Need help finding the volume. 
Not sure how to find the volume to this problem.
 A: Hint:  it is a solid of revolution, a piece of a circle revolved around the vertical axis.  Use the disc method.
A: An alternative answer. 

We will estimate the volume of water by approximating it as the volume of discs stacked on top of each other. The radius of the sphere is $R$, the height of the water is $h$ and the corresponding radius of the disc at height $h$ is $r$.
Then the volume of water is given by:
$$ V = \int_{0}^{R/5} \pi [r(h)] ^{2} dh.$$
Clearly the radius of the discs increase as $h$ increases, in fact Pythagoras gives us:
$$ r^{2} = 2Rh  -h^2.$$
Substitute into the integral.
$$ V = \pi \int_{0}^{R/5}2Rh - h^2 dh $$
Evaluate this integral. We find that $V = \frac{14}{375}\pi R^{3}$ as in Michael Li's answer above.
A: The volume of water required is $\frac{14}{375}\pi R^3$. Consider the cross-section of the sphere, we denote the angle $a$.

Clearly $sin(a)=\frac{4}{5}$.
Recall the derivation of the volume equation.
$$V_{sphere}=\pi R^3\int^{0.5\pi}_{-0.5\pi} cos^3(\theta)d\theta=2\pi R^3\int^{0.5\pi}_0cos^3(\theta)d\theta=\frac{4}{3}\pi R^3$$
So volume of a hemisphere is $\frac{2}{3}\pi R^3$. Cut the hemisphere along plane $DE$, the volume of the solid below is,
$$V'=\pi R^3\int^a_0cos^3(\theta)d\theta=\pi R^3(\frac{3sin(\theta)-sin^3(\theta)}{3})|^a_0=\frac{236}{375}\pi R^3$$
Above the plane $DE$, the volume required is just the difference of the results above.
