|dw:1329557645302:dw| Due to symmetry we can integrate from 0 to r and just double the integral $V = 2\int\limits_{0}^{r}A(x) dx$ $V = 8\int\limits_{0}^{r}(r^{2}-x^{2}) dx$ $8 | r^{2}x - x^{3}/3 = 8(r^{3} - r^{3}/3) = \frac{16}{3}r^{3}$