^{10}=-3.10\cdot10^{-3}\end{align*}$

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Example Question #321 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{-6}^{-3}\int_{10}^{14.5}\int_{6}^{8.5}(\frac{(3sin(y + 2))}{(496xz)})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{-6}^{-3}\int_{10}^{14.5}\int_{6}^{8.5}(\frac{(3sin(y + 2))}{(496xz)})dxdydz\end{align*}$

Possible Answers:

$-1.13\cdot10^{-2}$ $\displaystyle -1.13\cdot10^{-2}$

$7.53\cdot10^{-4}$ $\displaystyle 7.53\cdot10^{-4}$

$-2.26\cdot10^{-3}$ $\displaystyle -2.26\cdot10^{-3}$

$2.26\cdot10^{-3}$ $\displaystyle 2.26\cdot10^{-3}$

Correct answer:

$-2.26\cdot10^{-3}$ $\displaystyle -2.26\cdot10^{-3}$

Explanation:

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

$\begin{align*}&\int_{-6}^{-3}\int_{10}^{14.5}\int_{6}^{8.5}(\frac{(3sin(y + 2))}{(496xz)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-6}^{-3}\int_{10}^{14.5}\int_{6}^{8.5}(\frac{(3sin(y + 2))}{(496xz)})dxdydz=\int_{-6}^{-3}\int_{10}^{14.5}(\frac{(3sin(y + 2)ln(x))}{(496z)})dydz|_{6}^{8.5}\\&\int_{-6}^{-3}\int_{10}^{14.5}(\frac{(3sin(y + 2)ln(\frac{17}{12}))}{(496z)})dydz=\int_{-6}^{-3}(-\frac{(3cos(y + 2)ln(\frac{17}{12}))}{(496z)})dz|_{10}^{14.5}\\&\int_{-6}^{-3}(\frac{(3ln(\frac{17}{12})\cdot (cos(12) - cos(\frac{33}{2})))}{(496z)})dz=ln(z)\cdot (\frac{(3cos(12)ln(\frac{17}{12}))}{496}-\frac{ (3cos(\frac{33}{2})ln(\frac{17}{12}))}{496})|_{-6}^{-3}=-2.26\cdot10^{-3}\end{align*}$ $\displaystyle \begin{align*}&\int_{-6}^{-3}\int_{10}^{14.5}\int_{6}^{8.5}(\frac{(3sin(y + 2))}{(496xz)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-6}^{-3}\int_{10}^{14.5}\int_{6}^{8.5}(\frac{(3sin(y + 2))}{(496xz)})dxdydz=\int_{-6}^{-3}\int_{10}^{14.5}(\frac{(3sin(y + 2)ln(x))}{(496z)})dydz|_{6}^{8.5}\\&\int_{-6}^{-3}\int_{10}^{14.5}(\frac{(3sin(y + 2)ln(\frac{17}{12}))}{(496z)})dydz=\int_{-6}^{-3}(-\frac{(3cos(y + 2)ln(\frac{17}{12}))}{(496z)})dz|_{10}^{14.5}\\&\int_{-6}^{-3}(\frac{(3ln(\frac{17}{12})\cdot (cos(12) - cos(\frac{33}{2})))}{(496z)})dz=ln(z)\cdot (\frac{(3cos(12)ln(\frac{17}{12}))}{496}-\frac{ (3cos(\frac{33}{2})ln(\frac{17}{12}))}{496})|_{-6}^{-3}=-2.26\cdot10^{-3}\end{align*}$

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Example Question #451 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{-7}^{-2}\int_{-3.5}^{1}\int_{-9}^{-5.5}(\frac{(29\cdot 3^{(\frac{y}{2})})}{(5x^{2}z^{2})})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{-7}^{-2}\int_{-3.5}^{1}\int_{-9}^{-5.5}(\frac{(29\cdot 3^{(\frac{y}{2})})}{(5x^{2}z^{2})})dxdydz\end{align*}$

Possible Answers:

$-7.05\cdot10^{-2}$ $\displaystyle -7.05\cdot10^{-2}$

0.85 $\displaystyle 0.85$

-0.85 $\displaystyle -0.85$

0.42 $\displaystyle 0.42$

Correct answer:

0.42 $\displaystyle 0.42$

Explanation:

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

$\begin{align*}&\int_{-7}^{-2}\int_{-3.5}^{1}\int_{-9}^{-5.5}(\frac{(29\cdot 3^{(\frac{y}{2})})}{(5x^{2}z^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-7}^{-2}\int_{-3.5}^{1}\int_{-9}^{-5.5}(\frac{(29\cdot 3^{(\frac{y}{2})})}{(5x^{2}z^{2})})dxdydz=\int_{-7}^{-2}\int_{-3.5}^{1}(-\frac{(29\cdot 3^{(\frac{y}{2})})}{(5xz^{2})})dydz|_{-9}^{-5.5}\\&\int_{-7}^{-2}\int_{-3.5}^{1}(\frac{(203\cdot 3^{(\frac{y}{2})})}{(495z^{2})})dydz=\int_{-7}^{-2}(\frac{(406\cdot 3^{(\frac{y}{2})})}{(495z^{2}ln(3))})dz|_{-3.5}^{1}\\&\int_{-7}^{-2}(\frac{(203\cdot (2\cdot 3^{(\frac{1}{2})} -\frac{ (2\cdot 3^{(\frac{1}{4})})}{9}))}{(495z^{2}ln(3))})dz=-\frac{(3654\cdot 3^{(\frac{1}{2})} - 406\cdot 3^{(\frac{1}{4})})}{(4455zln(3))}|_{-7}^{-2}=0.42\end{align*}$ $\displaystyle \begin{align*}&\int_{-7}^{-2}\int_{-3.5}^{1}\int_{-9}^{-5.5}(\frac{(29\cdot 3^{(\frac{y}{2})})}{(5x^{2}z^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-7}^{-2}\int_{-3.5}^{1}\int_{-9}^{-5.5}(\frac{(29\cdot 3^{(\frac{y}{2})})}{(5x^{2}z^{2})})dxdydz=\int_{-7}^{-2}\int_{-3.5}^{1}(-\frac{(29\cdot 3^{(\frac{y}{2})})}{(5xz^{2})})dydz|_{-9}^{-5.5}\\&\int_{-7}^{-2}\int_{-3.5}^{1}(\frac{(203\cdot 3^{(\frac{y}{2})})}{(495z^{2})})dydz=\int_{-7}^{-2}(\frac{(406\cdot 3^{(\frac{y}{2})})}{(495z^{2}ln(3))})dz|_{-3.5}^{1}\\&\int_{-7}^{-2}(\frac{(203\cdot (2\cdot 3^{(\frac{1}{2})} -\frac{ (2\cdot 3^{(\frac{1}{4})})}{9}))}{(495z^{2}ln(3))})dz=-\frac{(3654\cdot 3^{(\frac{1}{2})} - 406\cdot 3^{(\frac{1}{4})})}{(4455zln(3))}|_{-7}^{-2}=0.42\end{align*}$

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Example Question #325 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{-6}^{-4}\int_{3}^{7}\int_{4}^{9}(\frac{(9cos(4z))}{(53\cdot 3^{(\frac{x}{2})}y)})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{-6}^{-4}\int_{3}^{7}\int_{4}^{9}(\frac{(9cos(4z))}{(53\cdot 3^{(\frac{x}{2})}y)})dxdydz\end{align*}$

Possible Answers:

$-1.05\cdot10^{-3}$ $\displaystyle -1.05\cdot10^{-3}$

$4.21\cdot10^{-3}$ $\displaystyle 4.21\cdot10^{-3}$

$-4.21\cdot10^{-3}$ $\displaystyle -4.21\cdot10^{-3}$

$7.01\cdot10^{-4}$ $\displaystyle 7.01\cdot10^{-4}$

Correct answer:

$-4.21\cdot10^{-3}$ $\displaystyle -4.21\cdot10^{-3}$

Explanation:

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

$\begin{align*}&\int_{-6}^{-4}\int_{3}^{7}\int_{4}^{9}(\frac{(9cos(4z))}{(53\cdot 3^{(\frac{x}{2})}y)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-6}^{-4}\int_{3}^{7}\int_{4}^{9}(\frac{(9cos(4z))}{(53\cdot 3^{(\frac{x}{2})}y)})dxdydz=\int_{-6}^{-4}\int_{3}^{7}(-\frac{(18cos(4z))}{(53\cdot 3^{(\frac{x}{2})}yln(3))})dydz|_{4}^{9}\\&\int_{-6}^{-4}\int_{3}^{7}(-\frac{(cos(4z)\cdot (2\cdot 3^{(\frac{1}{2})} - 54))}{(1431yln(3))})dydz=\int_{-6}^{-4}(\frac{(ln(y)\cdot (54cos(4z) - 2\cdot 3^{(\frac{1}{2})}cos(4z)))}{(1431ln(3))})dz|_{3}^{7}\\&\int_{-6}^{-4}(\frac{(54cos(4z)ln(\frac{7}{3}) - 2\cdot 3^{(\frac{1}{2})}cos(4z)ln(\frac{7}{3}))}{(1431ln(3))})dz=\frac{(27sin(4z)ln(\frac{7}{3}) - 3^{(\frac{1}{2})}sin(4z)ln(\frac{7}{3}))}{(2862ln(3))}|_{-6}^{-4}=-4.21\cdot10^{-3}\end{align*}$ $\displaystyle \begin{align*}&\int_{-6}^{-4}\int_{3}^{7}\int_{4}^{9}(\frac{(9cos(4z))}{(53\cdot 3^{(\frac{x}{2})}y)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-6}^{-4}\int_{3}^{7}\int_{4}^{9}(\frac{(9cos(4z))}{(53\cdot 3^{(\frac{x}{2})}y)})dxdydz=\int_{-6}^{-4}\int_{3}^{7}(-\frac{(18cos(4z))}{(53\cdot 3^{(\frac{x}{2})}yln(3))})dydz|_{4}^{9}\\&\int_{-6}^{-4}\int_{3}^{7}(-\frac{(cos(4z)\cdot (2\cdot 3^{(\frac{1}{2})} - 54))}{(1431yln(3))})dydz=\int_{-6}^{-4}(\frac{(ln(y)\cdot (54cos(4z) - 2\cdot 3^{(\frac{1}{2})}cos(4z)))}{(1431ln(3))})dz|_{3}^{7}\\&\int_{-6}^{-4}(\frac{(54cos(4z)ln(\frac{7}{3}) - 2\cdot 3^{(\frac{1}{2})}cos(4z)ln(\frac{7}{3}))}{(1431ln(3))})dz=\frac{(27sin(4z)ln(\frac{7}{3}) - 3^{(\frac{1}{2})}sin(4z)ln(\frac{7}{3}))}{(2862ln(3))}|_{-6}^{-4}=-4.21\cdot10^{-3}\end{align*}$

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Example Question #326 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{8}^{11.5}\int_{10}^{14.5}\int_{6}^{9.5}(\frac{(5sin(4x)sin(z + 1))}{(6y)})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{8}^{11.5}\int_{10}^{14.5}\int_{6}^{9.5}(\frac{(5sin(4x)sin(z + 1))}{(6y)})dxdydz\end{align*}$

Possible Answers:

-0.31 $\displaystyle -0.31$

$-2.61\cdot10^{-2}$ $\displaystyle -2.61\cdot10^{-2}$

0.16 $\displaystyle 0.16$

$7.84\cdot10^{-2}$ $\displaystyle 7.84\cdot10^{-2}$

Correct answer:

$7.84\cdot10^{-2}$ $\displaystyle 7.84\cdot10^{-2}$

Explanation:

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

$\begin{align*}&\int_{8}^{11.5}\int_{10}^{14.5}\int_{6}^{9.5}(\frac{(5sin(4x)sin(z + 1))}{(6y)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{8}^{11.5}\int_{10}^{14.5}\int_{6}^{9.5}(\frac{(5sin(4x)sin(z + 1))}{(6y)})dxdydz=\int_{8}^{11.5}\int_{10}^{14.5}(-\frac{(5cos(4x)sin(z + 1))}{(24y)})dydz|_{6}^{9.5}\\&\int_{8}^{11.5}\int_{10}^{14.5}(\frac{(5sin(z + 1)\cdot (\frac{cos(24)}{4}-\frac{ cos(38)}{4}))}{(6y)})dydz=\int_{8}^{11.5}(ln(y)\cdot (\frac{(5sin(z + 1)cos(24))}{24}-\frac{ (5sin(z + 1)cos(38))}{24}))dz|_{10}^{14.5}\\&\int_{8}^{11.5}(\frac{(5sin(z + 1)ln(\frac{29}{20})\cdot (cos(24) - cos(38)))}{24})dz=-\frac{(5cos(z + 1)ln(\frac{29}{20})\cdot (cos(24) - cos(38)))}{24}|_{8}^{11.5}=7.84\cdot10^{-2}\end{align*}$ $\displaystyle \begin{align*}&\int_{8}^{11.5}\int_{10}^{14.5}\int_{6}^{9.5}(\frac{(5sin(4x)sin(z + 1))}{(6y)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{8}^{11.5}\int_{10}^{14.5}\int_{6}^{9.5}(\frac{(5sin(4x)sin(z + 1))}{(6y)})dxdydz=\int_{8}^{11.5}\int_{10}^{14.5}(-\frac{(5cos(4x)sin(z + 1))}{(24y)})dydz|_{6}^{9.5}\\&\int_{8}^{11.5}\int_{10}^{14.5}(\frac{(5sin(z + 1)\cdot (\frac{cos(24)}{4}-\frac{ cos(38)}{4}))}{(6y)})dydz=\int_{8}^{11.5}(ln(y)\cdot (\frac{(5sin(z + 1)cos(24))}{24}-\frac{ (5sin(z + 1)cos(38))}{24}))dz|_{10}^{14.5}\\&\int_{8}^{11.5}(\frac{(5sin(z + 1)ln(\frac{29}{20})\cdot (cos(24) - cos(38)))}{24})dz=-\frac{(5cos(z + 1)ln(\frac{29}{20})\cdot (cos(24) - cos(38)))}{24}|_{8}^{11.5}=7.84\cdot10^{-2}\end{align*}$

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Example Question #461 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{8}^{9.5}\int_{-4}^{0.5}\int_{10}^{12.5}(\frac{(3cos(3z)e^{(-2y)})}{(62x)})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{8}^{9.5}\int_{-4}^{0.5}\int_{10}^{12.5}(\frac{(3cos(3z)e^{(-2y)})}{(62x)})dxdydz\end{align*}$

Possible Answers:

3.66 $\displaystyle 3.66$

-0.61 $\displaystyle -0.61$

7.31 $\displaystyle 7.31$

-10.97 $\displaystyle -10.97$

Correct answer:

3.66 $\displaystyle 3.66$

Explanation:

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

$\begin{align*}&\int_{8}^{9.5}\int_{-4}^{0.5}\int_{10}^{12.5}(\frac{(3cos(3z)e^{(-2y)})}{(62x)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{8}^{9.5}\int_{-4}^{0.5}\int_{10}^{12.5}(\frac{(3cos(3z)e^{(-2y)})}{(62x)})dxdydz=\int_{8}^{9.5}\int_{-4}^{0.5}(\frac{(3cos(3z)e^{(-2y)}ln(x))}{62})dydz|_{10}^{12.5}\\&\int_{8}^{9.5}\int_{-4}^{0.5}(\frac{(3cos(3z)e^{(-2y)}ln(\frac{5}{4}))}{62})dydz=\int_{8}^{9.5}(-\frac{(3cos(3z)e^{(-2y)}ln(\frac{5}{4}))}{124})dz|_{-4}^{0.5}\\&\int_{8}^{9.5}(\frac{(3cos(3z)e^{(-1)}ln(\frac{5}{4})\cdot (e^{(9)} - 1))}{124})dz=\frac{(sin(3z)e^{(-1)}ln(\frac{5}{4})\cdot (e^{(9)} - 1))}{124}|_{8}^{9.5}=3.66\end{align*}$ $\displaystyle \begin{align*}&\int_{8}^{9.5}\int_{-4}^{0.5}\int_{10}^{12.5}(\frac{(3cos(3z)e^{(-2y)})}{(62x)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{8}^{9.5}\int_{-4}^{0.5}\int_{10}^{12.5}(\frac{(3cos(3z)e^{(-2y)})}{(62x)})dxdydz=\int_{8}^{9.5}\int_{-4}^{0.5}(\frac{(3cos(3z)e^{(-2y)}ln(x))}{62})dydz|_{10}^{12.5}\\&\int_{8}^{9.5}\int_{-4}^{0.5}(\frac{(3cos(3z)e^{(-2y)}ln(\frac{5}{4}))}{62})dydz=\int_{8}^{9.5}(-\frac{(3cos(3z)e^{(-2y)}ln(\frac{5}{4}))}{124})dz|_{-4}^{0.5}\\&\int_{8}^{9.5}(\frac{(3cos(3z)e^{(-1)}ln(\frac{5}{4})\cdot (e^{(9)} - 1))}{124})dz=\frac{(sin(3z)e^{(-1)}ln(\frac{5}{4})\cdot (e^{(9)} - 1))}{124}|_{8}^{9.5}=3.66\end{align*}$

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Example Question #462 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{3.5}^{4.5}\int_{-8}^{-3}\int_{-4.5}^{-2}(\frac{(37\cdot 2^{(\frac{x}{4})}cos(4y)e^{(-z)})}{4})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{3.5}^{4.5}\int_{-8}^{-3}\int_{-4.5}^{-2}(\frac{(37\cdot 2^{(\frac{x}{4})}cos(4y)e^{(-z)})}{4})dxdydz\end{align*}$

Possible Answers:

-0.14 $\displaystyle -0.14$

$3.45\cdot10^{-2}$ $\displaystyle 3.45\cdot10^{-2}$

$-1.15\cdot10^{-2}$ $\displaystyle -1.15\cdot10^{-2}$

$6.89\cdot10^{-2}$ $\displaystyle 6.89\cdot10^{-2}$

Correct answer:

$6.89\cdot10^{-2}$ $\displaystyle 6.89\cdot10^{-2}$

Explanation:

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

$\begin{align*}&\int_{3.5}^{4.5}\int_{-8}^{-3}\int_{-4.5}^{-2}(\frac{(37\cdot 2^{(\frac{x}{4})}cos(4y)e^{(-z)})}{4})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{3.5}^{4.5}\int_{-8}^{-3}\int_{-4.5}^{-2}(\frac{(37\cdot 2^{(\frac{x}{4})}cos(4y)e^{(-z)})}{4})dxdydz=\int_{3.5}^{4.5}\int_{-8}^{-3}(\frac{(37\cdot 2^{(\frac{x}{4})}cos(4y)e^{(-z)})}{ln(2)})dydz|_{-4.5}^{-2}\\&\int_{3.5}^{4.5}\int_{-8}^{-3}(\frac{(37cos(4y)e^{(-z)}\cdot (2\cdot 2^{(\frac{1}{2})} - 2^{(\frac{7}{8})}))}{(4ln(2))})dydz=\int_{3.5}^{4.5}(\frac{(37sin(4y)e^{(-z)}\cdot (2\cdot 2^{(\frac{1}{2})} - 2^{(\frac{7}{8})}))}{(16ln(2))})dz|_{-8}^{-3}\\&\int_{3.5}^{4.5}(\frac{(37\cdot 2^{(\frac{1}{2})}e^{(-z)}\cdot (sin(12) - sin(32))\cdot (2^{(\frac{3}{8})} - 2))}{(16ln(2))})dz=-\frac{(37\cdot 2^{(\frac{1}{2})}e^{(-z)}\cdot (sin(12) - sin(32))\cdot (2^{(\frac{3}{8})} - 2))}{(16ln(2))}|_{3.5}^{4.5}=6.89\cdot10^{-2}\end{align*}$ $\displaystyle \begin{align*}&\int_{3.5}^{4.5}\int_{-8}^{-3}\int_{-4.5}^{-2}(\frac{(37\cdot 2^{(\frac{x}{4})}cos(4y)e^{(-z)})}{4})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{3.5}^{4.5}\int_{-8}^{-3}\int_{-4.5}^{-2}(\frac{(37\cdot 2^{(\frac{x}{4})}cos(4y)e^{(-z)})}{4})dxdydz=\int_{3.5}^{4.5}\int_{-8}^{-3}(\frac{(37\cdot 2^{(\frac{x}{4})}cos(4y)e^{(-z)})}{ln(2)})dydz|_{-4.5}^{-2}\\&\int_{3.5}^{4.5}\int_{-8}^{-3}(\frac{(37cos(4y)e^{(-z)}\cdot (2\cdot 2^{(\frac{1}{2})} - 2^{(\frac{7}{8})}))}{(4ln(2))})dydz=\int_{3.5}^{4.5}(\frac{(37sin(4y)e^{(-z)}\cdot (2\cdot 2^{(\frac{1}{2})} - 2^{(\frac{7}{8})}))}{(16ln(2))})dz|_{-8}^{-3}\\&\int_{3.5}^{4.5}(\frac{(37\cdot 2^{(\frac{1}{2})}e^{(-z)}\cdot (sin(12) - sin(32))\cdot (2^{(\frac{3}{8})} - 2))}{(16ln(2))})dz=-\frac{(37\cdot 2^{(\frac{1}{2})}e^{(-z)}\cdot (sin(12) - sin(32))\cdot (2^{(\frac{3}{8})} - 2))}{(16ln(2))}|_{3.5}^{4.5}=6.89\cdot10^{-2}\end{align*}$

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Example Question #463 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{-9}^{-4}\int_{-10}^{-8}\int_{4}^{5.5}(\frac{(10cos(z + 2)sin(y + 1)e^{(-2x)})}{13})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{-9}^{-4}\int_{-10}^{-8}\int_{4}^{5.5}(\frac{(10cos(z + 2)sin(y + 1)e^{(-2x)})}{13})dxdydz\end{align*}$

Possible Answers:

$-2.58\cdot10^{-5}$ $\displaystyle -2.58\cdot10^{-5}$

$2.58\cdot10^{-5}$ $\displaystyle 2.58\cdot10^{-5}$

$-2.58\cdot10^{-4}$ $\displaystyle -2.58\cdot10^{-4}$

$5.15\cdot10^{-5}$ $\displaystyle 5.15\cdot10^{-5}$

Correct answer:

$5.15\cdot10^{-5}$ $\displaystyle 5.15\cdot10^{-5}$

Explanation:

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

$\begin{align*}&\int_{-9}^{-4}\int_{-10}^{-8}\int_{4}^{5.5}(\frac{(10cos(z + 2)sin(y + 1)e^{(-2x)})}{13})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-9}^{-4}\int_{-10}^{-8}\int_{4}^{5.5}(\frac{(10cos(z + 2)sin(y + 1)e^{(-2x)})}{13})dxdydz=\int_{-9}^{-4}\int_{-10}^{-8}(-\frac{(5cos(z + 2)sin(y + 1)e^{(-2x)})}{13})dydz|_{4}^{5.5}\\&\int_{-9}^{-4}\int_{-10}^{-8}(\frac{(5cos(z + 2)sin(y + 1)e^{(-11)}\cdot (e^{(3)} - 1))}{13})dydz=\int_{-9}^{-4}(-\frac{(5cos(y + 1)cos(z + 2)e^{(-11)}\cdot (e^{(3)} - 1))}{13})dz|_{-10}^{-8}\\&\int_{-9}^{-4}(-\frac{(5cos(z + 2)e^{(-11)}\cdot (cos(7) - cos(9))\cdot (e^{(3)} - 1))}{13})dz=-\frac{(5sin(z + 2)e^{(-11)}\cdot (cos(7) - cos(9))\cdot (e^{(3)} - 1))}{13}|_{-9}^{-4}=5.15\cdot10^{-5}\end{align*}$ $\displaystyle \begin{align*}&\int_{-9}^{-4}\int_{-10}^{-8}\int_{4}^{5.5}(\frac{(10cos(z + 2)sin(y + 1)e^{(-2x)})}{13})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-9}^{-4}\int_{-10}^{-8}\int_{4}^{5.5}(\frac{(10cos(z + 2)sin(y + 1)e^{(-2x)})}{13})dxdydz=\int_{-9}^{-4}\int_{-10}^{-8}(-\frac{(5cos(z + 2)sin(y + 1)e^{(-2x)})}{13})dydz|_{4}^{5.5}\\&\int_{-9}^{-4}\int_{-10}^{-8}(\frac{(5cos(z + 2)sin(y + 1)e^{(-11)}\cdot (e^{(3)} - 1))}{13})dydz=\int_{-9}^{-4}(-\frac{(5cos(y + 1)cos(z + 2)e^{(-11)}\cdot (e^{(3)} - 1))}{13})dz|_{-10}^{-8}\\&\int_{-9}^{-4}(-\frac{(5cos(z + 2)e^{(-11)}\cdot (cos(7) - cos(9))\cdot (e^{(3)} - 1))}{13})dz=-\frac{(5sin(z + 2)e^{(-11)}\cdot (cos(7) - cos(9))\cdot (e^{(3)} - 1))}{13}|_{-9}^{-4}=5.15\cdot10^{-5}\end{align*}$

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Calculus 3 : Triple Integration in Cartesian Coordinates

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Example Question #321 : Triple Integration In Cartesian Coordinates

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Example Question #451 : Multiple Integration

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