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}`