Calculus--->Work?

Question

A trough is 5 feet long and 1 foot high. The vertical cross-section of the trough parallel to an end is shaped like the graph of y=x4 from x=−1 to x=1 . The trough is full of water. Find the amount of work in foot-pounds required to empty the trough by pumping the water over the top. Note: The weight of water is 62 pounds per cubic foot.

suppose to be x^4 NOT x4......

Answer 1

Consider a column of water of constant cross-sectional area A and height H. To lift a slice of water of infinitesimal height dh located at height h would take (H-h) * A * dh * 62 lbs. The sum of the work needed to lift all those slices (and thus the whole column) out of the trough is the integral [0, H]∫62A(H-h) dh = -31A(H-h)² | [0, H] = 31AH².

Now, consider dividing the trough lengthwise into thin slices of infinitesimal width dx. For a slice located at x, the water in this slice is of an approximately constant depth x⁴ and has a horizontal cross-sectional area of 5 dx. The energy needed to life that slice out of the trough is 31* (5 dx) * (x⁴)². The energy needed to lift all those slices (and thus all the water in the trough) out of the trough is [-1, 1]∫31* (5 dx) * (x⁴)² = [-1, 1]∫155x⁸ dx = 155x⁹/9 | [-1, 1] = 310/9 ft·lbs. ≈ 34.4444 ft.·lbs.