Question

Find the volume of the solid bounded by the planes x = y, x + y + z = 4,

y = 0, z = 0.

Answer 1

Let us find the volume of the solid bounded by the planes $x = y, x + y + z = 4, y = 0, z = 0.$ The plane $x+y+z=4$ intersects the plane $z=0$ by the line $x+y=4.$ The intesection in the plane $z=0$ of two lines $x=y$ and $x+y=4$ is the point $A(2,2).$ Therefore, the volume is equal to

$V=\int_0^2dy\int_y^{4-y}(4-x-y)dx=\int_0^2(4x-\frac{x^2}{2}-yx)|_y^{4-y}dy= \int_0^2(4(4-y)-\frac{(4-y)^2}{2}-y(4-y)-4y+\frac{y^2}{2}+y^2)dy= \int_0^2(16-4y-8+4y-\frac{y^2}{2}-4y+y^2-4y+\frac{y^2}{2}+y^2)dy= \int_0^2(2y^2-8y+8)dy=(\frac{2y^3}{3}-4y^2+8y)|_0^2=\frac{16}{3}-16+16=\frac{16}{3}.$