A trough is 3 feet long and 1 foot high. The trough is full of water... A trough is 3 feet long and 1 foot high. The vertical cross-section of the trough parallel to an end is shaped like the graph of $x^2$ from -1 to 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.
 A: Consider a small amount of water in the trough at depth $h$ below the top edge. The water has a volume $dV$ and a mass of $\rho\;dV$ where $\rho$ is the density of water.  The small amount of work $dW$ to remove this small amount of water from the trough is $h\rho\; dV$. You need to calculate the total work, which is $$\iiint_{\text{trough}} dW$$
and this is equal to $$\iiint_{\text{trough}} h\rho \;dV.$$
$dV$, the volume of each small amount of water, is $dx\;dy\;dz$, and there is a simple relationship between $h$ and $z$.
Is this enough to get you started?
A: Basically, the hard part of this question would be finding the surface area of the trough at a certain depth of the water. So let's say the height of the water is $y$ from the bottom, while the depth of the water is $1-y$. Now at whatever height $y$ you are at, the length will always be $3$. The width, however varies between $0$ and $2$. Notice that the equation of the parabola is $y=x^2$. If you were to put a vertical axis in between the parabola, you would see that the width of the trough at that height is twice the distance from the vertical axis to either side of the parabola. The distance between the vertical axis to the parabola is $x=\sqrt{y}$, so the total width of the parabola at height $y$ is $2\sqrt{y}$. That means at height $y$, the top surface area is $2\sqrt{y}*3 = 6\sqrt{y}$.
Now all that is left is your integral. You need to do an integral from 0 to 1 of (the weight of water)(the depth of the water)(the area of the water at the given depth)(d-height of water) =
$\int_0^1 (62)(1-y)(6\sqrt{y})dy = 99.2$ foot-pounds.
