Each cross section has a diameter equal the horizontal distance between the two "halves" of the parabola. We have
y = 4x² ⇒ x = ±√y/2
so the diameter for some given y would be √y/2 - (-√y/2) = √y.
The volume of a cross section with thickness ∆y is then
π/2 (√y/2)² ∆y = π/8 y ∆y
The total volume of the solid would then be
[tex]\displaystyle \int_0^4 \frac\pi8 y \, dy = \frac\pi8 \times \frac12 y^2 \bigg|_{y=0}^{y=4} = \boxed{\pi}[/tex]
