Finding the boundaries on $\rho$ in spherical coordinates

Question

For spherical coordinates, how would we find the boundary on $\rho$ for the volume enclosed by the surfaces $z = \sqrt{x^2 + y^2}$ and $x^2 + y^2 + (z-a)^2 = a^2$? I'm not sure how to algebraic show this, as I can't really do it visually either. Visually, I know it should depend on the value of $\phi$, but that's about it.

I was able to find the boundary on $\rho$ visually, when it was constant bounds as in this question I had submitted earlier: Spherical coordinates and finding the boundaries for $\rho$.

Answer 1

I disagree with the answer posted, and believe the correct answer is indeed $2a\cos\phi$. Here is why. Let us view this region in the $x$-$z$ plane, and since it is symmetric about the $z$ axis, this is all we need to look at. Taking some line in some direction specified by $(\theta,\phi)$, how far does $\rho$ go? Below I have plotted what this looks like with $a=2$:

$\rho$ goes as far as the surface $x^2+y^2+(z-a)^2=a^2$, so we solve for this. $$\rho^2\sin^2\phi\cos^2\theta+\rho^2\sin^2\phi\sin^2\theta+(\rho\cos\phi-a)^2=a^2$$ $$\implies\rho^2\sin^2\phi+\rho^2\cos^2\phi-2a\rho\cos\phi=0$$ $$\implies\rho=2a\cos\phi$$

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Also, we can check that this is correct. Take the angle $\pm\pi/4$. This is along one of the red lines in my plot. These lines go from $(0,0)$ to $(\pm a,a)$, and so have length $\sqrt{(\pm a)^2+a^2}=\sqrt2a$. Meanwhile using our formula, $2a\cos(\pi/4)=2a\times\frac1{\sqrt2}=\sqrt2a$ as well.

We can also check what happens when we take the angle $\phi=0$. In the image, this of course goes from $(0,0)$ to $(0,2a)$, which has length $2a$. This agrees with $2a\cos(0)=2a$. So our answer seems to be right.

Answer 2

This surface is just a cone bounded from 0, at the origin, to $2a$, in the positive z direction. In spherical coordinates, the equation of an right circular cone is $\phi = \frac{\pi}{ 4}$. The minimum value for $\rho$ can be calculated by dropping a perpendicular line from the intersection point of the cone and the sphere. For this right triangle, $\rho^2 = a^2 + a^2$, so $\rho = \sqrt{2}a$ here.

For the volume integral, the angle $\phi$ will go from $- \frac{\pi}{ 4} $ to $\frac{\pi}{ 4}$, and in doing so will hit $\phi = 0$. At this point, $\rho$ is simply a vertical line up the $z$ axis through the sphere. So, it is evident that at this point $\rho$ is at its maximum value: $2a$.