Please check my work:

Find the hydrostatic pressure on one end of a water trough full of water, the end of which is a trapezoid with given dimensions: top of trapezoid = 20 feet, sides of trapezoid both = 8 feet, bottom of trapezoid = 12 feet.

Depth of water = 8 feet
Density of water = 62.4 lb/ft^3
gravity = 32.15 ft/s^2

a/(8-xi*) = (4ft)/(8ft)
a = (8-xi*)/2 = 4 - (xi/2)
Wi = 2(6+a) = 2(6+4-(1/2)xi*)=20 - xi*
Ai = Wi*delta x = (20-xi*)delta x
Pi = rho*g*d
Pi = (62.4 lb/ft^3)(32.15 ft/s^2)xi
Fi = Pi*Ai
Fi = (62.4 b/ft^3)(32.1/s^2)xi*(20-xi*) delta x
Fi = Integral from 0 to 8 of:
(62.4)(32.15)x(20-x)dx
Fi = 2006.16*Integral 0 to 8 of: (20x-x^2)dx
=2006.16[10x^2-(x^3)/3] evaluated at 0 and 8
=9.42 x 10^5 (what are the units here? lb/(ft^3*s^2)???

Your density of water in lbs/ft^3 is the weight of water at sea level. It is not mass density. So no need to multiply by 32 ft/sec to turn into a force. That is one error.

I don't follow your work, I probably need a diagram to see what your x and dx is. But I know the area of the end is appx 16*6 or appx 100 ft squared (engineering estimate)and the average depth is about 3 ft, so the area*density water*depth is about E2*62*3 appx 18K lbs, which is about half of your adjusted answer (adjusted by dividing by 32). Recheck your calcs.

also, your depth is wrong. If the slanted sides are 8 ft, the depth is about six feet.

Hmm. . . Let me think about this one.

16049.28

To find the hydrostatic pressure on one end of the water trough, we can use the formula:

P = ρgh

where P is the hydrostatic pressure, ρ is the density of water, g is the acceleration due to gravity, and h is the height of the water column.

Given that the depth of water in the trough is 8 feet, the density of water is 62.4 lb/ft^3, and the acceleration due to gravity is 32.15 ft/s^2, we can calculate the hydrostatic pressure:

P = (62.4 lb/ft^3) * (32.15 ft/s^2) * 8 feet
P ≈ 15934.08 lb/ft^2 (or psi)

Therefore, the hydrostatic pressure on one end of the water trough is approximately 15934.08 lb/ft^2 (or psi).

It seems that there are a few errors in the provided work. Let me explain how to correctly find the hydrostatic pressure on one end of the water trough.

To find the hydrostatic pressure, we can use the formula: P = ρ * g * h

Where:
P is the hydrostatic pressure
ρ (rho) is the density of water
g is the acceleration due to gravity
h is the height of the water column

Given:
Density of water (ρ) = 62.4 lb/ft³
Acceleration due to gravity (g) = 32.15 ft/s²
Depth of water (h) = 6 ft (since the slanted sides are 8 ft, the depth is approximately 6 ft)

Step 1: Calculate the height of the water column.
Since the water trough is trapezoidal, we need to calculate the average width of the top and bottom bases.
Average width = (top width + bottom width) / 2
Average width = (20 ft + 12 ft) / 2 = 16 ft

Step 2: Calculate the hydrostatic pressure.
P = ρ * g * h
P = 62.4 lb/ft³ * 32.15 ft/s² * 6 ft
P ≈ 12047.68 lb/ft²

The resulting hydrostatic pressure is approximately 12047.68 lb/ft².

Please note that I've used an engineering estimate for the area of the end of the trough as approximately 16 ft² based on the dimensions you provided. If you have the exact dimensions available, you can use those to get a more precise result.