A trough is 4 feet long and 1 foot high. The vertical cross-section of the trough parallel to an end is shaped like the graph of y=x^10 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.
Here is what I did:
wrote y=x^10 in terms of y ---> x=y^(1/10)
found the area in terms of y ---> A
=2*(y^(1/10))*4
Set up my integral ---> W = ∫(from 0 to 1) 8y^(1/10)*y dy
came out to be ---> W = 4960/21 (which was wrong)
what did I screw up?
Here is what I did:
wrote y=x^10 in terms of y ---> x=y^(1/10)
found the area in terms of y ---> A
=2*(y^(1/10))*4Set up my integral ---> W = ∫(from 0 to 1) 8y^(1/10)*y dy
came out to be ---> W = 4960/21 (which was wrong)
what did I screw up?