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I have to get the volume of the set $E=\{(x,y,z) : x^2+y^2+z^2 \leq r^2, x^2+y^2-rx \geq 0, x^2+y^2+rx \geq 0\}$.

Are spherical coordinates the best way to go?, because if i put $$x = p\cos\theta\sin\psi \\ y= p\sin\theta\sin\psi \\ z = p\cos\psi$$ i know that because of $x^2+y^2+z^2 \leq r^2$ must be $|p| \leq |r|$, but how do i get the limits of integration for $\theta$ and $\psi$?.

I have that $x^2+y^2-rx \geq 0 \iff p^2\cos^2\theta\sin^2\psi + p^2\sin^2\theta\sin^2\psi - rpcos \theta\sin\psi \geq 0 \iff p\sin\psi(p\sin\psi - r\cos\theta) \geq 0$

And also $x^2+y^2+rx \geq 0 \iff p\sin\psi(p\sin\psi + r\cos\theta) \geq 0$.

Does that mean that i should find $\theta$ and $\psi$ such that $\displaystyle\frac{p\sin\psi - r\cos\theta}{p\sin\psi + r\cos\theta} \geq 0$ ?.

How can i solve an equation like the last one?

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  • $\begingroup$ My first thought on the last two is to complete the square: $(x-\frac r2)^2+y^2 \ge \frac {r^2}4$ and $(x+\frac r2)^2+y^2 \ge \frac {r^2}4$. You are looking at the intersection of two cylinders and a sphere. The cylinders destroy the spherical symmetry, so I would stay Cartesian. $\endgroup$ – Ross Millikan Jul 19 '13 at 4:05
  • $\begingroup$ @trash: Are you forced to evaluate the Volume using spherical coordinates? I think using cylindrical ones would be easier to understand. $\endgroup$ – mrs Jul 19 '13 at 10:59
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It seems to me that using Cylindrical coordinates will make our job easier. As you see, our region is symmetric so it is enough to consider $1/8$ of whole volume such that $x\ge 0, y\ge 0,z\ge 0$:

enter image description here

Now let's look at the bottom of above region which is on $z=0$:

enter image description here

So the limits are as follows: $$\theta|_0^{\pi/2},~~r|_{a\cos\theta}^a,~~z|_0^{\sqrt{a^2-r^2}}$$

Note: for plots I assumed that $a=2$ and $a$ is $r$ in your original equations.

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  • $\begingroup$ Thanks! It works great with cylindrical coordinates, and i wasn't forced to use spherical coords, it just happens that i still have some troubles to picture the best way to go -and because of the $x^2+y^2+z^2\leq r^2$ spherical coordinates was the first that came to my mind-. $\endgroup$ – Cure Jul 19 '13 at 14:10
  • $\begingroup$ I love your graphs, dear Babak! (+) $\endgroup$ – Namaste Jul 19 '13 at 20:00
  • $\begingroup$ @amWhy: I will be lucky if mathematicians accept me in Maths world just cause of this job. Thanks :-) $\endgroup$ – mrs Jul 19 '13 at 20:02
  • $\begingroup$ Hello, dear friend. Good to see you 'round! ;-) $\endgroup$ – Namaste Jul 20 '13 at 18:15
  • $\begingroup$ Are those Mathematica graphics, Babak? $\endgroup$ – Alexander Gruber Jul 27 '13 at 7:04

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