How to find bounds for $\phi$ in spherical coordinates with a cone
I have a sphere and a cone making up a region.
Sphere $x^{2}+y^{2}+z^{2}= a^{2}$Cone $z=c \sqrt{x^{2}+y^{2}}$
Where $c$ and $a$ are positive constants.
I need to find the integral of $x^{2}+y^{2}+z^{2}$ over the region that is above the cone and below the sphere. Using spherical coordinates I have set up the following bounds:
$ 0 ≤ ρ ≤ a $
$ 0 ≤ θ ≤ 2\pi $
$0 ≤ φ ≤ ? $
I don't know how to find the bounds for phi. If there were no constants I would have plugged in
$x^{2}+y^{2}+{(c\sqrt{x^{2}+y^{2}})}^{2} = a^{2}$then
$ (c^{2} +1) (x^{2}+y^{2})= a^{2}$then $x^{2}+y^{2} = r^{2} $so $ (c^{2} +1) (r^2)= a^{2}$
$ r= \frac{a}{\sqrt{c^{2}+1}}$
we also know the relationship $ r= ρ\sin\phi$
So finally
$ \frac{1}{\sqrt{c^{2}+1}} =\sin\phi$Then I would take inverse sign which would give me the max angle for $φ$ and therefore would know the bound, in this case I don't know and I don't know how to find it, I was considering using cylindrical coordinates instead but it led to a tedious integral. Please your advice on how to solve this and the thought process behind. Thank you :)
$\endgroup$1 Answer
$\begingroup$Since you are given $$z=c\sqrt {x^2+y^2}= cr$$ and you know that $$\tan \phi = \frac {r}{z}$$
We get $$\tan \phi =1/c$$
Thus the limits for $\phi$ are $$0\le \phi \le \tan ^{-1} (1/c)$$
$\endgroup$ 2
