I am trying to find the volume of intersection of a cone with two spheres:

$\displaystyle x^2+y^2+z^2=R_1^2 $

$\displaystyle x^2+y^2+(z-z_0)^2=R_2^2 $

$\displaystyle z=\sqrt{\frac{x^2+y^2}{r}} $

Can anyone help me out?

Thank you!

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- Jan 4th 2010, 12:27 AMandra7Find the volume of a solid bound by two spheres and cone
I am trying to find the volume of intersection of a cone with two spheres:

$\displaystyle x^2+y^2+z^2=R_1^2 $

$\displaystyle x^2+y^2+(z-z_0)^2=R_2^2 $

$\displaystyle z=\sqrt{\frac{x^2+y^2}{r}} $

Can anyone help me out?

Thank you!

- Jan 4th 2010, 04:35 AMHallsofIvy
I'm not clear on what you mean by "the volume bounded by". What volume is bounded and whether or not there is any volume bounded by these figures depends upon specific values. One sphere has center at the origin and the other has center at $\displaystyle (0, 0, z_0)$. Whether there is any volume bounded by them and the cone depends upon the relation between $\displaystyle R_1$ and $\displaystyle R_2$, the radii of the two spheres, as well as the relation between them and $\displaystyle z_0$.

- Jan 4th 2010, 05:14 AMandra7
Of course, if there are no solid bordered with these three surfaces, then the volume of intersection will be zero.

But, if we assume that there is such a volume that is greater than zero, what will be the equation of it? What are the limits of integration?

Tnx

- Jan 4th 2010, 05:21 AMshawsend
Right off the bat I'd plug numbers in it and see what's going on. Here's three:

$\displaystyle x^2+y^2+z^2=1$

$\displaystyle x^2+y^2+(z-1)^2=2$

$\displaystyle z=\sqrt{\frac{x^2+y^2}{2}}$

The plot below is a cross-section of the three along the y-plane. From this, I'd assume you mean the volume element in purple. Then how about using volume by discs:

$\displaystyle \mathop\int\limits_{\hspace{-6pt}\text{Purple}} \pi r^2 dz=\mathop\int\limits_{p_1}+\mathop\int\limits_{p_ 2}$

. . . or maybe it's some other piece in there that you want. - Jan 4th 2010, 06:10 AMandra7
Actually, I mean the volume element in red on the plot below.

- Jan 4th 2010, 06:49 AMshawsend
Very good then. Do a few with real numbers, then figure out how to generalize it to arbitrary constants. So, let's look first at the set:

$\displaystyle x^2+y^2+z^2=1$

$\displaystyle x^2+y^2+(z-2)^2=2$

$\displaystyle z=\sqrt{\frac{x^2+y^2}{1/10}}$

the cross-section of which is shown in the first plot. And I guess if my life dependent on it, I could figure out how to color in that section you want but it's not easy for me so you know which part we're lookin' at. Ain't that still circles in there because of the cone? Then can't I still use as a first effort, the disc metod:

$\displaystyle V=\mathop\int\limits_{\hspace{-8pt}\substack{\text{low}\\ \text{section}}} \pi r^2 dz+\mathop\int\limits_{\hspace{-8pt}\substack{\text{mid}\\ \text{section}}} \pi r^2 dz+\mathop\int\limits_{\hspace{-8pt}\substack{\text{top}\\ \text{section}}} \pi r^2 dz$

where I've drawn horizontal lines to separate the low, mid, and top sections which I've tried to show in the second plot but still hard to see the small low section. You know what I mean though. - Jan 4th 2010, 10:10 AMandra7
I think I can figure out from here.

Thanks so much!!