Consider the region bounded by x=4-y^2, x-axis and y-axis...?

The curve (x = 4 - y²) meets x-axis (y = 0) at (4, 0)
The curve: x = 4 - y²   ⇒  y² = 4 - x

Volume of the solid
    ₄
= ∫ πy²dx
   ⁰
    ₄
= ∫ π(4 - x)dx
   ⁰
                        ₄
= π [4x - (x²/2)]
                        ⁰

= π [4*4 - (4²/2)]

= 8π

There are two distinct closed regions having those boundaries, one below the x-axis and one above. They generate the same solid of revolution. It is a paraboloid. Its base radius is 2, and its height is 4.

volume = (1/2)π(2)²(4) = 8π