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 answer in the question is very odd, since it says the volume about the $y$ axis but that is the integral of $y^2$, which would be correct if we were rotating about the $x$ axis. For rotating about the $y$ axis, we get the volume V by integrating shells of $\pi x^2y$ from 0 to 2, so we need
$$
V=\pi\int_0^2 x^3 \sqrt{4-x^2} dx
$$
This integral may be performed by substituting $u=x^2$, $du=2x dx$ to give
$$
\frac12\int_0^4 u \sqrt{4-u} du
$$
then substituting $s=4-u$ to give
$$
\frac12\int_0^44s^{1/2}-s^{3/2} ds=
\frac12\left\{4\times\frac23s^{3/2}-\frac25s^{5/2}\right\}\Biggr|_0^4={64\over 15}
$$
