Question

Find the volume of the region in the first octant bounded by coordinate planes and the planes x+z=1,y+2z=2

Answer 1

$z= 1-x; z= \frac{2-y}{2}\\ 1-x = \frac{2-y}{2}\\ 2-2x=2-y\\ y=2x\\ z= 0 \implies y= 2\\ Volume = \int\int_D\int _{z=0}^{\frac{2-y}{2}}dzdA = \int\int_D \frac{2-y}{2}dA\\ =\int_0^1 \int_{2x}^{2} \frac{2-y}{2}dydx\\ =x^2-2x+1\\ =\int _0^1\left(x^2-2x+1\right)dx\\ =\int _0^1x^2- \int _0^12x+\int _0^11dx\\ =\frac{1}{3}$