For clarity of question, I'll pose an example.

Consider the region bounded by $z = \sqrt{x^2 + y^2}$ and $x^2 + y^2 + z^2 = 4$. Upon using spherical coordinates, my source says that $\rho$ should range from $0$ to $2$. I don't understand this, because when I draw the diagram, I see that the distance from origin to the surface enclosing this region is NOT always 2, as opposed to a sphere which would always have radius 2, and would allow $\rho$ to range from 0 to 2.
Why is $\rho$ from 0 to 2 and what would be the general method to finding the variation in $\rho$?

I think in 2-D and polar coordinates, the general method was to sub in $x = rcos(\theta)$ and $y = rsin(\theta)$ to find the relationship between $r$ and $\theta$. Is this similar for $\rho$?

  • $\begingroup$ This is analogous to having a sector of a circle in 2D Polar co-ordinates - you would keep the radius 2 and the angle would be restricted $\endgroup$ – John Doe Apr 29 '17 at 12:42
  • $\begingroup$ Ah I see it now. Thanks. What would a general method be (if there is one) to find $\rho$? $\endgroup$ – Twenty-six colours Apr 29 '17 at 12:45
  • $\begingroup$ It'd depend on the particular example, but it may be useful so set $\rho$ to be the maximum distance from the origin, and then if for some $\theta$, the actual distance is less, then have some $\rho$ dependence in your $\theta$ range. This would probably make more sense when you see more examples, I don't think I worded that very well haha $\endgroup$ – John Doe Apr 29 '17 at 12:55
  • $\begingroup$ actually i'm not sure if I understand. For the region bounded by $z=2$ and $z = \sqrt{x^2 + y^2}$, $\rho$ will range from $0$ to $2sec(\phi)$... can you please explain why that's the maximum? $\endgroup$ – Twenty-six colours Apr 29 '17 at 12:55
  • $\begingroup$ I have written an answer, let me know if it makes sense $\endgroup$ – John Doe Apr 29 '17 at 13:14

Your first surface is a cone where for each value $z=k$, you have a circle centred at $0$ in this plane of radius $|k|$. Your second surface is a sphere of radius $2$ centred at the origin. The region bounded by these two surfaces looks like this in the region $z>0$ (to find it for $z<0$, just reflect it in the plane $z=0$):

enter image description here

If you travel from the origin outwards, you will never hit any part of the cone. You will eventually hit the surface of the sphere, by which point you will have travelled a distance of $2$. This is the case no matter what angle you travel at, as long as it is inside the cone. But this is dependent completely on the angle, and not on $\rho$, and so you don't need to worry about it for the range of $\rho$, which we can now conclude is $[0,2]$.

  • $\begingroup$ Cheers! I also used this explanation for my other question and I think I have it. Is there an algebraic way of showing that result (in your post) as opposed to just by inspection? Probably just solving $x^2 + y^2 + z^2 = 4$ for each respective $x,y,z$ spherical substitution? $\endgroup$ – Twenty-six colours Apr 29 '17 at 14:01
  • 1
    $\begingroup$ There may be a way, but you'd also have to include the cone as well as the sphere - if you had a cylinder instead then things would have been different. In any case, when looking at integrals in $\Bbb R^3$, I always find it more useful to draw the region out, as it is usually easier this way $\endgroup$ – John Doe Apr 29 '17 at 14:06
  • $\begingroup$ Hi again, For surfaces like $x^2 + y^2 + (z-a)^2 = a^2$ (for $a >0$), with the same cone in the OP, we would have a different bound for $\rho$ right? How would we find this bound? $\endgroup$ – Twenty-six colours Apr 30 '17 at 4:50
  • $\begingroup$ @PaulWoch I was just about to go to sleep, so I won't be able to answer it right now. I'd suggest making another question, and perhaps mentioning this one in it, since in the surface you mention, the limits on $\rho$ do depend on the angle, and so it is a different kind of question. I will have a look if it is not yet answered in the morning. Also, to answer that, I'd probably need a diagram, which can't be done in comments. $\endgroup$ – John Doe Apr 30 '17 at 5:02
  • $\begingroup$ That's okay, I'll post the link to the new question here as well when I submit it. The new post is: math.stackexchange.com/questions/2258639/… $\endgroup$ – Twenty-six colours Apr 30 '17 at 5:23

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