# Answer in Calculus for loligny #206311

October 21st, 2022

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

y = 0, z = 0.

Let us find the volume of the solid bounded by the planes “x = y, x + y + z = 4,nny = 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_0^2dyint_y^{4-y}(4-x-y)dx=int_0^2(4x-frac{x^2}{2}-yx)|_y^{4-y}dy=nint_0^2(4(4-y)-frac{(4-y)^2}{2}-y(4-y)-4y+frac{y^2}{2}+y^2)dy=nint_0^2(16-4y-8+4y-frac{y^2}{2}-4y+y^2-4y+frac{y^2}{2}+y^2)dy=nint_0^2(2y^2-8y+8)dy=(frac{2y^3}{3}-4y^2+8y)|_0^2=frac{16}{3}-16+16=frac{16}{3}.”