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Volume of overlapping cones.

I have a problem I hope someone can help me with! :)

I have two cones. In the beginning they are more than 2r apart from eachother and the volume are just the volume of two cones. But as they are moved closer together some of the volume are overlapping and the total volume are less than two cones. In the end the total volume are the volume of one cone.

How do I find this total volume for when the cones are less than 2r apart?

Re: Volume of overlapping cones.

The closest I can see to your set up would be

$\displaystyle \displaystyle{\int_c^R \ \int_0^{\sqrt{R^2 - x^2}} \ \ H\ -\ \frac{H}{R} \sqrt{x^2 + y^2}\ \ dy\ dx}$

i.e. prettly close, but a minus in the y limit, and the cone inverted and dragged up z-wise by H units. I mean translated positively!

Integrating twice does get hairy. The first...

integrate sqrt(x^2 + y^2) dy - Wolfram|Alpha

is ok... click on 'show steps', also here: http://mathhelpforum.com/calculus/18...-question.html

Then wrt x, integrate sqrt(R^2 - x^2) dx - Wolfram|Alpha also e.g. http://mathhelpforum.com/calculus/15...-integral.html

And integrate x^2 ln(R + sqrt(R^2 - x^2)) dx - Wolfram|Alpha

And some simple parts integrate x^2 ln x - Wolfram|Alpha and e.g. http://mathhelpforum.com/calculus/18...ion-parts.html

Which is all going to be messy, no denying.

Maybe

$\displaystyle \displaystyle{\int_\Big{0}^{\Big{ (1 - \frac{c}{R}}\Big{)}\Big{H} } \ \int_\Big{c}^{\Big{(1 - \frac{z}{H}\Big{)}\Big{R}}} \ \ \sqrt{R^2 - x^2} \ \ dx\ dz}$

... will be better.