find the volume of the solid when the region under the curve $y=x\sqrt{4-x^2}$ from $x=0$ to $x=2$ is rotated about the $y$-axis I tried:
$$y = x\sqrt{4-x^2}$$
When the curve is rotated from $x = 0$ to $x = 2$, the radius of the curve obtained is:
$$A = \pi \ r^2 = \pi(x\sqrt{4-x^2})^2 = \pi(x^2(4-x^2)) = \pi(4x^2-x^4)$$
So, the volume is:
$$V = \pi\int_0^2(4x^2-x^4)dx = \pi\frac{64}{15} = 13.4$$
When I put the answer in webwork, it said it was wrong.

The curve has maximum value at $ (\sqrt2,2) $. So need to only cover volume of external shells as the inside shells overlap, they should be excluded.
$$ V = \int_{\sqrt2}^2 \pi x^2 \frac{dy}{dx} dx $$
