# Work - integral problem

1. Mar 28, 2004

### drklrdbill

A hemispherical tank of radius 5 feet is situated so that its flat face is on top. It is full of water. Water weighs 62.5 pounds per cubic foot. The work needed to pump the water out of the lip of the tank is ? foot-pounds.

I tried evaluating the integral of pi(125x/3)(5-x)^2 from 0 to 5.

What am I doign wrong? I cannot figure this out for the life of me, spent about the last hour on a seemingly simple problem that continues to stump me.

2. Mar 28, 2004

### matt grime

It's a volume integral you'll need. Suppose the hemisphere is situated so that it is the bottom half of the sphere centred on the origin.

Let the deinsity by p. A tiny volume dV at point (x,y,z) will require you to do work -zpdV to get it to the top of the lip. Now integrate over the volume of the sphere. Note, I can never get my minus signs correct, so watch out for that.

3. Mar 28, 2004

### ShawnD

woops, my work was based on an incorrect assumption.
sorry.

Last edited: Mar 28, 2004
4. Mar 28, 2004

### Staff: Mentor

Here's how I would calculate the work done. First imagine the volume as a stack of circular disks of area $\pi r^2$ and thickness dx. The work needed to lift each disk is $\rho \pi r^2 x dx$, where $\rho$ is the weight per unit volume and $x$ is the height it needs to be lifted. Of course, $r^2 = R^2 - x^2$, so the integral you need is:
$$W = \pi \rho \int_{0}^{R} (R^2 - x^2)x dx$$

5. Mar 28, 2004

### ShawnD

Where are you getting this relation from?

*edit
is that because of the circle formula x^2 + y^2 = r^2?

Last edited: Mar 28, 2004
6. Mar 28, 2004

### Staff: Mentor

The Pythagorean theorem!

It's a right triangle: R is hypotenuse, x is vertical side, r is horizontal side.

7. Mar 28, 2004

Think about it a bit, Shawn. You can answer that on your own. =]