A trough is 8 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^2 from x=−1 to x=1. The trough is full of water. Find the amount of work required to empty the trough by pumping the water over the top. Note: The weight of water is 62 pounds per cubic foot.
Okay so for the first one, I got as far as slab has volume = 8 [ 2x dy ] = 16 √y dy , y in [0,1] , force = 62 volumes ; distance moved is [ 1 – y ], but I think I keep doing it wrong.
I'm not sure how tp continue
Best Answer
Draw a vertical cross-section of the trough parallel to one end. So we draw the segment of the curve $y=x^2$ from $x=-1$ to $x=1$.
Draw a thin horizontal strip at height $y$, of thickness "$dy$."
This strip will have to be lifted, as you pointed out, through a distance $1-y$.
The strip in essence has thickness $dy$, and "width" $2x$, where $x^2=y$. If we take into account the length of the trough, the volume of the thin layer of water is $(8)(2x)dy$, so has weight $(62)(8)(2x)dy$, which is $(62)(8)(2\sqrt{y})dy$.
Lifting this through a distance $1-y$ means work done is $(62)(8)(2\sqrt{y})(1-y)dy$.
"Add up" (integrate) from $y=0$ to $y=1$. We get $$\int_0^1 (62)(8)(2\sqrt{y})(1-y)\,dy.$$