The following is the exercise I'm trying to do:
Find the volume above the cone $z=c\sqrt{x^2+y^2}$ and within the sphere $x^2+y^2+z^2=a^2$ using spherical coordinates.
Actually computing is trivial, so my question is concerning finding the bounds. I denote the spherical coordinates by
\begin{align}
x &=R\cos{\theta}\sin{\phi} \\
y &=R\sin{\theta}\sin{\phi}\\
z &=R\cos{\phi}
\end{align}
Finding the $\theta$ bounds is easy ($0\leq\theta\leq2\pi$), and using some algebraic manipulation to find the intersection of the surfaces, we can find the integration bound for $\phi$ to be $0\leq \phi \leq\arctan{\frac{1}{c}}$, for example using:
$$z=R\cos{\phi}=cR\sin{\phi} \Rightarrow \phi=\arctan{\frac{1}{c}}$$
Now, this is where I get confused. To me, to find the integration bound for $R$, we ought to consider the values of $R$ (which is the distance from the origin to the points which we are integrating over) given the restrictions by the other bounds. Thus, I drew this curde picture:
In my mind, the rightmost sketch (or $(1)$) is what we want to integrate, thus we have to integrate from some point $R>0$ (indicated by the small point) to $R=2a$. If we were to integrate from $R=0$ to $R=2a$, then in my mind, we would obtain the region in the rightmost picture. However, the answer states that the integration bound ought to be from $R=0$ to $R=2a$. Where is my thinking incorrect/ how should I go about reasoning to find the integration bounds?
The sources I have been looking over, and my professors methods have been very "hand wavy", so input would be much appriciated!
"Find the volume above the cone" means the volume above the infinite conic surface $z=c\sqrt{x^2+y^2}$. This infinite cone has no "base". Therefore the range for $R$ is $[0,a]$ where $a$ is the radius of the bounding sphere and the correct picture is (2).
Hence, from your work, we find that the volume of the given solid is $$V=\int_{R=0}^a\int_{\phi=0}^{\arctan(1/c)}\int_{\theta=0}^{2\pi}R^2\sin\phi \, d\theta d\phi dR.$$