Evaluating triple integral with tetrahedron

Question

∫∫∫xyz dV where T is is the solid tetrahedron with vertices (0,0,0) (1,0,0) (1,1,0) (1,0,1).

How do I figure out what z=f(x,y) is with these points? The equation of a tetrahedron is x/a+y/b+z/c=1

But with these points I don't know how to use the formula

Answer 1

Denote the vertices of the tetrehedron T by O(0,0,0), A(1,0,0), B(1,1,0), and C(1,0,1). Then the sides OAB and OAC are on the xy and xz planes, with equations z=0 and y=0, respectively, ABC is parallel to the yz-plane, with equation x=1. To figure out the equation of the side OBC, we need to take the cross (vector) product of the vectors b=vec(OB)=i+j and c=vec(OC)=i+k. Then bxc = i–j–k = (1,–1,–1). Thus the equation of the plane OBC is x–y–z=0 or z = x–y. [Notice that since the plane passes through the origin, it has zero intercept, so the intercept form of the equation x/a+y/b+z/c=1 does not apply.]

To evaluate the integral, we turn it into an iterated integral:

First consider the region D on the xy-plane which is the side OAB. This is enclosed by the lines y=x, x=1, and y=0. We integrate with respect to z first, with z from 0 to x–y, then to y, with y from 0 to x; and finally to x, with x from 0 to 1. Thus we have

∫∫∫_Dxyzdxdydz = ∫₀¹xdx∫₀^xydy∫₀^x-yzdz = ∫₀¹xdx∫₀^xydy(x-y)²/2

=(1/2)∫₀¹xdx∫₀^xy(x-y)²dy = (1/2)∫₀¹xdx∫₀^x(x²y-2xy²-y³)dy

=(1/2)∫₀¹xdx(x²∫₀^xydy-2x∫₀^xy²dy+∫₀^xy³dy)

= (1/2)∫₀¹x(x⁴/2-2x⁴/3+x⁴/4)dx

=(1/24)∫₀¹x⁵dx=(1/24)(1/6)=1/144.