## Thomas' Calculus 13th Edition

V=$\int\int f(x,y)dA$ =$\int^{1}_{0} \int^{1}_{0}(2-x-y)dy dx$ =$\int^{1}_{0}[2y-xy-\frac{1}{2}y^2]^1_0dx$ =$\int^{1}_{0}(\frac{3}{2}-x)dx$ =$[\frac{3}{2}x-\frac{1}{2}x^2]^1_0$ =1