Hydrostatic pressure on an equilateral triangle The Triangle is point downward at water level, all sides are 2 feet.
I have no idea what to do, I know that the interval for my integration depends on the height of the triangle but I do not know how to find that and I am pretty certain this is some weird geometry trick that I do not know.
 A: 
Apply the Pithagorean theorem and find $c$ (see sketch). The width $w(h)$ of a rectangular strip drawn on the triangle at a distance $h$ from the top varies linearly with $h$ from $l$ to $0$ as $h$ goes from $0$ to $c$. Deduce the following formula
$$
w(h)=l\left( 1-\frac{h}{c}\right).\tag{1}
$$
Since the hydrostatic pressure is given by 
$$P(h)=\rho gh,\tag{2}$$  the hydrostatic force exerted on the triangle is the definite integral
$$
F=\int_{0}^{c}\rho ghw(h)\; dh=l\rho g\int_{0}^{c}h\left( 1-\frac{h}{c}\right)
dh=\ldots =l\rho g\times \frac{1}{6}c^{2}\tag{3}
$$
As for the meaning, numeric values and units of the symbols they are as follows:


*

*$P$ is the hydrostatic pressure at a generic point $H(\text{measured in }\textrm{Pa } \equiv $ Pascal above the atmospheric pressure)

*$\rho $ is the water density ($\approx 1000 \textrm{kg/m}^{3}$),

*$g$ is the gravitational acceleration ($9.81 \textrm{m/s}^{2}$),

*$h$ is the height of the fluid column above $H\; (\textrm{m}).$ $^1$

*$F$ is the hydrostatic force $(\text{ }\mathrm{N})$


$^1$ $1\text{ }\mathrm{ft\ }=0.3048\text{ }\mathrm{m}$
(See this answer.)
