Triple integral of a generic right tetrahedron

Setup and evaluate the triple integral whose bound is the right tetrahedron where the vertices are (0,0,0), (A,0,0), (0,B,0,), and (0,0,C). I know the ultimate formula is (1/6)ABC and I need to show that by using triple integrals.

Please help, need an answer by tonight! Thank You!

I'll assume that a, b, c > 0.

The equation of the plane passing through the three nonzero points is given by
x/a + y/b + z/c = 1.

Projecting the tetrahedral region onto the xy-plane is the region bounded by x/a + y/b = 1 in the first quadrant ==> y = 0 to y = b(1 - x/a) with x in [0, a].

Thus, the volume equals ∫∫∫ 1 dV
= ∫(x = 0 to a) ∫(y = 0 to b(1 - x/a)) ∫(z = 0 to c(1 - x/a - y/b)) 1 dz dy dx.

Evaluating this yields
∫(x = 0 to a) ∫(y = 0 to b(1 - x/a)) c(1 - x/a - y/b) dy dx
= c ∫(x = 0 to a) [(1 - x/a)y - (1/b) y^2/2] {for y = 0 to b(1 - x/a)} dx
= c ∫(x = 0 to a) [(1 - x/a)(1 - x/a) - (1/b) * b^2 (1 - x/a)^2/2] dx
= (bc/2) ∫(x = 0 to a) (1 - x/a)^2 dx
= (bc/2) * -a(1 - x/a)^3 / 3 {for x = 0 to a}
= abc / 6.

I hope this helps!