Force due to fluid pressure

A window on the side of a 5 foot deep swimming pool is in the shape of an inverted parabola $\displaystyle y=\frac{(9-x^2)}{3}$ located at the bottom of the pool 2 feet below the surface. Find the force F of water on this window.

This formula is used: $\displaystyle F=\int_a^b whA$ $\displaystyle dy$ where w is the density of the fluid, h is the strip depth and A is $\displaystyle L(y)dy$.

$\displaystyle x =\sqrt{9 - 3y}$

$\displaystyle F=\int_0^3 62.4(5-y)(2\sqrt{9-3y})$ $\displaystyle dy$

$\displaystyle F= 124.8\left[\int_0^3 5\sqrt{9-3y} - \int_0^3 y\sqrt{9-3y}\right]$ $\displaystyle dy$

I'm not sure if its supposed to be $\displaystyle 2\sqrt{9-3y}$ or just $\displaystyle \sqrt{9-3y}$ if someone can answer this part than I should be able to do the rest. However my friend says the first is u substitution but the 2nd one is integration by parts. What exactly is integration by parts?