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.

  • $\begingroup$ A trough in the shape of a parabolic prism? Seems pretty far-fetched to me. I think your teacher is pulling your leg. $\endgroup$ – MJD Sep 29 '12 at 17:08
  • $\begingroup$ @MJD Calculus makes that not an issue. $\endgroup$ – Sidd Singal Sep 29 '12 at 17:09
  • $\begingroup$ I want to know what jerk-ass farmer is watering his livestock out of a parabolic trough. Everyone knows that cattle drink optimally from catenaries. $\endgroup$ – MJD Sep 29 '12 at 17:15

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.


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?

  • $\begingroup$ Look at my answer. This could be a way to do it, but in physics they teach you a different, easier way. $\endgroup$ – Sidd Singal Sep 29 '12 at 17:31
  • $\begingroup$ @mathguy It is exactly the same! Two of the three integrals in my expression turn out to be trivial multiplications. $\endgroup$ – MJD Sep 29 '12 at 17:45
  • $\begingroup$ oh is it? My bad. I just didn't know how to interpret the triple integrals yet and I wasn't sure how g would fit into this problem. $\endgroup$ – Sidd Singal Sep 29 '12 at 17:49
  • 1
    $\begingroup$ The $g$ is my mistake. I was thinking that the 62 pounds was pounds mass. But of course it isn't. I will correct. The triple integral just means you integrate over the three dimensions of the trough, but the two horizontal dimensions are trivial and just multiply by the horizontal measurements of the trough, leaving you with one interesting integral over the height, just as in your solution. $\endgroup$ – MJD Sep 29 '12 at 17:54

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