Use the given formula :
[tex]V=\frac{1}3\pi r^2h=\frac{1}3\pi (\frac{D}2)^2h[/tex] where [tex]D[/tex] is the diameter.
By Pythagoras, [tex]h=\sqrt{17^2-(30/2)^2}=8\text{ in}[/tex]
Hence [tex]V\approx\frac{3.14}3*15^2*8=\boxed{1884.0\text{ in}^3}[/tex]