Question

use double integrals to find the volume of: the region beneath the plane y+z=1 and above the triangle with vertices (0,0), (1,1), and (0,1)

Answer 1

The volume will be given by

$\iint\limits_Df(x,y)dA$

Here D is the triangle, formed by y = x, y = 1, and the y-axis. $z=f(x,y)=1-y$

Hence

$\iint\limits_Df(x,y)dA=\int\limits_0^1dy\int\limits_0^y(1-y)dx=\int\limits_0^1y(1-y)dy=\left(\frac{y^2}{2}-\frac{y^3}{3}\right)_0^1=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$