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Calculus 3 : Triple Integrals

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Example Questions

Example Question #311 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{-3.5}^{1}\int_{-9}^{-7}\int_{-6}^{-1}(\frac{(2^{(\frac{z}{2})}sin(y + 1))}{11})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{-3.5}^{1}\int_{-9}^{-7}\int_{-6}^{-1}(\frac{(2^{(\frac{z}{2})}sin(y + 1))}{11})dxdydz\end{align*}$

Possible Answers:

-0.4 $\displaystyle -0.4$

3.24 $\displaystyle 3.24$

-1.62 $\displaystyle -1.62$

0.32 $\displaystyle 0.32$

Correct answer:

-1.62 $\displaystyle -1.62$

Explanation:

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

$\begin{align*}&\int_{-3.5}^{1}\int_{-9}^{-7}\int_{-6}^{-1}(\frac{(2^{(\frac{z}{2})}sin(y + 1))}{11})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-3.5}^{1}\int_{-9}^{-7}\int_{-6}^{-1}(\frac{(2^{(\frac{z}{2})}sin(y + 1))}{11})dxdydz=\int_{-3.5}^{1}\int_{-9}^{-7}(\frac{(2^{(\frac{z}{2})}xsin(y + 1))}{11})dydz|_{-6}^{-1}\\&\int_{-3.5}^{1}\int_{-9}^{-7}(\frac{(5\cdot2^{(\frac{z}{2})}sin(y + 1))}{11})dydz=\int_{-3.5}^{1}(-\frac{(5\cdot2^{(\frac{z}{2})}cos(y + 1))}{11})dz|_{-9}^{-7}\\&\int_{-3.5}^{1}(-\frac{(5\cdot2^{(\frac{z}{2})}\cdot(cos(6) - cos(8)))}{11})dz=-\frac{(10\cdot2^{(\frac{z}{2})}\cdot(cos(6) - cos(8)))}{(11ln(2))}|_{-3.5}^{1}=-1.62\end{align*}$ $\displaystyle \begin{align*}&\int_{-3.5}^{1}\int_{-9}^{-7}\int_{-6}^{-1}(\frac{(2^{(\frac{z}{2})}sin(y + 1))}{11})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-3.5}^{1}\int_{-9}^{-7}\int_{-6}^{-1}(\frac{(2^{(\frac{z}{2})}sin(y + 1))}{11})dxdydz=\int_{-3.5}^{1}\int_{-9}^{-7}(\frac{(2^{(\frac{z}{2})}xsin(y + 1))}{11})dydz|_{-6}^{-1}\\&\int_{-3.5}^{1}\int_{-9}^{-7}(\frac{(5\cdot2^{(\frac{z}{2})}sin(y + 1))}{11})dydz=\int_{-3.5}^{1}(-\frac{(5\cdot2^{(\frac{z}{2})}cos(y + 1))}{11})dz|_{-9}^{-7}\\&\int_{-3.5}^{1}(-\frac{(5\cdot2^{(\frac{z}{2})}\cdot(cos(6) - cos(8)))}{11})dz=-\frac{(10\cdot2^{(\frac{z}{2})}\cdot(cos(6) - cos(8)))}{(11ln(2))}|_{-3.5}^{1}=-1.62\end{align*}$

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Example Question #312 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{-4.5}^{2.25}\int_{-8}^{-7}\int_{7}^{8.5}(\frac{(61e^{(-2z)})}{(5x^{2}y^{2})})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{-4.5}^{2.25}\int_{-8}^{-7}\int_{7}^{8.5}(\frac{(61e^{(-2z)})}{(5x^{2}y^{2})})dxdydz\end{align*}$

Possible Answers:

-89.01 $\displaystyle -89.01$

22.25 $\displaystyle 22.25$

89.01 $\displaystyle 89.01$

-7.42 $\displaystyle -7.42$

Correct answer:

22.25 $\displaystyle 22.25$

Explanation:

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

$\begin{align*}&\int_{-4.5}^{2.25}\int_{-8}^{-7}\int_{7}^{8.5}(\frac{(61e^{(-2z)})}{(5x^{2}y^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-4.5}^{2.25}\int_{-8}^{-7}\int_{7}^{8.5}(\frac{(61e^{(-2z)})}{(5x^{2}y^{2})})dxdydz=\int_{-4.5}^{2.25}\int_{-8}^{-7}(-\frac{(61e^{(-2z)})}{(5xy^{2})})dydz|_{7}^{8.5}\\&\int_{-4.5}^{2.25}\int_{-8}^{-7}(\frac{(183e^{(-2z)})}{(595y^{2})})dydz=\int_{-4.5}^{2.25}(-\frac{(183e^{(-2z)})}{(595y)})dz|_{-8}^{-7}\\&\int_{-4.5}^{2.25}(\frac{(183e^{(-2z)})}{33320})dz=-\frac{(183e^{(-2z)})}{66640}|_{-4.5}^{2.25}=22.25\end{align*}$ $\displaystyle \begin{align*}&\int_{-4.5}^{2.25}\int_{-8}^{-7}\int_{7}^{8.5}(\frac{(61e^{(-2z)})}{(5x^{2}y^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-4.5}^{2.25}\int_{-8}^{-7}\int_{7}^{8.5}(\frac{(61e^{(-2z)})}{(5x^{2}y^{2})})dxdydz=\int_{-4.5}^{2.25}\int_{-8}^{-7}(-\frac{(61e^{(-2z)})}{(5xy^{2})})dydz|_{7}^{8.5}\\&\int_{-4.5}^{2.25}\int_{-8}^{-7}(\frac{(183e^{(-2z)})}{(595y^{2})})dydz=\int_{-4.5}^{2.25}(-\frac{(183e^{(-2z)})}{(595y)})dz|_{-8}^{-7}\\&\int_{-4.5}^{2.25}(\frac{(183e^{(-2z)})}{33320})dz=-\frac{(183e^{(-2z)})}{66640}|_{-4.5}^{2.25}=22.25\end{align*}$

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Example Question #313 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{10}^{14.5}\int_{2}^{7}\int_{5}^{8}(\frac{(53z^{2})}{(9\cdot3^{(\frac{x}{2})}y)})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{10}^{14.5}\int_{2}^{7}\int_{5}^{8}(\frac{(53z^{2})}{(9\cdot3^{(\frac{x}{2})}y)})dxdydz\end{align*}$

Possible Answers:

-950.2 $\displaystyle -950.2$

475.1 $\displaystyle 475.1$

950.2 $\displaystyle 950.2$

-79.19 $\displaystyle -79.19$

Correct answer:

475.1 $\displaystyle 475.1$

Explanation:

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

$\begin{align*}&\int_{10}^{14.5}\int_{2}^{7}\int_{5}^{8}(\frac{(53z^{2})}{(9\cdot3^{(\frac{x}{2})}y)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{10}^{14.5}\int_{2}^{7}\int_{5}^{8}(\frac{(53z^{2})}{(9\cdot3^{(\frac{x}{2})}y)})dxdydz=\int_{10}^{14.5}\int_{2}^{7}(-\frac{(106z^{2})}{(9\cdot3^{(\frac{x}{2})}yln(3))})dydz|_{5}^{8}\\&\int_{10}^{14.5}\int_{2}^{7}(\frac{(53z^{2}\cdot(6\cdot3^{(\frac{1}{2})} - 2))}{(729yln(3))})dydz=\int_{10}^{14.5}(\frac{(ln(y)\cdot(318\cdot3^{(\frac{1}{2})}z^{2} - 106z^{2}))}{(729ln(3))})dz|_{2}^{7}\\&\int_{10}^{14.5}(\frac{(106z^{2}ln(\frac{7}{2})\cdot(3\cdot3^{(\frac{1}{2})} - 1))}{(729ln(3))})dz=-\frac{(z^{3}\cdot(106ln(\frac{7}{2}) - 318\cdot3^{(\frac{1}{2})}ln(\frac{7}{2})))}{(2187ln(3))}|_{10}^{14.5}=475.1\end{align*}$ $\displaystyle \begin{align*}&\int_{10}^{14.5}\int_{2}^{7}\int_{5}^{8}(\frac{(53z^{2})}{(9\cdot3^{(\frac{x}{2})}y)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{10}^{14.5}\int_{2}^{7}\int_{5}^{8}(\frac{(53z^{2})}{(9\cdot3^{(\frac{x}{2})}y)})dxdydz=\int_{10}^{14.5}\int_{2}^{7}(-\frac{(106z^{2})}{(9\cdot3^{(\frac{x}{2})}yln(3))})dydz|_{5}^{8}\\&\int_{10}^{14.5}\int_{2}^{7}(\frac{(53z^{2}\cdot(6\cdot3^{(\frac{1}{2})} - 2))}{(729yln(3))})dydz=\int_{10}^{14.5}(\frac{(ln(y)\cdot(318\cdot3^{(\frac{1}{2})}z^{2} - 106z^{2}))}{(729ln(3))})dz|_{2}^{7}\\&\int_{10}^{14.5}(\frac{(106z^{2}ln(\frac{7}{2})\cdot(3\cdot3^{(\frac{1}{2})} - 1))}{(729ln(3))})dz=-\frac{(z^{3}\cdot(106ln(\frac{7}{2}) - 318\cdot3^{(\frac{1}{2})}ln(\frac{7}{2})))}{(2187ln(3))}|_{10}^{14.5}=475.1\end{align*}$

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Example Question #314 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{9}^{11.5}\int_{10}^{13.5}\int_{6}^{9}(\frac{(11sin(x + 1))}{35})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{9}^{11.5}\int_{10}^{13.5}\int_{6}^{9}(\frac{(11sin(x + 1))}{35})dxdydz\end{align*}$

Possible Answers:

4.38 $\displaystyle 4.38$

13.14 $\displaystyle 13.14$

-1.46 $\displaystyle -1.46$

-8.76 $\displaystyle -8.76$

Correct answer:

4.38 $\displaystyle 4.38$

Explanation:

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

$\begin{align*}&\int_{9}^{11.5}\int_{10}^{13.5}\int_{6}^{9}(\frac{(11sin(x + 1))}{35})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{9}^{11.5}\int_{10}^{13.5}\int_{6}^{9}(\frac{(11sin(x + 1))}{35})dxdydz=\int_{9}^{11.5}\int_{10}^{13.5}(-\frac{(11cos(x + 1))}{35})dydz|_{6}^{9}\\&\int_{9}^{11.5}\int_{10}^{13.5}(\frac{(11cos(7))}{35}-\frac{ (11cos(10))}{35})dydz=\int_{9}^{11.5}(y\cdot(\frac{(11cos(7))}{35}-\frac{ (11cos(10))}{35}))dz|_{10}^{13.5}\\&\int_{9}^{11.5}(\frac{(11cos(7))}{10}-\frac{ (11cos(10))}{10})dz=z\cdot(\frac{(11cos(7))}{10}-\frac{ (11cos(10))}{10})|_{9}^{11.5}=4.38\end{align*}$ $\displaystyle \begin{align*}&\int_{9}^{11.5}\int_{10}^{13.5}\int_{6}^{9}(\frac{(11sin(x + 1))}{35})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{9}^{11.5}\int_{10}^{13.5}\int_{6}^{9}(\frac{(11sin(x + 1))}{35})dxdydz=\int_{9}^{11.5}\int_{10}^{13.5}(-\frac{(11cos(x + 1))}{35})dydz|_{6}^{9}\\&\int_{9}^{11.5}\int_{10}^{13.5}(\frac{(11cos(7))}{35}-\frac{ (11cos(10))}{35})dydz=\int_{9}^{11.5}(y\cdot(\frac{(11cos(7))}{35}-\frac{ (11cos(10))}{35}))dz|_{10}^{13.5}\\&\int_{9}^{11.5}(\frac{(11cos(7))}{10}-\frac{ (11cos(10))}{10})dz=z\cdot(\frac{(11cos(7))}{10}-\frac{ (11cos(10))}{10})|_{9}^{11.5}=4.38\end{align*}$

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Example Question #315 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{-4.5}^{-3.5}\int_{5}^{7.5}\int_{9}^{11.5}(\frac{(34e^{(-y)}e^{(-z)})}{(3x)})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{-4.5}^{-3.5}\int_{5}^{7.5}\int_{9}^{11.5}(\frac{(34e^{(-y)}e^{(-z)})}{(3x)})dxdydz\end{align*}$

Possible Answers:

-3.91 $\displaystyle -3.91$

0.98 $\displaystyle 0.98$

0.16 $\displaystyle 0.16$

-0.49 $\displaystyle -0.49$

Correct answer:

0.98 $\displaystyle 0.98$

Explanation:

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

$\begin{align*}&\int_{-4.5}^{-3.5}\int_{5}^{7.5}\int_{9}^{11.5}(\frac{(34e^{(-y)}e^{(-z)})}{(3x)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-4.5}^{-3.5}\int_{5}^{7.5}\int_{9}^{11.5}(\frac{(34e^{(-y)}e^{(-z)})}{(3x)})dxdydz=\int_{-4.5}^{-3.5}\int_{5}^{7.5}(\frac{(34e^{(- y - z)}ln(x))}{3})dydz|_{9}^{11.5}\\&\int_{-4.5}^{-3.5}\int_{5}^{7.5}(\frac{(34e^{(- y - z)}ln(\frac{23}{18}))}{3})dydz=\int_{-4.5}^{-3.5}(-\frac{(34e^{(- y - z)}ln(\frac{23}{18}))}{3})dz|_{5}^{7.5}\\&\int_{-4.5}^{-3.5}(\frac{(34e^{(- z -\frac{ 15}{2})}ln(\frac{23}{18})\cdot(e^{(\frac{5}{2})} - 1))}{3})dz=-\frac{(34e^{(- z -\frac{ 15}{2})}ln(\frac{23}{18})\cdot(e^{(\frac{5}{2})} - 1))}{3}|_{-4.5}^{-3.5}=0.98\end{align*}$ $\displaystyle \begin{align*}&\int_{-4.5}^{-3.5}\int_{5}^{7.5}\int_{9}^{11.5}(\frac{(34e^{(-y)}e^{(-z)})}{(3x)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-4.5}^{-3.5}\int_{5}^{7.5}\int_{9}^{11.5}(\frac{(34e^{(-y)}e^{(-z)})}{(3x)})dxdydz=\int_{-4.5}^{-3.5}\int_{5}^{7.5}(\frac{(34e^{(- y - z)}ln(x))}{3})dydz|_{9}^{11.5}\\&\int_{-4.5}^{-3.5}\int_{5}^{7.5}(\frac{(34e^{(- y - z)}ln(\frac{23}{18}))}{3})dydz=\int_{-4.5}^{-3.5}(-\frac{(34e^{(- y - z)}ln(\frac{23}{18}))}{3})dz|_{5}^{7.5}\\&\int_{-4.5}^{-3.5}(\frac{(34e^{(- z -\frac{ 15}{2})}ln(\frac{23}{18})\cdot(e^{(\frac{5}{2})} - 1))}{3})dz=-\frac{(34e^{(- z -\frac{ 15}{2})}ln(\frac{23}{18})\cdot(e^{(\frac{5}{2})} - 1))}{3}|_{-4.5}^{-3.5}=0.98\end{align*}$

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Example Question #316 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{6}^{8.5}\int_{-6}^{-4.5}\int_{-9}^{-7}(\frac{(38x^{2}y^{2}z)}{3})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{6}^{8.5}\int_{-6}^{-4.5}\int_{-9}^{-7}(\frac{(38x^{2}y^{2}z)}{3})dxdydz\end{align*}$

Possible Answers:

$1.23\cdot10^{6}$ $\displaystyle 1.23\cdot10^{6}$

$\displaystyle -409900$

$3.689\cdot10^{6}$ $\displaystyle 3.689\cdot10^{6}$

$-4.918\cdot10^{6}$ $\displaystyle -4.918\cdot10^{6}$

Correct answer:

$1.23\cdot10^{6}$ $\displaystyle 1.23\cdot10^{6}$

Explanation:

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

$\begin{align*}&\int_{6}^{8.5}\int_{-6}^{-4.5}\int_{-9}^{-7}(\frac{(38x^{2}y^{2}z)}{3})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{6}^{8.5}\int_{-6}^{-4.5}\int_{-9}^{-7}(\frac{(38x^{2}y^{2}z)}{3})dxdydz=\int_{6}^{8.5}\int_{-6}^{-4.5}(\frac{(38x^{3}y^{2}z)}{9})dydz|_{-9}^{-7}\\&\int_{6}^{8.5}\int_{-6}^{-4.5}(\frac{(14668y^{2}z)}{9})dydz=\int_{6}^{8.5}(\frac{(14668y^{3}z)}{27})dz|_{-6}^{-4.5}\\&\int_{6}^{8.5}(\frac{(135679z)}{2})dz=\frac{(135679z^{2})}{4}|_{6}^{8.5}=1.23\cdot10^{6}\end{align*}$ $\displaystyle \begin{align*}&\int_{6}^{8.5}\int_{-6}^{-4.5}\int_{-9}^{-7}(\frac{(38x^{2}y^{2}z)}{3})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{6}^{8.5}\int_{-6}^{-4.5}\int_{-9}^{-7}(\frac{(38x^{2}y^{2}z)}{3})dxdydz=\int_{6}^{8.5}\int_{-6}^{-4.5}(\frac{(38x^{3}y^{2}z)}{9})dydz|_{-9}^{-7}\\&\int_{6}^{8.5}\int_{-6}^{-4.5}(\frac{(14668y^{2}z)}{9})dydz=\int_{6}^{8.5}(\frac{(14668y^{3}z)}{27})dz|_{-6}^{-4.5}\\&\int_{6}^{8.5}(\frac{(135679z)}{2})dz=\frac{(135679z^{2})}{4}|_{6}^{8.5}=1.23\cdot10^{6}\end{align*}$

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Example Question #317 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{6}^{9.5}\int_{9}^{10}\int_{8}^{9}(\frac{(9cos(z + 1)sin(y + 2))}{49})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{6}^{9.5}\int_{9}^{10}\int_{8}^{9}(\frac{(9cos(z + 1)sin(y + 2))}{49})dxdydz\end{align*}$

Possible Answers:

-0.95 $\displaystyle -0.95$

0.24 $\displaystyle 0.24$

0.06 $\displaystyle 0.06$

-0.06 $\displaystyle -0.06$

Correct answer:

0.24 $\displaystyle 0.24$

Explanation:

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

$\begin{align*}&\int_{6}^{9.5}\int_{9}^{10}\int_{8}^{9}(\frac{(9cos(z + 1)sin(y + 2))}{49})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{6}^{9.5}\int_{9}^{10}\int_{8}^{9}(\frac{(9cos(z + 1)sin(y + 2))}{49})dxdydz=\int_{6}^{9.5}\int_{9}^{10}(\frac{(9xcos(z + 1)sin(y + 2))}{49})dydz|_{8}^{9}\\&\int_{6}^{9.5}\int_{9}^{10}(\frac{(9cos(z + 1)sin(y + 2))}{49})dydz=\int_{6}^{9.5}(-\frac{(9cos(y + 2)cos(z + 1))}{49})dz|_{9}^{10}\\&\int_{6}^{9.5}(\frac{(9cos(z + 1)\cdot(cos(11) - cos(12)))}{49})dz=sin(z + 1)\cdot(\frac{(9cos(11))}{49}-\frac{ (9cos(12))}{49})|_{6}^{9.5}=0.24\end{align*}$ $\displaystyle \begin{align*}&\int_{6}^{9.5}\int_{9}^{10}\int_{8}^{9}(\frac{(9cos(z + 1)sin(y + 2))}{49})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{6}^{9.5}\int_{9}^{10}\int_{8}^{9}(\frac{(9cos(z + 1)sin(y + 2))}{49})dxdydz=\int_{6}^{9.5}\int_{9}^{10}(\frac{(9xcos(z + 1)sin(y + 2))}{49})dydz|_{8}^{9}\\&\int_{6}^{9.5}\int_{9}^{10}(\frac{(9cos(z + 1)sin(y + 2))}{49})dydz=\int_{6}^{9.5}(-\frac{(9cos(y + 2)cos(z + 1))}{49})dz|_{9}^{10}\\&\int_{6}^{9.5}(\frac{(9cos(z + 1)\cdot(cos(11) - cos(12)))}{49})dz=sin(z + 1)\cdot(\frac{(9cos(11))}{49}-\frac{ (9cos(12))}{49})|_{6}^{9.5}=0.24\end{align*}$

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Example Question #318 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{9}^{10.5}\int_{-3.5}^{-1.5}\int_{10}^{15}(\frac{(19cos(z + 2)sin(4x))}{(6\cdot3^y)})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{9}^{10.5}\int_{-3.5}^{-1.5}\int_{10}^{15}(\frac{(19cos(z + 2)sin(4x))}{(6\cdot3^y)})dxdydz\end{align*}$

Possible Answers:

3.99 $\displaystyle 3.99$

-1.6 $\displaystyle -1.6$

-23.95 $\displaystyle -23.95$

7.98 $\displaystyle 7.98$

Correct answer:

7.98 $\displaystyle 7.98$

Explanation:

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

$\begin{align*}&\int_{9}^{10.5}\int_{-3.5}^{-1.5}\int_{10}^{15}(\frac{(19cos(z + 2)sin(4x))}{(6\cdot3^y)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{9}^{10.5}\int_{-3.5}^{-1.5}\int_{10}^{15}(\frac{(19cos(z + 2)sin(4x))}{(6\cdot3^y)})dxdydz=\int_{9}^{10.5}\int_{-3.5}^{-1.5}(-\frac{(19cos(4x)cos(z + 2))}{(24\cdot3^y)})dydz|_{10}^{15}\\&\int_{9}^{10.5}\int_{-3.5}^{-1.5}(\frac{(19cos(z + 2)\cdot(\frac{cos(40)}{4}-\frac{ cos(60)}{4}))}{(6\cdot3^y)})dydz=\int_{9}^{10.5}(-\frac{(19cos(z + 2)\cdot(cos(40) - cos(60)))}{(24\cdot3^yln(3))})dz|_{-3.5}^{-1.5}\\&\int_{9}^{10.5}(\frac{(19\cdot3^{(\frac{1}{2})}cos(z + 2)\cdot(cos(40) - cos(60)))}{ln(3)})dz=\frac{(19\cdot3^{(\frac{1}{2})}sin(z + 2)\cdot(cos(40) - cos(60)))}{ln(3)}|_{9}^{10.5}=7.98\end{align*}$ $\displaystyle \begin{align*}&\int_{9}^{10.5}\int_{-3.5}^{-1.5}\int_{10}^{15}(\frac{(19cos(z + 2)sin(4x))}{(6\cdot3^y)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{9}^{10.5}\int_{-3.5}^{-1.5}\int_{10}^{15}(\frac{(19cos(z + 2)sin(4x))}{(6\cdot3^y)})dxdydz=\int_{9}^{10.5}\int_{-3.5}^{-1.5}(-\frac{(19cos(4x)cos(z + 2))}{(24\cdot3^y)})dydz|_{10}^{15}\\&\int_{9}^{10.5}\int_{-3.5}^{-1.5}(\frac{(19cos(z + 2)\cdot(\frac{cos(40)}{4}-\frac{ cos(60)}{4}))}{(6\cdot3^y)})dydz=\int_{9}^{10.5}(-\frac{(19cos(z + 2)\cdot(cos(40) - cos(60)))}{(24\cdot3^yln(3))})dz|_{-3.5}^{-1.5}\\&\int_{9}^{10.5}(\frac{(19\cdot3^{(\frac{1}{2})}cos(z + 2)\cdot(cos(40) - cos(60)))}{ln(3)})dz=\frac{(19\cdot3^{(\frac{1}{2})}sin(z + 2)\cdot(cos(40) - cos(60)))}{ln(3)}|_{9}^{10.5}=7.98\end{align*}$

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Example Question #319 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{7}^{10.5}\int_{8}^{11.5}\int_{10}^{13}(\frac{(58z^{2}cos(x + 1)cos(3y))}{11})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{7}^{10.5}\int_{8}^{11.5}\int_{10}^{13}(\frac{(58z^{2}cos(x + 1)cos(3y))}{11})dxdydz\end{align*}$

Possible Answers:

305 $\displaystyle 305$

-228.7 $\displaystyle -228.7$

-914.9 $\displaystyle -914.9$

914.9 $\displaystyle 914.9$

Correct answer:

914.9 $\displaystyle 914.9$

Explanation:

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

$\begin{align*}&\int_{7}^{10.5}\int_{8}^{11.5}\int_{10}^{13}(\frac{(58z^{2}cos(x + 1)cos(3y))}{11})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{7}^{10.5}\int_{8}^{11.5}\int_{10}^{13}(\frac{(58z^{2}cos(x + 1)cos(3y))}{11})dxdydz=\int_{7}^{10.5}\int_{8}^{11.5}(\frac{(58z^{2}cos(3y)sin(x + 1))}{11})dydz|_{10}^{13}\\&\int_{7}^{10.5}\int_{8}^{11.5}(-\frac{(58z^{2}cos(3y)\cdot(sin(11) - sin(14)))}{11})dydz=\int_{7}^{10.5}(-\frac{(58z^{2}sin(3y)\cdot(sin(11) - sin(14)))}{33})dz|_{8}^{11.5}\\&\int_{7}^{10.5}(\frac{(58z^{2}\cdot(sin(11) - sin(14))\cdot(sin(24) - sin(\frac{69}{2})))}{33})dz=z^{3}\cdot(\frac{(58sin(11)sin(24))}{99}-\frac{ (58sin(14)sin(24))}{99}-\frac{ (58sin(11)sin(\frac{69}{2}))}{99}+\frac{ (58sin(14)sin(\frac{69}{2}))}{99})|_{7}^{10.5}=914.9\end{align*}$ $\displaystyle \begin{align*}&\int_{7}^{10.5}\int_{8}^{11.5}\int_{10}^{13}(\frac{(58z^{2}cos(x + 1)cos(3y))}{11})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{7}^{10.5}\int_{8}^{11.5}\int_{10}^{13}(\frac{(58z^{2}cos(x + 1)cos(3y))}{11})dxdydz=\int_{7}^{10.5}\int_{8}^{11.5}(\frac{(58z^{2}cos(3y)sin(x + 1))}{11})dydz|_{10}^{13}\\&\int_{7}^{10.5}\int_{8}^{11.5}(-\frac{(58z^{2}cos(3y)\cdot(sin(11) - sin(14)))}{11})dydz=\int_{7}^{10.5}(-\frac{(58z^{2}sin(3y)\cdot(sin(11) - sin(14)))}{33})dz|_{8}^{11.5}\\&\int_{7}^{10.5}(\frac{(58z^{2}\cdot(sin(11) - sin(14))\cdot(sin(24) - sin(\frac{69}{2})))}{33})dz=z^{3}\cdot(\frac{(58sin(11)sin(24))}{99}-\frac{ (58sin(14)sin(24))}{99}-\frac{ (58sin(11)sin(\frac{69}{2}))}{99}+\frac{ (58sin(14)sin(\frac{69}{2}))}{99})|_{7}^{10.5}=914.9\end{align*}$

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Example Question #320 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{6}^{7.5}\int_{-4}^{0.5}\int_{-3.5}^{-2.5}(\frac{(7e^{(-2x)}e^{(-y)})}{(6z^{2})})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{6}^{7.5}\int_{-4}^{0.5}\int_{-3.5}^{-2.5}(\frac{(7e^{(-2x)}e^{(-y)})}{(6z^{2})})dxdydz\end{align*}$

Possible Answers:

$\displaystyle -3982$

995.5 $\displaystyle 995.5$

248.9 $\displaystyle 248.9$

-165.9 $\displaystyle -165.9$

Correct answer:

995.5 $\displaystyle 995.5$

Explanation:

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

$\begin{align*}&\int_{6}^{7.5}\int_{-4}^{0.5}\int_{-3.5}^{-2.5}(\frac{(7e^{(-2x)}e^{(-y)})}{(6z^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{6}^{7.5}\int_{-4}^{0.5}\int_{-3.5}^{-2.5}(\frac{(7e^{(-2x)}e^{(-y)})}{(6z^{2})})dxdydz=\int_{6}^{7.5}\int_{-4}^{0.5}(-\frac{(7e^{(- 2x - y)})}{(12z^{2})})dydz|_{-3.5}^{-2.5}\\&\int_{6}^{7.5}\int_{-4}^{0.5}(\frac{(7e^{(5)}e^{(-y)}\cdot(e^{(2)} - 1))}{(12z^{2})})dydz=\int_{6}^{7.5}(-\frac{(7e^{(5 - y)}\cdot(e^{(2)} - 1))}{(12z^{2})})dz|_{-4}^{0.5}\\&\int_{6}^{7.5}(\frac{(7e^{(\frac{9}{2})}\cdot(e^{(2)} - 1)\cdot(e^{(\frac{9}{2})} - 1))}{(12z^{2})})dz=\frac{(\frac{(7e^{(9)})}{12}-\frac{ (7e^{(\frac{9}{2})})}{12}-\frac{ (7e^{(11)})}{12}+\frac{ (7e^{(\frac{13}{2})})}{12})}{z}|_{6}^{7.5}=995.5\end{align*}$ $\displaystyle \begin{align*}&\int_{6}^{7.5}\int_{-4}^{0.5}\int_{-3.5}^{-2.5}(\frac{(7e^{(-2x)}e^{(-y)})}{(6z^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{6}^{7.5}\int_{-4}^{0.5}\int_{-3.5}^{-2.5}(\frac{(7e^{(-2x)}e^{(-y)})}{(6z^{2})})dxdydz=\int_{6}^{7.5}\int_{-4}^{0.5}(-\frac{(7e^{(- 2x - y)})}{(12z^{2})})dydz|_{-3.5}^{-2.5}\\&\int_{6}^{7.5}\int_{-4}^{0.5}(\frac{(7e^{(5)}e^{(-y)}\cdot(e^{(2)} - 1))}{(12z^{2})})dydz=\int_{6}^{7.5}(-\frac{(7e^{(5 - y)}\cdot(e^{(2)} - 1))}{(12z^{2})})dz|_{-4}^{0.5}\\&\int_{6}^{7.5}(\frac{(7e^{(\frac{9}{2})}\cdot(e^{(2)} - 1)\cdot(e^{(\frac{9}{2})} - 1))}{(12z^{2})})dz=\frac{(\frac{(7e^{(9)})}{12}-\frac{ (7e^{(\frac{9}{2})})}{12}-\frac{ (7e^{(11)})}{12}+\frac{ (7e^{(\frac{13}{2})})}{12})}{z}|_{6}^{7.5}=995.5\end{align*}$

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Calculus 3 : Triple Integrals

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