 # Determine the volume of the sold in the 1st octant bounded by x+y+2z=4 Rui Baldwin 2020-10-20 Answered
Determine the volume of the sold in the 1st octant bounded by $x+y+2z=4$
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Step 1
The given equation is:
$x+y+2z=4$
The above equation is solved as:
$2z=4-x-y$
$z=2-\frac{x}{2}-\frac{y}{2}$
$Thus,zvaries\in 0\le z\le 2-\frac{x}{2}-\frac{y}{2}$
In XY plane , the equation is:
x+y=4
Thus, y varies in $0\le y\le 4-x$
and x varies in $0\le x\le 4$
Step 2
The volume is given as:
$Volume={\int }_{0}^{4}{\int }_{0}^{4-x}{\int }_{0}^{2-\frac{x}{2}-\frac{y}{2}}dzdydx$
$={\int }_{0}^{4}{\int }_{0}^{4-x}{\left[z\right]}_{0}^{2-\frac{x}{2}-\frac{y}{2}}dydx$
$={\int }_{0}^{4}{\int }_{0}^{4-x}\left(2-\frac{x}{2}-\frac{y}{2}\right)dydx$
$={\int }_{0}^{4}{\left[2y-\frac{xy}{2}-\frac{{y}^{2}}{4}\right]}_{0}^{4-x}dx$
$={\int }_{0}^{4}\left[2\left(4-x\right)-\frac{x\left(4-x\right)}{2}-\frac{{\left(4-x\right)}^{2}}{4}\right]dx$
$={\int }_{0}^{4}\left[8-2x-2x+\frac{{x}^{2}}{2}-\frac{\left(16-8x+{x}^{2}\right)}{4}\right]dx$
$={\int }_{0}^{4}\left[8-2x-2x+\frac{{x}^{2}}{2}-4+2x-\frac{{x}^{2}}{4}\right]dx$
$={\int }_{0}^{4}\left[4-2x+\frac{{x}^{2}}{4}\right]dx$
$={\left[4x-\frac{2{x}^{2}}{2}+\frac{{x}^{3}}{12}\right]}_{0}^{4}$