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Calculus 3 : Calculus 3

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Example Questions

Example Question #151 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{6}^{7.5}\int_{4.5}^{8.5}\int_{-5}^{-0.5}(\frac{(11z^{2}e^{(-x)}e^{(-y)})}{38})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{6}^{7.5}\int_{4.5}^{8.5}\int_{-5}^{-0.5}(\frac{(11z^{2}e^{(-x)}e^{(-y)})}{38})dxdydz\end{align*}$

Possible Answers:

-7.95 $\displaystyle -7.95$

31.8 $\displaystyle 31.8$

5.3 $\displaystyle 5.3$

-190.8 $\displaystyle -190.8$

Correct answer:

31.8 $\displaystyle 31.8$

Explanation:

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

$\begin{align*}&\int_{6}^{7.5}\int_{4.5}^{8.5}\int_{-5}^{-0.5}(\frac{(11z^{2}e^{(-x)}e^{(-y)})}{38})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{6}^{7.5}\int_{4.5}^{8.5}\int_{-5}^{-0.5}(\frac{(11z^{2}e^{(-x)}e^{(-y)})}{38})dxdydz=\int_{6}^{7.5}\int_{4.5}^{8.5}(-\frac{(11z^{2}e^{(- x - y)})}{38})dydz|_{-5}^{-0.5}\\&\int_{6}^{7.5}\int_{4.5}^{8.5}(-\frac{(11z^{2}e^{(-y)}\cdot(e^{(\frac{1}{2})} - e^{(5)}))}{38})dydz=\int_{6}^{7.5}(\frac{(11z^{2}e^{(-y)}\cdot(e^{(\frac{1}{2})} - e^{(5)}))}{38})dz|_{4.5}^{8.5}\\&\int_{6}^{7.5}(-\frac{(11z^{2}\cdot(e^{(-8)} - e^{(-\frac{7}{2})})\cdot(e^{(4)} - 1))}{38})dz=z^{3}\cdot(\frac{(11e^{(\frac{1}{2})})}{114}-\frac{ (11e^{(-4)})}{114}+\frac{ (11e^{(-8)})}{114}-\frac{ (11e^{(-\frac{7}{2})})}{114})|_{6}^{7.5}=31.8\end{align*}$ $\displaystyle \begin{align*}&\int_{6}^{7.5}\int_{4.5}^{8.5}\int_{-5}^{-0.5}(\frac{(11z^{2}e^{(-x)}e^{(-y)})}{38})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{6}^{7.5}\int_{4.5}^{8.5}\int_{-5}^{-0.5}(\frac{(11z^{2}e^{(-x)}e^{(-y)})}{38})dxdydz=\int_{6}^{7.5}\int_{4.5}^{8.5}(-\frac{(11z^{2}e^{(- x - y)})}{38})dydz|_{-5}^{-0.5}\\&\int_{6}^{7.5}\int_{4.5}^{8.5}(-\frac{(11z^{2}e^{(-y)}\cdot(e^{(\frac{1}{2})} - e^{(5)}))}{38})dydz=\int_{6}^{7.5}(\frac{(11z^{2}e^{(-y)}\cdot(e^{(\frac{1}{2})} - e^{(5)}))}{38})dz|_{4.5}^{8.5}\\&\int_{6}^{7.5}(-\frac{(11z^{2}\cdot(e^{(-8)} - e^{(-\frac{7}{2})})\cdot(e^{(4)} - 1))}{38})dz=z^{3}\cdot(\frac{(11e^{(\frac{1}{2})})}{114}-\frac{ (11e^{(-4)})}{114}+\frac{ (11e^{(-8)})}{114}-\frac{ (11e^{(-\frac{7}{2})})}{114})|_{6}^{7.5}=31.8\end{align*}$

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Example Question #152 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{-6}^{-5}\int_{-8}^{-4.5}\int_{-4}^{-2}(\frac{(4cos(3y)e^{(-2x)})}{z^{2}})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{-6}^{-5}\int_{-8}^{-4.5}\int_{-4}^{-2}(\frac{(4cos(3y)e^{(-2x)})}{z^{2}})dxdydz\end{align*}$

Possible Answers:

555.8 $\displaystyle 555.8$

-333.5 $\displaystyle -333.5$

-111.2 $\displaystyle -111.2$

55.58 $\displaystyle 55.58$

Correct answer:

-111.2 $\displaystyle -111.2$

Explanation:

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

$\begin{align*}&\int_{-6}^{-5}\int_{-8}^{-4.5}\int_{-4}^{-2}(\frac{(4cos(3y)e^{(-2x)})}{z^{2}})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-6}^{-5}\int_{-8}^{-4.5}\int_{-4}^{-2}(\frac{(4cos(3y)e^{(-2x)})}{z^{2}})dxdydz=\int_{-6}^{-5}\int_{-8}^{-4.5}(-\frac{(2cos(3y)e^{(-2x)})}{z^{2}})dydz|_{-4}^{-2}\\&\int_{-6}^{-5}\int_{-8}^{-4.5}(\frac{(2cos(3y)e^{(4)}\cdot(e^{(4)} - 1))}{z^{2}})dydz=\int_{-6}^{-5}(\frac{(2sin(3y)e^{(4)}\cdot(e^{(4)} - 1))}{(3z^{2})})dz|_{-8}^{-4.5}\\&\int_{-6}^{-5}(\frac{(2e^{(4)}\cdot(sin(24) - sin(\frac{27}{2}))\cdot(e^{(4)} - 1))}{(3z^{2})})dz=-\frac{(2e^{(4)}\cdot(sin(24) - sin(\frac{27}{2}))\cdot(e^{(4)} - 1))}{(3z)}|_{-6}^{-5}=-111.2\end{align*}$ $\displaystyle \begin{align*}&\int_{-6}^{-5}\int_{-8}^{-4.5}\int_{-4}^{-2}(\frac{(4cos(3y)e^{(-2x)})}{z^{2}})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-6}^{-5}\int_{-8}^{-4.5}\int_{-4}^{-2}(\frac{(4cos(3y)e^{(-2x)})}{z^{2}})dxdydz=\int_{-6}^{-5}\int_{-8}^{-4.5}(-\frac{(2cos(3y)e^{(-2x)})}{z^{2}})dydz|_{-4}^{-2}\\&\int_{-6}^{-5}\int_{-8}^{-4.5}(\frac{(2cos(3y)e^{(4)}\cdot(e^{(4)} - 1))}{z^{2}})dydz=\int_{-6}^{-5}(\frac{(2sin(3y)e^{(4)}\cdot(e^{(4)} - 1))}{(3z^{2})})dz|_{-8}^{-4.5}\\&\int_{-6}^{-5}(\frac{(2e^{(4)}\cdot(sin(24) - sin(\frac{27}{2}))\cdot(e^{(4)} - 1))}{(3z^{2})})dz=-\frac{(2e^{(4)}\cdot(sin(24) - sin(\frac{27}{2}))\cdot(e^{(4)} - 1))}{(3z)}|_{-6}^{-5}=-111.2\end{align*}$

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Example Question #153 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{-6}^{-1.5}\int_{-4}^{-2}\int_{4}^{8}(\frac{(13\cdot2^{(\frac{x}{3})}\cdot3^y)}{(5z^{2})})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{-6}^{-1.5}\int_{-4}^{-2}\int_{4}^{8}(\frac{(13\cdot2^{(\frac{x}{3})}\cdot3^y)}{(5z^{2})})dxdydz\end{align*}$

Possible Answers:

-0.32 $\displaystyle -0.32$

1.94 $\displaystyle 1.94$

11.62 $\displaystyle 11.62$

-11.62 $\displaystyle -11.62$

Correct answer:

1.94 $\displaystyle 1.94$

Explanation:

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

$\begin{align*}&\int_{-6}^{-1.5}\int_{-4}^{-2}\int_{4}^{8}(\frac{(13\cdot2^{(\frac{x}{3})}\cdot3^y)}{(5z^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-6}^{-1.5}\int_{-4}^{-2}\int_{4}^{8}(\frac{(13\cdot2^{(\frac{x}{3})}\cdot3^y)}{(5z^{2})})dxdydz=\int_{-6}^{-1.5}\int_{-4}^{-2}(\frac{(39\cdot2^{(\frac{x}{3})}\cdot3^y)}{(5z^{2}ln(2))})dydz|_{4}^{8}\\&\int_{-6}^{-1.5}\int_{-4}^{-2}(-\frac{(13\cdot3^y\cdot(6\cdot2^{(\frac{1}{3})} - 12\cdot2^{(\frac{2}{3})}))}{(5z^{2}ln(2))})dydz=\int_{-6}^{-1.5}(-\frac{(78\cdot3^y\cdot(2^{(\frac{1}{3})} - 2\cdot2^{(\frac{2}{3})}))}{(5z^{2}ln(2)ln(3))})dz|_{-4}^{-2}\\&\int_{-6}^{-1.5}(-\frac{(208\cdot2^{(\frac{1}{3})} - 416\cdot2^{(\frac{2}{3})})}{(135z^{2}ln(2)ln(3))})dz=\frac{(208\cdot2^{(\frac{1}{3})} - 416\cdot4^{(\frac{1}{3})})}{(135zln(2)ln(3))}|_{-6}^{-1.5}=1.94\end{align*}$ $\displaystyle \begin{align*}&\int_{-6}^{-1.5}\int_{-4}^{-2}\int_{4}^{8}(\frac{(13\cdot2^{(\frac{x}{3})}\cdot3^y)}{(5z^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-6}^{-1.5}\int_{-4}^{-2}\int_{4}^{8}(\frac{(13\cdot2^{(\frac{x}{3})}\cdot3^y)}{(5z^{2})})dxdydz=\int_{-6}^{-1.5}\int_{-4}^{-2}(\frac{(39\cdot2^{(\frac{x}{3})}\cdot3^y)}{(5z^{2}ln(2))})dydz|_{4}^{8}\\&\int_{-6}^{-1.5}\int_{-4}^{-2}(-\frac{(13\cdot3^y\cdot(6\cdot2^{(\frac{1}{3})} - 12\cdot2^{(\frac{2}{3})}))}{(5z^{2}ln(2))})dydz=\int_{-6}^{-1.5}(-\frac{(78\cdot3^y\cdot(2^{(\frac{1}{3})} - 2\cdot2^{(\frac{2}{3})}))}{(5z^{2}ln(2)ln(3))})dz|_{-4}^{-2}\\&\int_{-6}^{-1.5}(-\frac{(208\cdot2^{(\frac{1}{3})} - 416\cdot2^{(\frac{2}{3})})}{(135z^{2}ln(2)ln(3))})dz=\frac{(208\cdot2^{(\frac{1}{3})} - 416\cdot4^{(\frac{1}{3})})}{(135zln(2)ln(3))}|_{-6}^{-1.5}=1.94\end{align*}$

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Example Question #152 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{8}^{13}\int_{10}^{12}\int_{-4}^{-2.5}(\frac{(z^{2}cos(3y))}{(7\cdot3^x)})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{8}^{13}\int_{10}^{12}\int_{-4}^{-2.5}(\frac{(z^{2}cos(3y))}{(7\cdot3^x)})dxdydz\end{align*}$

Possible Answers:

-1.19 $\displaystyle -1.19$

-5.97 $\displaystyle -5.97$

5.97 $\displaystyle 5.97$

1.99 $\displaystyle 1.99$

Correct answer:

-5.97 $\displaystyle -5.97$

Explanation:

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

$\begin{align*}&\int_{8}^{13}\int_{10}^{12}\int_{-4}^{-2.5}(\frac{(z^{2}cos(3y))}{(7\cdot3^x)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{8}^{13}\int_{10}^{12}\int_{-4}^{-2.5}(\frac{(z^{2}cos(3y))}{(7\cdot3^x)})dxdydz=\int_{8}^{13}\int_{10}^{12}(-\frac{(z^{2}cos(3y))}{(7\cdot3^xln(3))})dydz|_{-4}^{-2.5}\\&\int_{8}^{13}\int_{10}^{12}(-\frac{(9z^{2}cos(3y)\cdot(3^{(\frac{1}{2})} - 9))}{(7ln(3))})dydz=\int_{8}^{13}(-\frac{(3z^{2}sin(3y)\cdot(3^{(\frac{1}{2})} - 9))}{(7ln(3))})dz|_{10}^{12}\\&\int_{8}^{13}(\frac{(3z^{2}\cdot(sin(30) - sin(36))\cdot(3^{(\frac{1}{2})} - 9))}{(7ln(3))})dz=-\frac{(z^{3}\cdot(27sin(30) - 27sin(36) - 3\cdot3^{(\frac{1}{2})}sin(30) + 3\cdot3^{(\frac{1}{2})}sin(36)))}{(21ln(3))}|_{8}^{13}=-5.97\end{align*}$ $\displaystyle \begin{align*}&\int_{8}^{13}\int_{10}^{12}\int_{-4}^{-2.5}(\frac{(z^{2}cos(3y))}{(7\cdot3^x)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{8}^{13}\int_{10}^{12}\int_{-4}^{-2.5}(\frac{(z^{2}cos(3y))}{(7\cdot3^x)})dxdydz=\int_{8}^{13}\int_{10}^{12}(-\frac{(z^{2}cos(3y))}{(7\cdot3^xln(3))})dydz|_{-4}^{-2.5}\\&\int_{8}^{13}\int_{10}^{12}(-\frac{(9z^{2}cos(3y)\cdot(3^{(\frac{1}{2})} - 9))}{(7ln(3))})dydz=\int_{8}^{13}(-\frac{(3z^{2}sin(3y)\cdot(3^{(\frac{1}{2})} - 9))}{(7ln(3))})dz|_{10}^{12}\\&\int_{8}^{13}(\frac{(3z^{2}\cdot(sin(30) - sin(36))\cdot(3^{(\frac{1}{2})} - 9))}{(7ln(3))})dz=-\frac{(z^{3}\cdot(27sin(30) - 27sin(36) - 3\cdot3^{(\frac{1}{2})}sin(30) + 3\cdot3^{(\frac{1}{2})}sin(36)))}{(21ln(3))}|_{8}^{13}=-5.97\end{align*}$

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Example Question #155 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{10}^{15}\int_{-10}^{-7}\int_{8}^{9}(\frac{(23cos(z + 2)sin(3y))}{(18x)})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{10}^{15}\int_{-10}^{-7}\int_{8}^{9}(\frac{(23cos(z + 2)sin(3y))}{(18x)})dxdydz\end{align*}$

Possible Answers:

0.12 $\displaystyle 0.12$

-0.06 $\displaystyle -0.06$

0.09 $\displaystyle 0.09$

-0.01 $\displaystyle -0.01$

Correct answer:

-0.01 $\displaystyle -0.01$

Explanation:

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

$\begin{align*}&\int_{10}^{15}\int_{-10}^{-7}\int_{8}^{9}(\frac{(23cos(z + 2)sin(3y))}{(18x)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{10}^{15}\int_{-10}^{-7}\int_{8}^{9}(\frac{(23cos(z + 2)sin(3y))}{(18x)})dxdydz=\int_{10}^{15}\int_{-10}^{-7}(\frac{(23cos(z + 2)sin(3y)ln(x))}{18})dydz|_{8}^{9}\\&\int_{10}^{15}\int_{-10}^{-7}(\frac{(23cos(z + 2)sin(3y)ln(\frac{9}{8}))}{18})dydz=\int_{10}^{15}(-\frac{(23cos(3y)cos(z + 2)ln(\frac{9}{8}))}{54})dz|_{-10}^{-7}\\&\int_{10}^{15}(-\frac{(23cos(z + 2)ln(\frac{9}{8})\cdot(\frac{cos(21)}{3}-\frac{ cos(30)}{3}))}{18})dz=-\frac{(23sin(z + 2)ln(\frac{9}{8})\cdot(\frac{cos(21)}{3}-\frac{ cos(30)}{3}))}{18}|_{10}^{15}=-0.01\end{align*}$ $\displaystyle \begin{align*}&\int_{10}^{15}\int_{-10}^{-7}\int_{8}^{9}(\frac{(23cos(z + 2)sin(3y))}{(18x)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{10}^{15}\int_{-10}^{-7}\int_{8}^{9}(\frac{(23cos(z + 2)sin(3y))}{(18x)})dxdydz=\int_{10}^{15}\int_{-10}^{-7}(\frac{(23cos(z + 2)sin(3y)ln(x))}{18})dydz|_{8}^{9}\\&\int_{10}^{15}\int_{-10}^{-7}(\frac{(23cos(z + 2)sin(3y)ln(\frac{9}{8}))}{18})dydz=\int_{10}^{15}(-\frac{(23cos(3y)cos(z + 2)ln(\frac{9}{8}))}{54})dz|_{-10}^{-7}\\&\int_{10}^{15}(-\frac{(23cos(z + 2)ln(\frac{9}{8})\cdot(\frac{cos(21)}{3}-\frac{ cos(30)}{3}))}{18})dz=-\frac{(23sin(z + 2)ln(\frac{9}{8})\cdot(\frac{cos(21)}{3}-\frac{ cos(30)}{3}))}{18}|_{10}^{15}=-0.01\end{align*}$

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Example Question #156 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{-3}^{-0.5}\int_{-4.5}^{-1.5}\int_{5}^{10}(\frac{(21\cdot2^{(\frac{x}{4})}\cdot3^{(\frac{y}{2})}\cdot3^{(\frac{z}{2})})}{2})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{-3}^{-0.5}\int_{-4.5}^{-1.5}\int_{5}^{10}(\frac{(21\cdot2^{(\frac{x}{4})}\cdot3^{(\frac{y}{2})}\cdot3^{(\frac{z}{2})})}{2})dxdydz\end{align*}$

Possible Answers:

264.7 $\displaystyle 264.7$

132.3 $\displaystyle 132.3$

-26.47 $\displaystyle -26.47$

$\displaystyle -794$

Correct answer:

132.3 $\displaystyle 132.3$

Explanation:

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

$\begin{align*}&\int_{-3}^{-0.5}\int_{-4.5}^{-1.5}\int_{5}^{10}(\frac{(21\cdot2^{(\frac{x}{4})}\cdot3^{(\frac{y}{2})}\cdot3^{(\frac{z}{2})})}{2})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-3}^{-0.5}\int_{-4.5}^{-1.5}\int_{5}^{10}(\frac{(21\cdot2^{(\frac{x}{4})}\cdot3^{(\frac{y}{2})}\cdot3^{(\frac{z}{2})})}{2})dxdydz=\int_{-3}^{-0.5}\int_{-4.5}^{-1.5}(\frac{(42\cdot2^{(\frac{x}{4})}\cdot3^{(\frac{y}{2}+\frac{ z}{2})})}{ln(2)})dydz|_{5}^{10}\end{align*}$ $\displaystyle \begin{align*}&\int_{-3}^{-0.5}\int_{-4.5}^{-1.5}\int_{5}^{10}(\frac{(21\cdot2^{(\frac{x}{4})}\cdot3^{(\frac{y}{2})}\cdot3^{(\frac{z}{2})})}{2})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-3}^{-0.5}\int_{-4.5}^{-1.5}\int_{5}^{10}(\frac{(21\cdot2^{(\frac{x}{4})}\cdot3^{(\frac{y}{2})}\cdot3^{(\frac{z}{2})})}{2})dxdydz=\int_{-3}^{-0.5}\int_{-4.5}^{-1.5}(\frac{(42\cdot2^{(\frac{x}{4})}\cdot3^{(\frac{y}{2}+\frac{ z}{2})})}{ln(2)})dydz|_{5}^{10}\end{align*}$

$\begin{align*}\\&\int_{-3}^{-0.5}\int_{-4.5}^{-1.5}(\frac{(84\cdot2^{(\frac{1}{4})}\cdot3^{(\frac{y}{2})}\cdot3^{(\frac{z}{2})}\cdot(2\cdot2^{(\frac{1}{4})} - 1))}{ln(2)})dydz=\int_{-3}^{-0.5}(\frac{(2\cdot(168\cdot2^{(\frac{1}{2})}\cdot3^{(\frac{y}{2})}\cdot3^{(\frac{z}{2})} - 84\cdot2^{(\frac{1}{4})}\cdot3^{(\frac{y}{2})}\cdot3^{(\frac{z}{2})}))}{(ln(2)ln(3))})dz|_{-4.5}^{-1.5}\\&\int_{-3}^{-0.5}(-\frac{(56\cdot2^{(\frac{1}{4})}\cdot3^{(\frac{1}{4})}\cdot3^{(\frac{z}{2})}\cdot(2\cdot2^{(\frac{1}{4})} - 1)\cdot(3^{(\frac{1}{2})} - 9))}{(9ln(2)ln(3))})dz=\frac{(2\cdot3^{(\frac{z}{2})}\cdot(1008\cdot2^{(\frac{1}{2})}\cdot3^{(\frac{1}{4})} - 112\cdot2^{(\frac{1}{2})}\cdot27^{(\frac{1}{4})} - 504\cdot6^{(\frac{1}{4})} + 56\cdot54^{(\frac{1}{4})}))}{(9ln(2)ln(3)^{2})}|_{-3}^{-0.5}=132.3\end{align*}$ $\displaystyle \begin{align*}\\&\int_{-3}^{-0.5}\int_{-4.5}^{-1.5}(\frac{(84\cdot2^{(\frac{1}{4})}\cdot3^{(\frac{y}{2})}\cdot3^{(\frac{z}{2})}\cdot(2\cdot2^{(\frac{1}{4})} - 1))}{ln(2)})dydz=\int_{-3}^{-0.5}(\frac{(2\cdot(168\cdot2^{(\frac{1}{2})}\cdot3^{(\frac{y}{2})}\cdot3^{(\frac{z}{2})} - 84\cdot2^{(\frac{1}{4})}\cdot3^{(\frac{y}{2})}\cdot3^{(\frac{z}{2})}))}{(ln(2)ln(3))})dz|_{-4.5}^{-1.5}\\&\int_{-3}^{-0.5}(-\frac{(56\cdot2^{(\frac{1}{4})}\cdot3^{(\frac{1}{4})}\cdot3^{(\frac{z}{2})}\cdot(2\cdot2^{(\frac{1}{4})} - 1)\cdot(3^{(\frac{1}{2})} - 9))}{(9ln(2)ln(3))})dz=\frac{(2\cdot3^{(\frac{z}{2})}\cdot(1008\cdot2^{(\frac{1}{2})}\cdot3^{(\frac{1}{4})} - 112\cdot2^{(\frac{1}{2})}\cdot27^{(\frac{1}{4})} - 504\cdot6^{(\frac{1}{4})} + 56\cdot54^{(\frac{1}{4})}))}{(9ln(2)ln(3)^{2})}|_{-3}^{-0.5}=132.3\end{align*}$

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Example Question #157 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{-6}^{-4}\int_{-10}^{-7.5}\int_{4}^{8.5}(\frac{(cos(3y)cos(z + 1))}{(2\cdot2^{(\frac{x}{2})})})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{-6}^{-4}\int_{-10}^{-7.5}\int_{4}^{8.5}(\frac{(cos(3y)cos(z + 1))}{(2\cdot2^{(\frac{x}{2})})})dxdydz\end{align*}$

Possible Answers:

0.01 $\displaystyle 0.01$

-0.01 $\displaystyle -0.01$

-0.31 $\displaystyle -0.31$

0.05 $\displaystyle 0.05$

Correct answer:

0.05 $\displaystyle 0.05$

Explanation:

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

$\begin{align*}&\int_{-6}^{-4}\int_{-10}^{-7.5}\int_{4}^{8.5}(\frac{(cos(3y)cos(z + 1))}{(2\cdot2^{(\frac{x}{2})})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-6}^{-4}\int_{-10}^{-7.5}\int_{4}^{8.5}(\frac{(cos(3y)cos(z + 1))}{(2\cdot2^{(\frac{x}{2})})})dxdydz=\int_{-6}^{-4}\int_{-10}^{-7.5}(-\frac{(cos(3y)cos(z + 1))}{(2^{(\frac{x}{2})}ln(2))})dydz|_{4}^{8.5}\\&\int_{-6}^{-4}\int_{-10}^{-7.5}(-\frac{(cos(3y)cos(z + 1)\cdot(2^{(\frac{3}{4})} - 8))}{(32ln(2))})dydz=\int_{-6}^{-4}(-\frac{(cos(z + 1)sin(3y)\cdot(2^{(\frac{3}{4})} - 8))}{(96ln(2))})dz|_{-10}^{-7.5}\\&\int_{-6}^{-4}(-\frac{(cos(z + 1)\cdot(sin(30) - sin(\frac{45}{2}))\cdot(2^{(\frac{3}{4})} - 8))}{(96ln(2))})dz=-\frac{(sin(z + 1)\cdot(sin(30) - sin(\frac{45}{2}))\cdot(2^{(\frac{3}{4})} - 8))}{(96ln(2))}|_{-6}^{-4}=0.05\end{align*}$ $\displaystyle \begin{align*}&\int_{-6}^{-4}\int_{-10}^{-7.5}\int_{4}^{8.5}(\frac{(cos(3y)cos(z + 1))}{(2\cdot2^{(\frac{x}{2})})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-6}^{-4}\int_{-10}^{-7.5}\int_{4}^{8.5}(\frac{(cos(3y)cos(z + 1))}{(2\cdot2^{(\frac{x}{2})})})dxdydz=\int_{-6}^{-4}\int_{-10}^{-7.5}(-\frac{(cos(3y)cos(z + 1))}{(2^{(\frac{x}{2})}ln(2))})dydz|_{4}^{8.5}\\&\int_{-6}^{-4}\int_{-10}^{-7.5}(-\frac{(cos(3y)cos(z + 1)\cdot(2^{(\frac{3}{4})} - 8))}{(32ln(2))})dydz=\int_{-6}^{-4}(-\frac{(cos(z + 1)sin(3y)\cdot(2^{(\frac{3}{4})} - 8))}{(96ln(2))})dz|_{-10}^{-7.5}\\&\int_{-6}^{-4}(-\frac{(cos(z + 1)\cdot(sin(30) - sin(\frac{45}{2}))\cdot(2^{(\frac{3}{4})} - 8))}{(96ln(2))})dz=-\frac{(sin(z + 1)\cdot(sin(30) - sin(\frac{45}{2}))\cdot(2^{(\frac{3}{4})} - 8))}{(96ln(2))}|_{-6}^{-4}=0.05\end{align*}$

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Example Question #158 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{2}^{6.5}\int_{7}^{10.5}\int_{10}^{13}(\frac{(10cos(x + 2))}{(7y^{2})})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{2}^{6.5}\int_{7}^{10.5}\int_{10}^{13}(\frac{(10cos(x + 2))}{(7y^{2})})dxdydz\end{align*}$

Possible Answers:

0.73 $\displaystyle 0.73$

-0.12 $\displaystyle -0.12$

0.36 $\displaystyle 0.36$

-1.82 $\displaystyle -1.82$

Correct answer:

0.36 $\displaystyle 0.36$

Explanation:

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

$\begin{align*}&\int_{2}^{6.5}\int_{7}^{10.5}\int_{10}^{13}(\frac{(10cos(x + 2))}{(7y^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{2}^{6.5}\int_{7}^{10.5}\int_{10}^{13}(\frac{(10cos(x + 2))}{(7y^{2})})dxdydz=\int_{2}^{6.5}\int_{7}^{10.5}(\frac{(10sin(x + 2))}{(7y^{2})})dydz|_{10}^{13}\\&\int_{2}^{6.5}\int_{7}^{10.5}(-\frac{(10\cdot(sin(12) - sin(15)))}{(7y^{2})})dydz=\int_{2}^{6.5}(\frac{(\frac{(10sin(12))}{7}-\frac{ (10sin(15))}{7})}{y})dz|_{7}^{10.5}\\&\int_{2}^{6.5}(\frac{(10sin(15))}{147}-\frac{ (10sin(12))}{147})dz=-z\cdot(\frac{(10sin(12))}{147}-\frac{ (10sin(15))}{147})|_{2}^{6.5}=0.36\end{align*}$ $\displaystyle \begin{align*}&\int_{2}^{6.5}\int_{7}^{10.5}\int_{10}^{13}(\frac{(10cos(x + 2))}{(7y^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{2}^{6.5}\int_{7}^{10.5}\int_{10}^{13}(\frac{(10cos(x + 2))}{(7y^{2})})dxdydz=\int_{2}^{6.5}\int_{7}^{10.5}(\frac{(10sin(x + 2))}{(7y^{2})})dydz|_{10}^{13}\\&\int_{2}^{6.5}\int_{7}^{10.5}(-\frac{(10\cdot(sin(12) - sin(15)))}{(7y^{2})})dydz=\int_{2}^{6.5}(\frac{(\frac{(10sin(12))}{7}-\frac{ (10sin(15))}{7})}{y})dz|_{7}^{10.5}\\&\int_{2}^{6.5}(\frac{(10sin(15))}{147}-\frac{ (10sin(12))}{147})dz=-z\cdot(\frac{(10sin(12))}{147}-\frac{ (10sin(15))}{147})|_{2}^{6.5}=0.36\end{align*}$

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Example Question #291 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{3}^{4.5}\int_{-5}^{-1}\int_{8}^{10.5}(\frac{(27cos(3x)e^{(-2y)}e^{(-z)})}{4})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{3}^{4.5}\int_{-5}^{-1}\int_{8}^{10.5}(\frac{(27cos(3x)e^{(-2y)}e^{(-z)})}{4})dxdydz\end{align*}$

Possible Answers:

$\displaystyle -3792$

3792 $\displaystyle 3792$

948.1 $\displaystyle 948.1$

$\displaystyle -237$

Correct answer:

948.1 $\displaystyle 948.1$

Explanation:

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

$\begin{align*}&\int_{3}^{4.5}\int_{-5}^{-1}\int_{8}^{10.5}(\frac{(27cos(3x)e^{(-2y)}e^{(-z)})}{4})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{3}^{4.5}\int_{-5}^{-1}\int_{8}^{10.5}(\frac{(27cos(3x)e^{(-2y)}e^{(-z)})}{4})dxdydz=\int_{3}^{4.5}\int_{-5}^{-1}(\frac{(9sin(3x)e^{(- 2y - z)})}{4})dydz|_{8}^{10.5}\\&\int_{3}^{4.5}\int_{-5}^{-1}(-e^{(- 2y - z)}\cdot(\frac{(9sin(24))}{4}-\frac{ (9sin(\frac{63}{2}))}{4}))dydz=\int_{3}^{4.5}(\frac{(9e^{(- 2y - z)}\cdot(sin(24) - sin(\frac{63}{2})))}{8})dz|_{-5}^{-1}\\&\int_{3}^{4.5}(-\frac{(9e^{(2 - z)}\cdot(sin(24) - sin(\frac{63}{2}))\cdot(e^{(8)} - 1))}{8})dz=\frac{(9e^{(2 - z)}\cdot(sin(24) - sin(\frac{63}{2}))\cdot(e^{(8)} - 1))}{8}|_{3}^{4.5}=948.1\end{align*}$ $\displaystyle \begin{align*}&\int_{3}^{4.5}\int_{-5}^{-1}\int_{8}^{10.5}(\frac{(27cos(3x)e^{(-2y)}e^{(-z)})}{4})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{3}^{4.5}\int_{-5}^{-1}\int_{8}^{10.5}(\frac{(27cos(3x)e^{(-2y)}e^{(-z)})}{4})dxdydz=\int_{3}^{4.5}\int_{-5}^{-1}(\frac{(9sin(3x)e^{(- 2y - z)})}{4})dydz|_{8}^{10.5}\\&\int_{3}^{4.5}\int_{-5}^{-1}(-e^{(- 2y - z)}\cdot(\frac{(9sin(24))}{4}-\frac{ (9sin(\frac{63}{2}))}{4}))dydz=\int_{3}^{4.5}(\frac{(9e^{(- 2y - z)}\cdot(sin(24) - sin(\frac{63}{2})))}{8})dz|_{-5}^{-1}\\&\int_{3}^{4.5}(-\frac{(9e^{(2 - z)}\cdot(sin(24) - sin(\frac{63}{2}))\cdot(e^{(8)} - 1))}{8})dz=\frac{(9e^{(2 - z)}\cdot(sin(24) - sin(\frac{63}{2}))\cdot(e^{(8)} - 1))}{8}|_{3}^{4.5}=948.1\end{align*}$

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Example Question #160 : Triple Integration In Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{3}^{7.5}\int_{-3}^{1.5}\int_{10}^{15}(\frac{(9\cdot3^{(\frac{y}{2})})}{(53\cdot2^zx)})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{3}^{7.5}\int_{-3}^{1.5}\int_{10}^{15}(\frac{(9\cdot3^{(\frac{y}{2})})}{(53\cdot2^zx)})dxdydz\end{align*}$

Possible Answers:

-0.18 $\displaystyle -0.18$

0.05 $\displaystyle 0.05$

-0.01 $\displaystyle -0.01$

0.01 $\displaystyle 0.01$

Correct answer:

0.05 $\displaystyle 0.05$

Explanation:

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

$\begin{align*}&\int_{3}^{7.5}\int_{-3}^{1.5}\int_{10}^{15}(\frac{(9\cdot3^{(\frac{y}{2})})}{(53\cdot2^zx)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{3}^{7.5}\int_{-3}^{1.5}\int_{10}^{15}(\frac{(9\cdot3^{(\frac{y}{2})})}{(53\cdot2^zx)})dxdydz=\int_{3}^{7.5}\int_{-3}^{1.5}(\frac{(9\cdot(\frac{1}{2})^z\cdot3^{(\frac{y}{2})}ln(x))}{53})dydz|_{10}^{15}\\&\int_{3}^{7.5}\int_{-3}^{1.5}(\frac{(9\cdot3^{(\frac{y}{2})}ln(\frac{3}{2}))}{(53\cdot2^z)})dydz=\int_{3}^{7.5}(\frac{(18\cdot(\frac{1}{2})^z\cdot3^{(\frac{y}{2})}ln(\frac{3}{2}))}{(53ln(3))})dz|_{-3}^{1.5}\\&\int_{3}^{7.5}(-\frac{(ln(\frac{3}{2})\cdot(2\cdot3^{(\frac{1}{2})} - 18\cdot3^{(\frac{3}{4})}))}{(53\cdot2^zln(3))})dz=\frac{(2\cdot(3^{(\frac{1}{2})}ln(\frac{3}{2}) - 9\cdot27^{(\frac{1}{4})}ln(\frac{3}{2})))}{(53\cdot2^zln(2)ln(3))}|_{3}^{7.5}=0.05\end{align*}$ $\displaystyle \begin{align*}&\int_{3}^{7.5}\int_{-3}^{1.5}\int_{10}^{15}(\frac{(9\cdot3^{(\frac{y}{2})})}{(53\cdot2^zx)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{3}^{7.5}\int_{-3}^{1.5}\int_{10}^{15}(\frac{(9\cdot3^{(\frac{y}{2})})}{(53\cdot2^zx)})dxdydz=\int_{3}^{7.5}\int_{-3}^{1.5}(\frac{(9\cdot(\frac{1}{2})^z\cdot3^{(\frac{y}{2})}ln(x))}{53})dydz|_{10}^{15}\\&\int_{3}^{7.5}\int_{-3}^{1.5}(\frac{(9\cdot3^{(\frac{y}{2})}ln(\frac{3}{2}))}{(53\cdot2^z)})dydz=\int_{3}^{7.5}(\frac{(18\cdot(\frac{1}{2})^z\cdot3^{(\frac{y}{2})}ln(\frac{3}{2}))}{(53ln(3))})dz|_{-3}^{1.5}\\&\int_{3}^{7.5}(-\frac{(ln(\frac{3}{2})\cdot(2\cdot3^{(\frac{1}{2})} - 18\cdot3^{(\frac{3}{4})}))}{(53\cdot2^zln(3))})dz=\frac{(2\cdot(3^{(\frac{1}{2})}ln(\frac{3}{2}) - 9\cdot27^{(\frac{1}{4})}ln(\frac{3}{2})))}{(53\cdot2^zln(2)ln(3))}|_{3}^{7.5}=0.05\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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Example Question #151 : Triple Integration In Cartesian Coordinates

Example Question #152 : Triple Integration In Cartesian Coordinates

Example Question #153 : Triple Integration In Cartesian Coordinates

Example Question #152 : Triple Integration In Cartesian Coordinates

Example Question #155 : Triple Integration In Cartesian Coordinates

Example Question #156 : Triple Integration In Cartesian Coordinates

Example Question #157 : Triple Integration In Cartesian Coordinates

Example Question #158 : Triple Integration In Cartesian Coordinates

Example Question #291 : Multiple Integration

Example Question #160 : Triple Integration In Cartesian Coordinates

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