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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.

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    If you denote the height by $b$, then that is right. Since $s=2$ the equation for the height is $h^2+(2/1)^2=2^2$, which is what you have.2012-06-18

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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.)