# Evaluate the integral:int int int_E(xy+z^2)dV, where E={(x,y,z)|0<=x<=2,0<=y<=1,0<=z<3}

Evaluate the integral:
$\int \int {\int }_{E}\left(xy+{z}^{2}\right)dV$,
where $E=\left\{\left(x,y,z\right)\mid 0\le x\le 2,0\le y\le 1,0\le z<3\right\}$

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oppturf

Here we have to evalute the integral $\int \int \int \left(xy+{z}^{2}\right)dxdydz$ over the regin bounded by $0\le x\le 2,0\le y\le 1,0\le z<3$. Then
$\int \int \int \left(xy+{z}^{2}\right)dxdydz={\int }_{x=0}^{2}{\int }_{y=0}^{1}{\int }_{z=0}^{3}\left(xy+{z}^{2}\right)dxdydz$
$={\int }_{x=0}^{2}{\int }_{y=0}^{1}\left(xyz+\left(\frac{{z}^{3}}{3}\right){\mid }_{z=0}^{3}dxdy$
$={\int }_{x=0}^{2}{\int }_{y=0}^{1}\left(3xy+9\right)dxdy$
$={\int }_{x=0}^{2}\left(3x\frac{{y}^{2}}{2}+9y\right)\mid +{\left(y=0\right)}^{1}dx$
$={\int }_{x=0}^{2}\left(\frac{3x}{2}+9\right)dx$
$=\left(\frac{3{x}^{2}}{4}+9x\right){\mid }_{x=0}^{2}$
=(3+18)
=21