The ends of a water trough have the shape of the region bounded by the graphs y=x^2 and y=4. to what depth must the trough be filled with water so that the force of the water exerted on either end of the trough is 779.423 lbs? (water has a density of 62.5 lbs/ft^3. both x and y are measured in feet.

 4 is the maximum depth of the trough so we expect the answer to be less than 4. As the cross section of the trough has the shape y = x², then the width of the water level at any depth y is 2x. 

Let a to b be the water depth, σ is the weight density of the fluid (water), y is the depth of water at any level and the width of water at that level is w(x). 

The fluid force against the walls of the trough equals :- 

b
 ∫σyw(x) dx. . . . a = 0 and b is the desired depth, y = x², w(x) = 2x a 

b
 ∫62.5 * x² *2x dx = 62.5 * 2 ∫x³ dx = 125 * ¼x⁴ = 31.25x⁴ 0 

As the lower level as zero, we now have :-
31.25b⁴ = 779.423 
b⁴ = 24.941536 
b = 2.23476 ft . . . which is the required depth of water.

I’d say letter a..

Arhombus is a quadrilateral.