How to integrate hydrostatic force on a two dimensional shape? I'm so confused this question is very different from the other hydrostatic force questions and I think I am misunderstanding the question.

I am primarily concerned with 15 because I somehow managed to get 13 correct. Am I correct in assuming this shape has a z-axis depth that is not given and does not matter? I am used to modeling Work = $\int_a^b Force$ (Force is in most cases weight density * height * area of cross-section)
I could really use some pointers on how to integrate this. I realize the sides can be modeled with the equations y=$\frac 83$ x
My attempt: $\int_0^5 62.4*2* {\frac 83} x*(5-x) dx$ and I got 6933 when the correct answer is ~3705. I don't know where I went wrong..
 A: Decide which direction is "up" and which axis points in which direction.
Evidently the narrowest angle of the triangle is "up"
and the side of length $6$ is "down".
Based on the formula $y = \frac83 x$, I thought you had decided
that the $y$-axis points either up or down
(not clear which; this would be important) and the $x$-axis
points to the right or left (it does not really matter which).
Looking at your integral, however, I began to suspect that you decided
your $x$-axis would point straight up, with $x=0$ at the bottom of the tank
(so that the depth at any point is $5 - x$)
and that the $y$-axis points left or right.
If you meant for the $x$-axis to point upward then you have the wrong
formula for the sides of the triangle.
Both the slope and the $y$-intercept are incorrect.
You did remember to multiply by $2$ in order to count both sides
of the triangle (left or right of the centerline), so I think
if you replace $\frac83 x$ with the correct formula for $y$
along one side of the triangle then you will be OK.
