## Calculus (3rd Edition)

$9 \pi cm^3/s$
The flow rate $(Q)$ through the pipe can be calculated as: $Q=2 \pi \int_{0}^{a} r \ v(r) \ dr$ Now, $Q= 2 \pi \int_{0}^{3} r (3-r) \ dr \\=2 \pi \int_0^3 (-r^2+3r) \ dr \\=2 \pi [-\dfrac{r^3}{3}+\dfrac{3r^2}{2}]_0^{3} \\=2 \pi \times [-\dfrac{(3)^3}{3}+\dfrac{3(3^2)}{2}] \\=2 \pi [-\dfrac{27}{2} -9] \\=9 \pi cm^3/s$