Calculus problem #2030

$z$ bounds: $0 \leq z \leq 2 - x - 2y$

Projection into xy-plane: $x + 2y = 2, x = 2y, x = 0$ .

$\rightarrow \frac{x}{2} \leq y \leq 1 - \frac{x}{2}$ with $x$ in $[0, 1]$ .

Therefore, the volume equals :

$=\int_{x=0}^{x=1}\int_{y=\frac{x}{2}}^{y=1-\frac{x}{2}}\int_{z=0}^{z=2-x-2y} 1 dz.dy.dx$

$=\int_{x=0}^{x=1}\int_{y=\frac{x}{2}}^{y=1-\frac{x}{2}} (2 - x - 2y) dy.dx$

$=\int_{x=0}^{x=1} [(2 - x)y - y^2]$ {for $y = \frac{x}{2}$ to $1 - \frac{x}{2}$ } $dx$

$=\int_{x=0}^{x=1} [(2 - x)(1 - \frac{x}{2}) - (1 - \frac{x}{2})^2] - [(2 - x) \frac{x}{2} - (\frac{x}{2})^2] dx$

$=\int_{x=0}^{x=1} (x^2 - 2x + 1) dx$

$=\int_{x=0}^{x=1} (x - 1)^2 dx$

$= \frac{(x - 1)^3}{3}$ {for $x = 0$ to $1$ }

$= \frac{1}{3}$

$=\boxed{0.333}$

solving we get vertices as O(0,0,0) , A(0,1,0), B(0,0,2) and C(1,0.5,0) therefore vol= 1/6 *[ a, b c]=0.3333