Math question for a real genius?

Question

A trough is filled with a liquid of density 840 kg/m^3. The ends of the trough are equilateral triangles with sides 8 m long and vertex at the bottom. Find the hydrostatic force on one end of the trough.

Answer 1

The total force is 527401 Newtons.

Calculations are as follows:
Pressure at any point = (fluid depth) * (density) * (acc. of gravity)
This is a function of one variable, which is the fluid depth.
Pressure = d * (840 kg/m^3) * (9.81 m/s^2), where d is the depth in meters.

To get the total force on the end of the trough, we need to integrate the pressure function over the area of the end of the trough.

From geometry, the height of the trough is 4 * sqrt(3) meters = 6.9282 m.

Let z be the height above the bottom of the trough.
Therefore, d = 6.9282 - z
The width of the trough at height z is:
width = z * 2 / sqrt(3) = 1.1547 z

We must calculate the following:
Integral[0 to 6.9282] p(z) w(z) dz,
where p(z) = (6.9282 - z)(840)(9.81)
and w(z) = 1.1547 z

That's equivalent to:
Integral[0 to 6.9282] (65922 z - 9515 z^2) dz

The indefinite integral is:
32961 z^2 - 3171.6 z^3

Do the remaining arithmetic by inserting 6.9282 for z.
The result is 527401 Newtons.