Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

So I'm about to finish my initial course on Integral Calculus when I come across this problem:

Let there be a spherical container full of water with a 2 meter radius. How much work would it take to raise this volume of water 10 meters up from the top of the container?

I can't figure out how to even write this down as a definite integral. I do know how integrating to get Work goes:

A definite integral of the force times distance.

What I don't get is how to input the sphere (x = sqrt(4 - y^2)) in the equation or how to even write it so it's a definite integral. Help is very much appreciated.

share|cite|improve this question
up vote 1 down vote accepted

It doesn't matter much, but let the sphere have centre the origin. So the sphere has equation $x^2+y^2+z^2=4$.

For any $z$ (naturally between $-2$ and $2$), the cross-section at height $z$ parallel to the $x$-$y$ plane is a circle the square of whose radius is $4-z^2$.

So any thin slice of thickness "$dz$" at height $z$ has volume about $\pi(4-z^2)dz$. We have to lift this slice to height $12$ above the $x$-$y$ plane.

"Add up" over all slices, that is, integrate. The work done is therefore $$\int_{-2}^2 K(12-z)\pi(4-z^2)\,dz,$$ where $K$ is a constant that depends on the units used.

Another way: We set up an integral, because you asked for that. And the idea will be useful for other problrms.

But there is a much easier way. It takes $0$ work to bring the water to the $x$-$y$ plane. This is because the work done in lifting the lower part is cancelled by the work done by the water moving from the upper part. Then we have to lift the whole mass, if you wish concentrated at the origin, through a distance $12$. Thus the work is the volume times $12$ times the same constant $K$ as above. The volume is $\frac{4\pi}{3}(2^3)$. Now multiply by $12K$.

share|cite|improve this answer
I assume that this K, in my case, would be the value of gravity accelaration (9.81 kg*m/s^2) and the density of water (1000 kg/m^3) . This way I'll be able to integrate force with distance knowing only the volume. – Eduardo de Luna Nov 25 '12 at 3:44
@EduardodeLuna: In MKS units, it would be (kg)(m$^2$)/(sec$^2$), that is, joules. – André Nicolas Nov 25 '12 at 4:04
Thank you André. – Eduardo de Luna Nov 30 '12 at 13:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.