May be it's too late, for you...but I've just found the same integral studying heat transmission in electro-heating...so don't become mad, i think it's easy:
\int_0^R x^3 J_0(ax)dx=\frac{1}{a^2}\left[R^2\left\{2J_2(aR)-aRJ_3(aR)\right\}\right]
J_3(aR)=\frac{4}{aR}J_2(aR)-J_1(aR)
So...