Question

Find the volume of the solid in the first octant (x ≥ 0, y ≥ 0, z ≥ 0) bounded by z = 8 - x², z = x², and y = 3.​

Answer 1

Setting up and computing the integral is straightforward. The two parabolic cylinders $z=8-x^2$ and $z=x^2$ meet when

$x^2 = 8 - x^2 \implies 2x^2 = 8 \implies x = \pm2$

Then the volume is

$\displaystyle \int_0^3 \int_{-2}^2 \int_{x^2}^{8-x^2} dz \, dx \, dy = 3 \int_{-2}^2 ((8-x^2) - x^2) \, dx \\\\ ~~~~~~~~ = 6 \int_0^2 (8 - 2x^2) \, dx = \boxed{64}$