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

\(\displaystyle \begin{align*}&\int_{-6}^{-5}\int_{-10}^{-6}\int_{9}^{14}(\frac{(9x^{2})}{(160yz^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-6}^{-5}\int_{-10}^{-6}\int_{9}^{14}(\frac{(9x^{2})}{(160yz^{2})})dxdydz=\int_{-6}^{-5}\int_{-10}^{-6}(\frac{(3x^{3})}{(160yz^{2})})dydz|_{9}^{14}\\&\int_{-6}^{-5}\int_{-10}^{-6}(\frac{1209}{(32yz^{2})})dydz=\int_{-6}^{-5}(\frac{(1209ln(y))}{(32z^{2})})dz|_{-10}^{-6}\\&\int_{-6}^{-5}(\frac{(1209ln(\frac{3}{5}))}{(32z^{2})})dz=-\frac{(1209ln(\frac{3}{5}))}{(32z)}|_{-6}^{-5}=-0.64\end{align*}\)

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

\(\displaystyle \begin{align*}&\int_{10}^{15}\int_{7}^{9.5}\int_{-7}^{-2.5}(\frac{(11cos(x + 2)sin(y + 1))}{(76z)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{10}^{15}\int_{7}^{9.5}\int_{-7}^{-2.5}(\frac{(11cos(x + 2)sin(y + 1))}{(76z)})dxdydz=\int_{10}^{15}\int_{7}^{9.5}(\frac{(11sin(x + 2)sin(y + 1))}{(76z)})dydz|_{-7}^{-2.5}\\&\int_{10}^{15}\int_{7}^{9.5}(-\frac{(11sin(y + 1)\cdot(sin(\frac{1}{2}) - sin(5)))}{(76z)})dydz=\int_{10}^{15}(\frac{(11cos(y + 1)\cdot(sin(\frac{1}{2}) - sin(5)))}{(76z)})dz|_{7}^{9.5}\\&\int_{10}^{15}(-\frac{(11\cdot(cos(8) - cos(\frac{21}{2}))\cdot(sin(\frac{1}{2}) - sin(5)))}{(76z)})dz=-ln(z)\cdot(\frac{(11cos(8)sin(\frac{1}{2}))}{76}-\frac{ (11cos(8)sin(5))}{76}-\frac{ (11cos(\frac{21}{2})sin(\frac{1}{2}))}{76}+\frac{ (11cos(\frac{21}{2})sin(5))}{76})|_{10}^{15}=-0.03\end{align*}\)

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

\(\displaystyle \begin{align*}&\int_{4}^{5.5}\int_{-9}^{-5.5}\int_{10}^{13}(\frac{(3\cdot3^z)}{(65x^{2}y^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{4}^{5.5}\int_{-9}^{-5.5}\int_{10}^{13}(\frac{(3\cdot3^z)}{(65x^{2}y^{2})})dxdydz=\int_{4}^{5.5}\int_{-9}^{-5.5}(-\frac{(3\cdot3^z)}{(65xy^{2})})dydz|_{10}^{13}\\&\int_{4}^{5.5}\int_{-9}^{-5.5}(\frac{(9\cdot3^z)}{(8450y^{2})})dydz=\int_{4}^{5.5}(-\frac{(9\cdot3^z)}{(8450y)})dz|_{-9}^{-5.5}\\&\int_{4}^{5.5}(\frac{(7\cdot3^z)}{92950})dz=\frac{(7\cdot3^z)}{(92950ln(3))}|_{4}^{5.5}=0.02\end{align*}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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