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Calculus 3 : Calculus 3

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Example Questions

Example Question #304 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{6}^{7}\int_{3}^{7.5}\int_{-3.5}^{-0.5}(10cos(3z)e^{(-2x)}e^{(y)})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{6}^{7}\int_{3}^{7.5}\int_{-3.5}^{-0.5}(10cos(3z)e^{(-2x)}e^{(y)})dxdydz\end{align*}$

Possible Answers:

$1.035\cdot10^{7}$ $\displaystyle 1.035\cdot10^{7}$

$5.175\cdot10^{6}$ $\displaystyle 5.175\cdot10^{6}$

$-3.105\cdot10^{7}$ $\displaystyle -3.105\cdot10^{7}$

$-1.725\cdot10^{6}$ $\displaystyle -1.725\cdot10^{6}$

Correct answer:

$5.175\cdot10^{6}$ $\displaystyle 5.175\cdot10^{6}$

Explanation:

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

$\begin{align*}&\int_{6}^{7}\int_{3}^{7.5}\int_{-3.5}^{-0.5}(10cos(3z)e^{(-2x)}e^{(y)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{6}^{7}\int_{3}^{7.5}\int_{-3.5}^{-0.5}(10cos(3z)e^{(-2x)}e^{(y)})dxdydz=\int_{6}^{7}\int_{3}^{7.5}(-5cos(3z)e^{(y - 2x)})dydz|_{-3.5}^{-0.5}\\&\int_{6}^{7}\int_{3}^{7.5}(5cos(3z)e^{(1)}e^{(y)}\cdot(e^{(6)} - 1))dydz=\int_{6}^{7}(5cos(3z)e^{(y + 1)}\cdot(e^{(6)} - 1))dz|_{3}^{7.5}\\&\int_{6}^{7}(5cos(3z)e^{(1)}\cdot(e^{(3)} - e^{(\frac{15}{2})}) - 5cos(3z)e^{(7)}\cdot(e^{(3)} - e^{(\frac{15}{2})}))dz=sin(3z)\cdot(\frac{(5e^{(4)})}{3}-\frac{ (5e^{(10)})}{3}-\frac{ (5e^{(\frac{17}{2})})}{3}+\frac{ (5e^{(\frac{29}{2})})}{3})|_{6}^{7}=5.175\cdot10^{6}\end{align*}$ $\displaystyle \begin{align*}&\int_{6}^{7}\int_{3}^{7.5}\int_{-3.5}^{-0.5}(10cos(3z)e^{(-2x)}e^{(y)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{6}^{7}\int_{3}^{7.5}\int_{-3.5}^{-0.5}(10cos(3z)e^{(-2x)}e^{(y)})dxdydz=\int_{6}^{7}\int_{3}^{7.5}(-5cos(3z)e^{(y - 2x)})dydz|_{-3.5}^{-0.5}\\&\int_{6}^{7}\int_{3}^{7.5}(5cos(3z)e^{(1)}e^{(y)}\cdot(e^{(6)} - 1))dydz=\int_{6}^{7}(5cos(3z)e^{(y + 1)}\cdot(e^{(6)} - 1))dz|_{3}^{7.5}\\&\int_{6}^{7}(5cos(3z)e^{(1)}\cdot(e^{(3)} - e^{(\frac{15}{2})}) - 5cos(3z)e^{(7)}\cdot(e^{(3)} - e^{(\frac{15}{2})}))dz=sin(3z)\cdot(\frac{(5e^{(4)})}{3}-\frac{ (5e^{(10)})}{3}-\frac{ (5e^{(\frac{17}{2})})}{3}+\frac{ (5e^{(\frac{29}{2})})}{3})|_{6}^{7}=5.175\cdot10^{6}\end{align*}$

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Example Question #305 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{9}^{14}\int_{-3.5}^{1}\int_{3.5}^{6.5}(\frac{(31\cdot3^{(\frac{x}{4})}cos(z + 2))}{(6\cdot3^{(\frac{y}{2})})})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{9}^{14}\int_{-3.5}^{1}\int_{3.5}^{6.5}(\frac{(31\cdot3^{(\frac{x}{4})}cos(z + 2))}{(6\cdot3^{(\frac{y}{2})})})dxdydz\end{align*}$

Possible Answers:

-255.4 $\displaystyle -255.4$

2044 $\displaystyle 2044$

510.9 $\displaystyle 510.9$

$\displaystyle -1022$

Correct answer:

510.9 $\displaystyle 510.9$

Explanation:

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

$\begin{align*}&\int_{9}^{14}\int_{-3.5}^{1}\int_{3.5}^{6.5}(\frac{(31\cdot3^{(\frac{x}{4})}cos(z + 2))}{(6\cdot3^{(\frac{y}{2})})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{9}^{14}\int_{-3.5}^{1}\int_{3.5}^{6.5}(\frac{(31\cdot3^{(\frac{x}{4})}cos(z + 2))}{(6\cdot3^{(\frac{y}{2})})})dxdydz=\int_{9}^{14}\int_{-3.5}^{1}(\frac{(62\cdot3^{(\frac{x}{4}-\frac{ y}{2})}cos(z + 2))}{(3ln(3))})dydz|_{3.5}^{6.5}\end{align*}$ $\displaystyle \begin{align*}&\int_{9}^{14}\int_{-3.5}^{1}\int_{3.5}^{6.5}(\frac{(31\cdot3^{(\frac{x}{4})}cos(z + 2))}{(6\cdot3^{(\frac{y}{2})})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{9}^{14}\int_{-3.5}^{1}\int_{3.5}^{6.5}(\frac{(31\cdot3^{(\frac{x}{4})}cos(z + 2))}{(6\cdot3^{(\frac{y}{2})})})dxdydz=\int_{9}^{14}\int_{-3.5}^{1}(\frac{(62\cdot3^{(\frac{x}{4}-\frac{ y}{2})}cos(z + 2))}{(3ln(3))})dydz|_{3.5}^{6.5}\end{align*}$

$\begin{align*}\\&\int_{9}^{14}\int_{-3.5}^{1}(-\frac{(62\cdot3^{(\frac{5}{8})}cos(z + 2)\cdot(3^{(\frac{1}{4})} - 3))}{(3\cdot3^{(\frac{y}{2})}ln(3))})dydz=\int_{9}^{14}(-\frac{(124\cdot(3\cdot243^{(\frac{1}{8})}cos(z + 2) - 2187^{(\frac{1}{8})}cos(z + 2)))}{(3\cdot3^{(\frac{y}{2})}ln(3)^{2})})dz|_{-3.5}^{1}\\&\int_{9}^{14}(-\frac{(124\cdot3^{(\frac{1}{8})}cos(z + 2)\cdot(9\cdot3^{(\frac{1}{2})} - 28\cdot3^{(\frac{1}{4})} + 3))}{(3ln(3)^{2})})dz=-\frac{(124\cdot3^{(\frac{1}{8})}sin(z + 2)\cdot(9\cdot3^{(\frac{1}{2})} - 28\cdot3^{(\frac{1}{4})} + 3))}{(3ln(3)^{2})}|_{9}^{14}=510.9\end{align*}$ $\displaystyle \begin{align*}\\&\int_{9}^{14}\int_{-3.5}^{1}(-\frac{(62\cdot3^{(\frac{5}{8})}cos(z + 2)\cdot(3^{(\frac{1}{4})} - 3))}{(3\cdot3^{(\frac{y}{2})}ln(3))})dydz=\int_{9}^{14}(-\frac{(124\cdot(3\cdot243^{(\frac{1}{8})}cos(z + 2) - 2187^{(\frac{1}{8})}cos(z + 2)))}{(3\cdot3^{(\frac{y}{2})}ln(3)^{2})})dz|_{-3.5}^{1}\\&\int_{9}^{14}(-\frac{(124\cdot3^{(\frac{1}{8})}cos(z + 2)\cdot(9\cdot3^{(\frac{1}{2})} - 28\cdot3^{(\frac{1}{4})} + 3))}{(3ln(3)^{2})})dz=-\frac{(124\cdot3^{(\frac{1}{8})}sin(z + 2)\cdot(9\cdot3^{(\frac{1}{2})} - 28\cdot3^{(\frac{1}{4})} + 3))}{(3ln(3)^{2})}|_{9}^{14}=510.9\end{align*}$

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Example Question #306 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{9}^{10.5}\int_{-9}^{-4.5}\int_{-4.5}^{-2}(\frac{(23cos(3z))}{(7\cdot3^xy^{3})})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{9}^{10.5}\int_{-9}^{-4.5}\int_{-4.5}^{-2}(\frac{(23cos(3z))}{(7\cdot3^xy^{3})})dxdydz\end{align*}$

Possible Answers:

-0.42 $\displaystyle -0.42$

-4.23 $\displaystyle -4.23$

6.34 $\displaystyle 6.34$

2.12 $\displaystyle 2.12$

Correct answer:

2.12 $\displaystyle 2.12$

Explanation:

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

$\begin{align*}&\int_{9}^{10.5}\int_{-9}^{-4.5}\int_{-4.5}^{-2}(\frac{(23cos(3z))}{(7\cdot3^xy^{3})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{9}^{10.5}\int_{-9}^{-4.5}\int_{-4.5}^{-2}(\frac{(23cos(3z))}{(7\cdot3^xy^{3})})dxdydz=\int_{9}^{10.5}\int_{-9}^{-4.5}(-\frac{(23cos(3z))}{(7\cdot3^xy^{3}ln(3))})dydz|_{-4.5}^{-2}\\&\int_{9}^{10.5}\int_{-9}^{-4.5}(\frac{(207cos(3z)\cdot(9\cdot3^{(\frac{1}{2})} - 1))}{(7y^{3}ln(3))})dydz=\int_{9}^{10.5}(\frac{(207cos(3z) - 1863\cdot3^{(\frac{1}{2})}cos(3z))}{(14y^{2}ln(3))})dz|_{-9}^{-4.5}\\&\int_{9}^{10.5}(\frac{(23cos(3z) - 207\cdot3^{(\frac{1}{2})}cos(3z))}{(42ln(3))})dz=\frac{(23sin(3z) - 207\cdot3^{(\frac{1}{2})}sin(3z))}{(126ln(3))}|_{9}^{10.5}=2.12\end{align*}$ $\displaystyle \begin{align*}&\int_{9}^{10.5}\int_{-9}^{-4.5}\int_{-4.5}^{-2}(\frac{(23cos(3z))}{(7\cdot3^xy^{3})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{9}^{10.5}\int_{-9}^{-4.5}\int_{-4.5}^{-2}(\frac{(23cos(3z))}{(7\cdot3^xy^{3})})dxdydz=\int_{9}^{10.5}\int_{-9}^{-4.5}(-\frac{(23cos(3z))}{(7\cdot3^xy^{3}ln(3))})dydz|_{-4.5}^{-2}\\&\int_{9}^{10.5}\int_{-9}^{-4.5}(\frac{(207cos(3z)\cdot(9\cdot3^{(\frac{1}{2})} - 1))}{(7y^{3}ln(3))})dydz=\int_{9}^{10.5}(\frac{(207cos(3z) - 1863\cdot3^{(\frac{1}{2})}cos(3z))}{(14y^{2}ln(3))})dz|_{-9}^{-4.5}\\&\int_{9}^{10.5}(\frac{(23cos(3z) - 207\cdot3^{(\frac{1}{2})}cos(3z))}{(42ln(3))})dz=\frac{(23sin(3z) - 207\cdot3^{(\frac{1}{2})}sin(3z))}{(126ln(3))}|_{9}^{10.5}=2.12\end{align*}$

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Example Question #307 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{9}^{10.5}\int_{7}^{9.5}\int_{8}^{13}(2sin(x + 2)sin(4y))dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{9}^{10.5}\int_{7}^{9.5}\int_{8}^{13}(2sin(x + 2)sin(4y))dxdydz\end{align*}$

Possible Answers:

0.02 $\displaystyle 0.02$

-0.57 $\displaystyle -0.57$

-0.03 $\displaystyle -0.03$

0.11 $\displaystyle 0.11$

Correct answer:

0.11 $\displaystyle 0.11$

Explanation:

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

$\begin{align*}&\int_{9}^{10.5}\int_{7}^{9.5}\int_{8}^{13}(2sin(x + 2)sin(4y))dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{9}^{10.5}\int_{7}^{9.5}\int_{8}^{13}(2sin(x + 2)sin(4y))dxdydz=\int_{9}^{10.5}\int_{7}^{9.5}(-2cos(x + 2)sin(4y))dydz|_{8}^{13}\\&\int_{9}^{10.5}\int_{7}^{9.5}(2sin(4y)\cdot(cos(10) - cos(15)))dydz=\int_{9}^{10.5}(-cos(4y)\cdot(\frac{cos(10)}{2}-\frac{ cos(15)}{2}))dz|_{7}^{9.5}\\&\int_{9}^{10.5}(\frac{((cos(10) - cos(15))\cdot(cos(28) - cos(38)))}{2})dz=\frac{(z\cdot(cos(10) - cos(15))\cdot(cos(28) - cos(38)))}{2}|_{9}^{10.5}=0.11\end{align*}$ $\displaystyle \begin{align*}&\int_{9}^{10.5}\int_{7}^{9.5}\int_{8}^{13}(2sin(x + 2)sin(4y))dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{9}^{10.5}\int_{7}^{9.5}\int_{8}^{13}(2sin(x + 2)sin(4y))dxdydz=\int_{9}^{10.5}\int_{7}^{9.5}(-2cos(x + 2)sin(4y))dydz|_{8}^{13}\\&\int_{9}^{10.5}\int_{7}^{9.5}(2sin(4y)\cdot(cos(10) - cos(15)))dydz=\int_{9}^{10.5}(-cos(4y)\cdot(\frac{cos(10)}{2}-\frac{ cos(15)}{2}))dz|_{7}^{9.5}\\&\int_{9}^{10.5}(\frac{((cos(10) - cos(15))\cdot(cos(28) - cos(38)))}{2})dz=\frac{(z\cdot(cos(10) - cos(15))\cdot(cos(28) - cos(38)))}{2}|_{9}^{10.5}=0.11\end{align*}$

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Example Question #308 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{9}^{14}\int_{-6}^{-1}\int_{4.5}^{9.5}(\frac{(sin(3y)e^{(2x)})}{(6z)})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{9}^{14}\int_{-6}^{-1}\int_{4.5}^{9.5}(\frac{(sin(3y)e^{(2x)})}{(6z)})dxdydz\end{align*}$

Possible Answers:

$\displaystyle -723000$

$-2.169\cdot10^{7}$ $\displaystyle -2.169\cdot10^{7}$

$1.807\cdot10^{6}$ $\displaystyle 1.807\cdot10^{6}$

$3.615\cdot10^{6}$ $\displaystyle 3.615\cdot10^{6}$

Correct answer:

$3.615\cdot10^{6}$ $\displaystyle 3.615\cdot10^{6}$

Explanation:

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

$\begin{align*}&\int_{9}^{14}\int_{-6}^{-1}\int_{4.5}^{9.5}(\frac{(sin(3y)e^{(2x)})}{(6z)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{9}^{14}\int_{-6}^{-1}\int_{4.5}^{9.5}(\frac{(sin(3y)e^{(2x)})}{(6z)})dxdydz=\int_{9}^{14}\int_{-6}^{-1}(\frac{(sin(3y)e^{(2x)})}{(12z)})dydz|_{4.5}^{9.5}\\&\int_{9}^{14}\int_{-6}^{-1}(\frac{(sin(3y)e^{(9)}\cdot(e^{(10)} - 1))}{(12z)})dydz=\int_{9}^{14}(-\frac{(cos(3y)e^{(9)}\cdot(e^{(10)} - 1))}{(36z)})dz|_{-6}^{-1}\\&\int_{9}^{14}(-\frac{(e^{(9)}\cdot(cos(3) - cos(18))\cdot(e^{(10)} - 1))}{(36z)})dz=ln(z)\cdot(\frac{(cos(3)e^{(9)})}{36}-\frac{ (cos(3)e^{(19)})}{36}-\frac{ (cos(18)e^{(9)})}{36}+\frac{ (cos(18)e^{(19)})}{36})|_{9}^{14}=3.615\cdot10^{6}\end{align*}$ $\displaystyle \begin{align*}&\int_{9}^{14}\int_{-6}^{-1}\int_{4.5}^{9.5}(\frac{(sin(3y)e^{(2x)})}{(6z)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{9}^{14}\int_{-6}^{-1}\int_{4.5}^{9.5}(\frac{(sin(3y)e^{(2x)})}{(6z)})dxdydz=\int_{9}^{14}\int_{-6}^{-1}(\frac{(sin(3y)e^{(2x)})}{(12z)})dydz|_{4.5}^{9.5}\\&\int_{9}^{14}\int_{-6}^{-1}(\frac{(sin(3y)e^{(9)}\cdot(e^{(10)} - 1))}{(12z)})dydz=\int_{9}^{14}(-\frac{(cos(3y)e^{(9)}\cdot(e^{(10)} - 1))}{(36z)})dz|_{-6}^{-1}\\&\int_{9}^{14}(-\frac{(e^{(9)}\cdot(cos(3) - cos(18))\cdot(e^{(10)} - 1))}{(36z)})dz=ln(z)\cdot(\frac{(cos(3)e^{(9)})}{36}-\frac{ (cos(3)e^{(19)})}{36}-\frac{ (cos(18)e^{(9)})}{36}+\frac{ (cos(18)e^{(19)})}{36})|_{9}^{14}=3.615\cdot10^{6}\end{align*}$

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Example Question #785 : Calculus 3

$\begin{align*}&\text{Evaluate the triple integral}\int_{9}^{10}\int_{10}^{14}\int_{-10}^{-9}(\frac{(9ycos(3z)sin(4x))}{61})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{9}^{10}\int_{10}^{14}\int_{-10}^{-9}(\frac{(9ycos(3z)sin(4x))}{61})dxdydz\end{align*}$

Possible Answers:

-3.09 $\displaystyle -3.09$

0.62 $\displaystyle 0.62$

-0.31 $\displaystyle -0.31$

0.21 $\displaystyle 0.21$

Correct answer:

0.62 $\displaystyle 0.62$

Explanation:

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

$\begin{align*}&\int_{9}^{10}\int_{10}^{14}\int_{-10}^{-9}(\frac{(9ycos(3z)sin(4x))}{61})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{9}^{10}\int_{10}^{14}\int_{-10}^{-9}(\frac{(9ycos(3z)sin(4x))}{61})dxdydz=\int_{9}^{10}\int_{10}^{14}(-\frac{(9ycos(4x)cos(3z))}{244})dydz|_{-10}^{-9}\\&\int_{9}^{10}\int_{10}^{14}(-\frac{(9ycos(3z)\cdot(\frac{cos(36)}{4}-\frac{ cos(40)}{4}))}{61})dydz=\int_{9}^{10}(-\frac{(9y^{2}cos(3z)\cdot(cos(36) - cos(40)))}{488})dz|_{10}^{14}\\&\int_{9}^{10}(-\frac{(432cos(3z)\cdot(\frac{cos(36)}{4}-\frac{ cos(40)}{4}))}{61})dz=-sin(3z)\cdot(\frac{(36cos(36))}{61}-\frac{ (36cos(40))}{61})|_{9}^{10}=0.62\end{align*}$ $\displaystyle \begin{align*}&\int_{9}^{10}\int_{10}^{14}\int_{-10}^{-9}(\frac{(9ycos(3z)sin(4x))}{61})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{9}^{10}\int_{10}^{14}\int_{-10}^{-9}(\frac{(9ycos(3z)sin(4x))}{61})dxdydz=\int_{9}^{10}\int_{10}^{14}(-\frac{(9ycos(4x)cos(3z))}{244})dydz|_{-10}^{-9}\\&\int_{9}^{10}\int_{10}^{14}(-\frac{(9ycos(3z)\cdot(\frac{cos(36)}{4}-\frac{ cos(40)}{4}))}{61})dydz=\int_{9}^{10}(-\frac{(9y^{2}cos(3z)\cdot(cos(36) - cos(40)))}{488})dz|_{10}^{14}\\&\int_{9}^{10}(-\frac{(432cos(3z)\cdot(\frac{cos(36)}{4}-\frac{ cos(40)}{4}))}{61})dz=-sin(3z)\cdot(\frac{(36cos(36))}{61}-\frac{ (36cos(40))}{61})|_{9}^{10}=0.62\end{align*}$

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Example Question #309 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{9}^{11}\int_{-10}^{-5}\int_{7}^{8.5}(\frac{(20cos(3x)cos(4y)sin(z + 2))}{3})dxdydz\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the triple integral}\int_{9}^{11}\int_{-10}^{-5}\int_{7}^{8.5}(\frac{(20cos(3x)cos(4y)sin(z + 2))}{3})dxdydz\end{align*}$

Possible Answers:

0.2 $\displaystyle 0.2$

-0.01 $\displaystyle -0.01$

0.01 $\displaystyle 0.01$

-0.04 $\displaystyle -0.04$

Correct answer:

-0.04 $\displaystyle -0.04$

Explanation:

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

$\begin{align*}&\int_{9}^{11}\int_{-10}^{-5}\int_{7}^{8.5}(\frac{(20cos(3x)cos(4y)sin(z + 2))}{3})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{9}^{11}\int_{-10}^{-5}\int_{7}^{8.5}(\frac{(20cos(3x)cos(4y)sin(z + 2))}{3})dxdydz=\int_{9}^{11}\int_{-10}^{-5}(\frac{(20cos(4y)sin(3x)sin(z + 2))}{9})dydz|_{7}^{8.5}\\&\int_{9}^{11}\int_{-10}^{-5}(-\frac{(20cos(4y)sin(z + 2)\cdot(\frac{sin(21)}{3}-\frac{ sin(\frac{51}{2})}{3}))}{3})dydz=\int_{9}^{11}(-\frac{(5sin(4y)sin(z + 2)\cdot(sin(21) - sin(\frac{51}{2})))}{9})dz|_{-10}^{-5}\\&\int_{9}^{11}(\frac{(5sin(z + 2)\cdot(sin(20) - sin(40))\cdot(sin(21) - sin(\frac{51}{2})))}{9})dz=-\frac{(5cos(z + 2)\cdot(sin(20) - sin(40))\cdot(sin(21) - sin(\frac{51}{2})))}{9}|_{9}^{11}=-0.04\end{align*}$ $\displaystyle \begin{align*}&\int_{9}^{11}\int_{-10}^{-5}\int_{7}^{8.5}(\frac{(20cos(3x)cos(4y)sin(z + 2))}{3})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{9}^{11}\int_{-10}^{-5}\int_{7}^{8.5}(\frac{(20cos(3x)cos(4y)sin(z + 2))}{3})dxdydz=\int_{9}^{11}\int_{-10}^{-5}(\frac{(20cos(4y)sin(3x)sin(z + 2))}{9})dydz|_{7}^{8.5}\\&\int_{9}^{11}\int_{-10}^{-5}(-\frac{(20cos(4y)sin(z + 2)\cdot(\frac{sin(21)}{3}-\frac{ sin(\frac{51}{2})}{3}))}{3})dydz=\int_{9}^{11}(-\frac{(5sin(4y)sin(z + 2)\cdot(sin(21) - sin(\frac{51}{2})))}{9})dz|_{-10}^{-5}\\&\int_{9}^{11}(\frac{(5sin(z + 2)\cdot(sin(20) - sin(40))\cdot(sin(21) - sin(\frac{51}{2})))}{9})dz=-\frac{(5cos(z + 2)\cdot(sin(20) - sin(40))\cdot(sin(21) - sin(\frac{51}{2})))}{9}|_{9}^{11}=-0.04\end{align*}$

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Example Question #311 : Multiple Integration

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

Possible Answers:

0.32 $\displaystyle 0.32$

-1.62 $\displaystyle -1.62$

3.24 $\displaystyle 3.24$

-0.4 $\displaystyle -0.4$

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 : Multiple Integration

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

Possible Answers:

-7.42 $\displaystyle -7.42$

-89.01 $\displaystyle -89.01$

22.25 $\displaystyle 22.25$

89.01 $\displaystyle 89.01$

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 : Multiple Integration

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

Possible Answers:

-79.19 $\displaystyle -79.19$

-950.2 $\displaystyle -950.2$

950.2 $\displaystyle 950.2$

475.1 $\displaystyle 475.1$

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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Calculus 3 : Calculus 3

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