Evaluate the Triple Integral?

Question

Evaluate the Triple Integral?

∫ ∫ ∫ z^2 dV
S

where S is the solid bounded by the cylinders x^2 + z =1 and y^2 +z =1 and the xy - plane.

Answer 1

It is convenient to rewrite the equations of the cylinders as z = 1 - x^2 (which intersects the xy-plane when x = +/- 1) and z = 1 - y^2 (which intersects the xy-plane when y = +/- 1). You can now picture this enclosed region as having a square base (-1 <= x <= 1 and -1 <= y <= 1) on the xy-plane (where z = 0) and arched walls, meeting in an apex at (0, 0, 1). The bounds of integration for z are thus [0, 1]. We may now want to rewrite the cylinders as x^2 = 1 - z and y^2 = 1 - z. Thus, for any given z, the bounds of integration for x are -sqrt(1 - z) and sqrt(1 - z), which are also the bounds of integration for y.

The integral is int(int(int dx)dy)z^2dz, which is hard to show here on Yahoo! Answers.

int dx = x, which when evaluated across the bounds of x is 2*sqrt(1 - z). int 2*sqrt(1 - z)dy = 2y*sqrt(1 - z), which when evaluated across the bounds of y is 4(1 - z). int 4(1 - z)z^2 dz = int 4z^2 - 4z^3 dz = (4/3)z^3 - z^4, which when evaluated across the bounds of z is (4/3) - 1 = 1/3.