Is the volume of the parabola $y=\frac{x^2}{4}$, rotated around the $y$-axis, bounded by the plane $y=2$, equal to $8\pi$?

Question

I tried doing this through integration(Washer method) and got $8\pi$, but I'm unsure if it is correct. If wrong, how do you do this correctly?

Answer 1

Cylindrical shells: $2\pi\int_0^{2\sqrt2}x(2-\dfrac {x^2}4)\operatorname dx=2\pi[x^2-\dfrac {x^4}{16}]_0^{2\sqrt2}=2\pi(8-\dfrac{64}{16})=2\pi(4)=8\pi$.

Disk method: $\pi\int_0^2(2\sqrt{y})^2\operatorname dy=\pi\int_0^2 4y\operatorname dy=4\pi[\dfrac {y^2}2]_0^2=4\pi(2)=8\pi$.

Answer 2

Your answer seems to be correct. Express $x$ in terms of $y$ for the first quadrant ($x=2\sqrt{y}$) and do integration with respect to $y$:

$$ V=\pi\int_{0}^{2}\left(2\sqrt{y}\right)^2\,dy=8\pi\ \text{cubic units}. $$