Finding $a$ s.t the cone $\sqrt{x^{2}+y^{2}}=za$ divides the upper half of the unit ball into two parts with the same volume

Question

My friend gave me the following question:

For which value of the parameter $a$ does the cone $\sqrt{x^{2}+y^{2}}=za$ divides $$\{(x,y,z):\, x^{2}+y^{2}+z^{2}\leq1,z\geq0\}$$ into two parts with the same volume ?

I am having some difficulties with the question.

What I did:

First, a ball with radius $R$ have the volume $\frac{4\pi}{3}R^{3}$ hence the volume of the upper half of the unit ball is $\frac{\pi}{3}$.

Secondly: I found where does the cone intersect with the boundary of the ball:

$$ \sqrt{x^{2}+y^{2}}=za $$

hence $$ z=\sqrt{\frac{x^{2}+y^{2}}{a^{2}}} $$

and $$ x^{2}+y^{2}+z^{2}=1 $$

setting $z$ we get $$ x^{2}(\frac{a^{2}+1}{a^{2}})+y^{2}(\frac{a^{2}+1}{a^{2}})=1 $$

hence $$ x^{2}+y^{2}=\frac{a^{2}}{a^{2}+1} $$

using the equation for the cone we get $$ z=\frac{1}{\sqrt{a^{2}+1}} $$

I then did (and I am unsure about the boundaries) : $0<z<\frac{1}{\sqrt{a^{2}+1}},0<r<az$ and using the coordinates $x=r\cos(\theta),y=r\sin(\theta),z=z$ I got that the volume that the cone enclose in the ball is $$ \int_{0}^{\frac{1}{\sqrt{a^{2}+1}}}dz\int_{0}^{az}dr\int_{0}^{2\pi} r d\theta $$

which evaluates to $$ \frac{\pi a^{2}}{3(a^{2}+1)^{\frac{3}{2}}} $$

I then required this will be equal to e the volume of the upper half of the unit ball : $$ \frac{\pi a^{2}}{3(a^{2}+1)^{\frac{3}{2}}}=\frac{\pi}{3} $$

and got $$ a^{2}=(a^{2}+1)^{\frac{3}{2}} $$

which have no real solution, according to WA.

Can someone please help me understand where I am wrong and how to solve this question ?

Answer 1

Check your bounds again. I believe they should be

$$\begin{align}0<&z<\frac{1}{\sqrt{a^2+1}}\\az<&r<\sqrt{1-z^2}\end{align}$$

Finishing the integral with these bounds should yield $a=\sqrt3$.

Answer 2

The way I look at this is to find the volume of the spherical segment bounded by the cone, which is the volume of the spherical sector minus the volume of the cone. The volume of the sector is straightforward in spherical coordinates once you realize that the parameter $a$ is nothing more than the tangent of the cone half-angle; thus

$$V_{\text{sector}} = 2 \pi \int_0^1 dr \, r^2 \, \int_0^{\arctan{a}} d\theta \, \sin{\theta} = \frac{2 \pi}{3} \left ( 1 - \frac{1}{\sqrt{a^2+1}}\right) $$

The volume of the cone is also straightforward; the radius of the base (at the plane of intersection with the sphere) is $1/\sqrt{a^2+1}$, the height is $a/\sqrt{a^2+1}$, and the volume is

$$V_{\text{cone}} = \frac{\pi}{3} \frac{a^2}{(a^2+1)^{3/2}}$$

We want the difference between these two volumes to be one-half the volume of the hemisphere, or $\pi/3$. The equation we must solve is then

$$2 \left ( 1 - \frac{1}{\sqrt{a^2+1}}\right) - \frac{a^2}{(a^2+1)^{3/2}} = 1$$

We may simplify this some by letting $y=a^2+1$; some algebraic manipulation (check me on this) reveals that $y$ satisfies a cubic:

$$y^3-9 y^2+6 y-1=0$$

This equation has three real roots, but only one that is greater than one. The root that is greater than one is

$$y_0 = 3+2 \sqrt{7} \cos \left(\frac{1}{3} \arctan\left(\frac{\sqrt{3}}{37}\right)\right)$$

The value of $a$ is then approximately $2.70016$.