evaluate the triple integral xyz where e is the solid z0 to

Solution

The base of the tethraedron on the x-y plane (i.e. on the plane z=0) goes through the points (0,0,0), (1,0,0) and (1,1,0) which confine an area limited by the lines y=0, x=1 and y=x The plane that goes through (0,0,0), (1,1,0) and (1,0,1) is x-y-z = 0 as can be easily checked and which we can also write as z = x-y. Therefore the integrals are Int (x=0 to 1) Int (y=0 to x) Int (z=0 to x-y) xyz dz dy dx = Int (x=0 to 1) Int (y=0 to x) x y (x-y)^2 / 2 dy dx = (1/2) Int (x=0 to 1) Int (y=0 to x) (x^3 y - 2 x^2 y^2 + x y^3) dy dx = (1/2) Int (x=0 to 1) [ (1/2) x^5 - (2/3) x^5 + (1/4) x^5 ] dx = (1/2) (1/12) Int (x=0 to 1) x^5 dx = (1/2) (1/12) (1/6) = 1 / 144