AP Calculus BC - Triple Integration in Cartesian Coordinates

$\begin{align*}&\text{Evaluate the triple integral}\int_{10}^{11.5}\int_{-3.5}^{1.5}\int_{9}^{13.5}(25sin(3x)sin(4z)e^{(2y)})dxdydz\end{align*}$

Explanation

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

$\begin{align*}&\int_{10}^{11.5}\int_{-3.5}^{1.5}\int_{9}^{13.5}(25sin(3x)sin(4z)e^{(2y)})dxdydz\&\text{The approach is simply to take it step by step:}\&\int_{10}^{11.5}\int_{-3.5}^{1.5}\int_{9}^{13.5}(25sin(3x)sin(4z)e^{(2y)})dxdydz=\int_{10}^{11.5}\int_{-3.5}^{1.5}(-\frac{(25cos(3x)sin(4z)e^{(2y)})}{3})dydz|_{9}^{13.5}\&\int_{10}^{11.5}\int_{-3.5}^{1.5}(25sin(4z)e^{(2y)}\cdot (\frac{cos(27)}{3}-\frac{ cos(\frac{81}{2})}{3}))dydz=\int_{10}^{11.5}(\frac{(25sin(4z)e^{(2y)}\cdot (\frac{cos(27)}{3}-\frac{ cos(\frac{81}{2})}{3}))}{2})dz|_{-3.5}^{1.5}\&\int_{10}^{11.5}(\frac{(25sin(4z)e^{(-7)}\cdot (cos(27) - cos(\frac{81}{2}))\cdot (e^{(10)} - 1))}{6})dz=-\frac{(25cos(4z)e^{(-7)}\cdot (cos(27) - cos(\frac{81}{2}))\cdot (e^{(10)} - 1))}{24}|_{10}^{11.5}=-3.19\end{align*}$

$\begin{align*}&\text{Evaluate the triple integral}\int_{9}^{12}\int_{3.5}^{4.5}\int_{4.5}^{8}(\frac{(12\cdot 3^{(\frac{x}{4})}\cdot 3^{(\frac{y}{2})})}{(7z^{2})})dxdydz\end{align*}$

Explanation

$\begin{align*}&\int_{9}^{12}\int_{3.5}^{4.5}\int_{4.5}^{8}(\frac{(12\cdot 3^{(\frac{x}{4})}\cdot 3^{(\frac{y}{2})})}{(7z^{2})})dxdydz\&\text{The approach is simply to take it step by step:}\&\int_{9}^{12}\int_{3.5}^{4.5}\int_{4.5}^{8}(\frac{(12\cdot 3^{(\frac{x}{4})}\cdot 3^{(\frac{y}{2})})}{(7z^{2})})dxdydz=\int_{9}^{12}\int_{3.5}^{4.5}(\frac{(48\cdot 3^{(\frac{x}{4}+\frac{ y}{2})})}{(7z^{2}ln(3))})dydz|_{4.5}^{8}\&\int_{9}^{12}\int_{3.5}^{4.5}(-\frac{(144\cdot 3^{(\frac{y}{2})}\cdot (3^{(\frac{1}{8})} - 3))}{(7z^{2}ln(3))})dydz=\int_{9}^{12}(-\frac{(288\cdot 3^{(\frac{y}{2})}\cdot (3^{(\frac{1}{8})} - 3))}{(7z^{2}ln(3)^{2})})dz|_{3.5}^{4.5}\&\int_{9}^{12}(\frac{(7776\cdot 3^{(\frac{1}{4})} - 2592\cdot 3^{(\frac{3}{4})} - 2592\cdot 3^{(\frac{3}{8})} + 864\cdot 3^{(\frac{7}{8})})}{(7z^{2}ln(3)^{2})})dz=-\frac{(7776\cdot 3^{(\frac{1}{4})} - 2592\cdot 27^{(\frac{1}{4})} - 2592\cdot 27^{(\frac{1}{8})} + 864\cdot 2187^{(\frac{1}{8})})}{(7zln(3)^{2})}|_{9}^{12}=8.78\end{align*}$

$\begin{align*}&\text{Evaluate the triple integral}\int_{8}^{10}\int_{9}^{13.5}\int_{-3.5}^{-1.5}(\frac{(17\cdot 3^{(\frac{x}{3})}sin(y + 2))}{10})dxdydz\end{align*}$

Explanation

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

$\begin{align*}&\int_{8}^{10}\int_{9}^{13.5}\int_{-3.5}^{-1.5}(\frac{(17\cdot 3^{(\frac{x}{3})}sin(y + 2))}{10})dxdydz\&\text{The approach is simply to take it step by step:}\&\int_{8}^{10}\int_{9}^{13.5}\int_{-3.5}^{-1.5}(\frac{(17\cdot 3^{(\frac{x}{3})}sin(y + 2))}{10})dxdydz=\int_{8}^{10}\int_{9}^{13.5}(\frac{(51\cdot 3^{(\frac{x}{3})}sin(y + 2))}{(10ln(3))})dydz|_{-3.5}^{-1.5}\&\int_{8}^{10}\int_{9}^{13.5}(\frac{(17sin(y + 2)\cdot (3\cdot 3^{(\frac{1}{2})} - 3^{(\frac{5}{6})}))}{(30ln(3))})dydz=\int_{8}^{10}(-\frac{(17cos(y + 2)\cdot (3\cdot 3^{(\frac{1}{2})} - 3^{(\frac{5}{6})}))}{(30ln(3))})dz|_{9}^{13.5}\&\int_{8}^{10}(-\frac{(17\cdot 3^{(\frac{1}{2})}\cdot (cos(11) - cos(\frac{31}{2}))\cdot (3^{(\frac{1}{3})} - 3))}{(30ln(3))})dz=-\frac{(17\cdot 3^{(\frac{1}{2})}z\cdot (cos(11) - cos(\frac{31}{2}))\cdot (3^{(\frac{1}{3})} - 3))}{(30ln(3))}|_{8}^{10}=2.74\end{align*}$

$\begin{align*}&\text{Evaluate the triple integral}\int_{-7}^{-3}\int_{8}^{12}\int_{10}^{13.5}(\frac{(10cos(4y)sin(x + 2))}{(9z^{3})})dxdydz\end{align*}$

Explanation

$\begin{align*}&\int_{-7}^{-3}\int_{8}^{12}\int_{10}^{13.5}(\frac{(10cos(4y)sin(x + 2))}{(9z^{3})})dxdydz\&\text{The approach is simply to take it step by step:}\&\int_{-7}^{-3}\int_{8}^{12}\int_{10}^{13.5}(\frac{(10cos(4y)sin(x + 2))}{(9z^{3})})dxdydz=\int_{-7}^{-3}\int_{8}^{12}(-\frac{(10cos(x + 2)cos(4y))}{(9z^{3})})dydz|_{10}^{13.5}\&\int_{-7}^{-3}\int_{8}^{12}(\frac{(10cos(4y)\cdot (cos(12) - cos(\frac{31}{2})))}{(9z^{3})})dydz=\int_{-7}^{-3}(\frac{(5sin(4y)\cdot (cos(12) - cos(\frac{31}{2})))}{(18z^{3})})dz|_{8}^{12}\&\int_{-7}^{-3}(-\frac{(5\cdot (cos(12) - cos(\frac{31}{2}))\cdot (sin(32) - sin(48)))}{(18z^{3})})dz=\frac{(\frac{(5cos(12)sin(32))}{36}-\frac{ (5cos(12)sin(48))}{36}-\frac{ (5cos(\frac{31}{2})sin(32))}{36}+\frac{ (5cos(\frac{31}{2})sin(48))}{36})}{z^{2}}|_{-7}^{-3}=0.03\end{align*}$

$\begin{align*}&\text{Evaluate the triple integral}\int_{3.5}^{7.5}\int_{-6}^{-4.5}\int_{9}^{13}(34xcos(3y)e^{(-2z)})dxdydz\end{align*}$

Explanation

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

$\begin{align*}&\int_{3.5}^{7.5}\int_{-6}^{-4.5}\int_{9}^{13}(34xcos(3y)e^{(-2z)})dxdydz\&\text{The approach is simply to take it step by step:}\&\int_{3.5}^{7.5}\int_{-6}^{-4.5}\int_{9}^{13}(34xcos(3y)e^{(-2z)})dxdydz=\int_{3.5}^{7.5}\int_{-6}^{-4.5}(17x^{2}cos(3y)e^{(-2z)})dydz|_{9}^{13}\&\int_{3.5}^{7.5}\int_{-6}^{-4.5}(1496cos(3y)e^{(-2z)})dydz=\int_{3.5}^{7.5}(\frac{(1496sin(3y)e^{(-2z)})}{3})dz|_{-6}^{-4.5}\&\int_{3.5}^{7.5}(\frac{(1496e^{(-2z)}\cdot (sin(18) - sin(\frac{27}{2})))}{3})dz=-\frac{(748e^{(-2z)}\cdot (sin(18) - sin(\frac{27}{2})))}{3}|_{3.5}^{7.5}=-0.35\end{align*}$

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

$4.35\cdot10^{-6}$

$-1.45\cdot10^{-6}$

$1.45\cdot10^{-6}$

$-1.31\cdot10^{-5}$

Explanation

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

$\begin{align*}&\text{Evaluate the triple integral}\int_{10}^{14}\int_{-6}^{-1}\int_{7}^{11}(\frac{(11cos(y + 1)sin(z + 1))}{(2x)})dxdydz\end{align*}$

Explanation

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

$\begin{align*}&\int_{10}^{14}\int_{-6}^{-1}\int_{7}^{11}(\frac{(11cos(y + 1)sin(z + 1))}{(2x)})dxdydz\&\text{The approach is simply to take it step by step:}\&\int_{10}^{14}\int_{-6}^{-1}\int_{7}^{11}(\frac{(11cos(y + 1)sin(z + 1))}{(2x)})dxdydz=\int_{10}^{14}\int_{-6}^{-1}(\frac{(11cos(y + 1)sin(z + 1)ln(x))}{2})dydz|_{7}^{11}\&\int_{10}^{14}\int_{-6}^{-1}(\frac{(11cos(y + 1)sin(z + 1)ln(\frac{11}{7}))}{2})dydz=\int_{10}^{14}(\frac{(11sin(y + 1)sin(z + 1)ln(\frac{11}{7}))}{2})dz|_{-6}^{-1}\&\int_{10}^{14}(\frac{(11sin(z + 1)ln(\frac{11}{7})sin(5))}{2})dz=-\frac{(11cos(z + 1)ln(\frac{11}{7})sin(5))}{2}|_{10}^{14}=-1.82\end{align*}$

$\begin{align*}&\text{Evaluate the triple integral}\int_{-3}^{2}\int_{8}^{11.5}\int_{10}^{11}(\frac{(3ye^{(-z)})}{(43x)})dxdydz\end{align*}$

Explanation

$\begin{align*}&\int_{-3}^{2}\int_{8}^{11.5}\int_{10}^{11}(\frac{(3ye^{(-z)})}{(43x)})dxdydz\&\text{The approach is simply to take it step by step:}\&\int_{-3}^{2}\int_{8}^{11.5}\int_{10}^{11}(\frac{(3ye^{(-z)})}{(43x)})dxdydz=\int_{-3}^{2}\int_{8}^{11.5}(\frac{(3ye^{(-z)}ln(x))}{43})dydz|_{10}^{11}\&\int_{-3}^{2}\int_{8}^{11.5}(\frac{(3ye^{(-z)}ln(\frac{11}{10}))}{43})dydz=\int_{-3}^{2}(\frac{(3y^{2}e^{(-z)}ln(\frac{11}{10}))}{86})dz|_{8}^{11.5}\&\int_{-3}^{2}(\frac{(819e^{(-z)}ln(\frac{11}{10}))}{344})dz=-\frac{(819e^{(-z)}ln(\frac{11}{10}))}{344}|_{-3}^{2}=4.53\end{align*}$

$\begin{align*}&\text{Evaluate the triple integral}\int_{4}^{7}\int_{8}^{12}\int_{9}^{10}(\frac{(17cos(x + 2))}{(4yz)})dxdydz\end{align*}$

Explanation

$\begin{align*}&\int_{4}^{7}\int_{8}^{12}\int_{9}^{10}(\frac{(17cos(x + 2))}{(4yz)})dxdydz\&\text{The approach is simply to take it step by step:}\&\int_{4}^{7}\int_{8}^{12}\int_{9}^{10}(\frac{(17cos(x + 2))}{(4yz)})dxdydz=\int_{4}^{7}\int_{8}^{12}(\frac{(17sin(x + 2))}{(4yz)})dydz|_{9}^{10}\&\int_{4}^{7}\int_{8}^{12}(-\frac{(17\cdot(sin(11) - sin(12)))}{(4yz)})dydz=\int_{4}^{7}(-\frac{(17ln(y)\cdot(sin(11) - sin(12)))}{(4z)})dz|_{8}^{12}\&\int_{4}^{7}(-\frac{(17ln(\frac{3}{2})\cdot(sin(11) - sin(12)))}{(4z)})dz=-ln(z)\cdot(\frac{(17ln(\frac{3}{2})sin(11))}{4}-\frac{ (17ln(\frac{3}{2})sin(12))}{4})|_{4}^{7}=0.45\end{align*}$

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

$3.615\cdot10^{6}$

$1.807\cdot10^{6}$

$-2.169\cdot10^{7}$

Explanation

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

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

Triple Integration in Cartesian Coordinates

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