What is the intersection of these two cylinders?

$$0\le x^2 + z^2 \le 1$$

$$0 \le y^2 + z^2 \le 1$$

I want to compute the volume of the intersection.

Sketching it out on paper is sort of nice: I see cross-sections that are disks, the first cylinder, the y-coordinate is free to vary, and for the second cylinder, the x-coordinate is free to vary.

The intersection, I would guess, seems to be something spherical.

So how can I pin down the actual set of points?

Well, one thing I thought of was to try to manipulate both inequalities to make use of the equation of a sphere, so I try looking at these inequalities instead:

$$y^2\le x^2 + y^2 + z^2 \le 1 +y^2$$

$$x^2 \le x^2 +y^2 + z^2 \le 1+x^2$$

Am I heading in the right direction? Where can I go from here?

Thanks,

• It's definitely not a sphere, not sure you can say much more about it than it's the intersection of two perpendicular cylinders. – Gregory Grant Dec 28 '15 at 23:51
• Hi Professor Grant, thanks for your comment, and especially for noting the orthogonality relationship. I think I am almost there... – user301446 Dec 29 '15 at 0:13
• This comment is to link this post as one of the (abstract) duplicates to the current choice of mother/target post. – Lee David Chung Lin Jan 22 at 10:29

The intersection of two cylinders is called a Steinmetz solid. You can give a description of the edges of the solid by

$$x = \pm z, \quad y = \pm \sqrt{1 - z^2}$$

and use these to give corresponding inequalities.

• Hi @user, thanks so much for your quick response. Well, using the above inequalities, I do see these coming out: $x = +/- \sqrt{1-z^2}$ and $y = +/- \sqrt{1-z^2}$. How come you don't mention this guy: $x = +/- \sqrt{1-z^2}$? ... is it already like implied somewhere in your description? Thanks, – user301446 Dec 28 '15 at 23:59
• Ooh @user, I think I see what you are saying ... – user301446 Dec 29 '15 at 0:00
• So my above comment, I have basically found my integration limits for both x and y...is that correct, @user? And now, I just have to find the integration limits for $z$ ... – user301446 Dec 29 '15 at 0:01
• But, solving for $z$ in each inequality gives us two upper and lower limits of integration, which is strange, @user ... – user301446 Dec 29 '15 at 0:03
• Perhaps the limits for $z$ is simply +/- 1, since the cross-sections are disks of radius 1, @user ... what do you think? Thanks, – user301446 Dec 29 '15 at 0:06

I could not resist to model this in GeoGebra. Zenith is $(0, 0, 1)$ and nadir $(0, 0, -1)$. A slice at height $z$ is a square with side length $$a = 2 \sqrt{1-z^2}$$ so $$dV = (2 \sqrt{1-z^2})^2 dz = 4(1-z^2) dz$$ • haha...@mvw, thanks so much. I am practicing for exams, so I think I had better pin down the algebra on paper for now. Hmm...so the intersection is indeed not spherical, as Professor Grant points out in his comment above... – user301446 Dec 29 '15 at 0:11
• I hoped its surface intersection operation could deal with this one, but that is either too much yet for GeoGebra or I am doing something wrong. – mvw Dec 29 '15 at 0:14