# Can someone help explain how to do these two work problems?

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

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 $$\int_0^1 (62)(8)(2\sqrt{y})(1-y)\,dy.$$

• That's what I keep doing and I get the answer 330.67 but my webwork keeps telling me I am wrong. – vd93 Apr 5 '14 at 1:25
• So we are integrating $(62)(8)(2)(y^{1/2}-y^{3/2})$. We get $(62)(8)(2)\left(\frac{2}{3}-\frac{2}{5}\right)$, about $264.5$. – André Nicolas Apr 5 '14 at 1:32
• Ok thank you! I was integrating it wrong. I kept taking out the (62*8) and multiplying it to end result! – vd93 Apr 5 '14 at 1:33
• Sorry about incompleteness in comment. An antiderivative is $(62)(8)(2)\left(\frac{2}{3}x^{3/2}-\frac{2}{5}y^{5/2} \right)$. And you are welcome, you had the idea right, which is the important thing. The rest is just a minor mechanical slip. – André Nicolas Apr 5 '14 at 1:36