How boring can you get?

I have a large, hemispherical piece of bread with a radius of $1$ foot. I make a bread bowl by boring out a cylindrical hole with radius $r,$ centered at the top of the hemisphere and extending all the way to the flat bottom crust.

What should the radius of my borehole be to maximize the volume of soup my bread bowl can hold?

N.B. I originally conceived of this week's Fiddler problem in my mind's eye with the hemispherical bread having a flat upper crust, which led to initially thinking that it was a very weird setup where you keep cutting your cylindrical hole downward until you ever so slightly bump up against the curved bottom crust at which point you stop since obviously otherwise your bread bowl would have a hole in it and your soup would leak out. This is essentially the inverted logic of the handwavy no-surface-tension argument I make below, so in the end I think the math ends up being roughly the same ....

So anyway, freely choosing the coordinate system that best suits me, assume that your bread fills the space $B = \{ (x,y,z) \mid x^2 + y^2 + z^2 \leq 1, z \geq 0 \}.$ Assume that a cylindrical borehole takes would remove the portion of the $B$ that satisfyies $x^2 + y^2 \leq r^2,$ for some radius $r \gt 0.$ Ultimately, since we cannot rely on molecular properties of the varying soups that might fill the bread bowl to postulate any additional volume of soup due to surface tension, let's assume that the bread bowl can only be filled up to its upper rim, that is, the cylindrical cavity is given by $C(r) = \{ (x,y,z) \mid x^2 + y^2 \leq r^2, 0 \lt z \leq \sqrt{1-r^2}\}.$ The volume of $C(r)$ is given by $V(r) = \pi r^2 \sqrt{1-r^2}.$

Differentiating $V(r)$ gives

$$V^\prime(r) = 2\pi r \sqrt{1-r^2} + \pi r^2 \left( \frac{-r}{\sqrt{1-r^2}} \right) = \frac{\pi r \left( 2(1-r^2) -r^2 \right)}{\sqrt{1-r^2}} = \frac{\pi r (2-3r^2)}{\sqrt{1-r^2}}.$$ Thus $V$ has critical points as $r_1 = 0$, $r_2 = \sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3},$ and $r_3 = 1.$ Since $V(0)=V(1)=0,$ the maximum possible volume of soup contained in this bread bowl is $$V^* = \frac{2\pi\sqrt{3}}{9} \approx 1.20919957616\dots$$ cubic feet, which occurs when choosing a radius of $$r^* = \frac{\sqrt{6}}{3} \approx 0.816496580928\dots$$ feet.

Instead of a hemisphere, now suppose my bread is a sphere with a radius of $1$ foot. Again, I make a bowl by boring out a cylindrical shape with radius $r,$ extending all the way to (but not through) the curved bottom crust of the bread. The central axis of the hole must pass through the center of the sphere.

What should the radius of my borehole be to maximize the volume of soup my bread bowl can hold?

Again hearkening back to my earlier spatial reasoning struggles, I could not for the life of me understand why this extra credit problem was in any way different from just having double the volume since you would then stop cutting as soon as you hit the curved bottom crust. Since I take Axiom of the Benevolent Fiddlermeister as a given, I have to assume that the extra credit problem is somehow different from the regular problem, so we will assume that I have a precision instrument that can bore a perfectly cylindrical hole in the spherical bread until I ever so slightly approach the curved lower crust and then somehow liquefy and extract the remaining bready portions all the way down to curved lower crust. So in this case the bowl takes the shape $\tilde{C}(r) = \{ (x,y,z) \mid x^2 + y^2 \leq r^2, -\sqrt{1-x^2-y^2} \lt z \leq \sqrt{1-r^2} \}.$ In this case, the volume is

$$\begin{align*}\tilde{V}(r) &= \int_{ x^2 + y^2 \leq r^2 } \int_{-\sqrt{1-x^2-y^2}}^{\sqrt{1-r^2}} \, dz \,dy \,dx\\ &= \int_{x^2 + y^2 \leq r^2} \left( \sqrt{1-r^2} + \sqrt{1-x^2 -y^2}\right) \,dy \,dx \\ &= \int_0^{2\pi} \int_0^r \left( \sqrt{1-r^2} + \sqrt{1 - \rho^2} \right) \rho \,d\rho \, d\theta \\ &= \pi r^2 \sqrt{1-r^2} + 2\pi \int_0^r \rho \sqrt{1-\rho^2} \,d\rho \\ &= \pi r^2 \sqrt{1-r^2} + \frac{2\pi}{3} \left( 1 - (1-r^2)^{3/2} \right).\end{align*}$$ We can write $(1-r^2)^{3/2} = (1-r^2) \sqrt{1-r^2}$ to then factor even further and see that $$\tilde{V}(r) = \pi r^2 \sqrt{1-r^2} + \frac{2\pi}{3} \left( 1 - (1-r^2) \sqrt{1-r^2} \right) = \frac{2\pi}{3} + \frac{\pi \sqrt{1-r^2}}{3} (5r^2 - 2).$$

Differentiating we get

$$\begin{align*}\tilde{V}^\prime (r) &= \frac{10\pi r}{3} \sqrt{1-r^2} + \frac{\pi}{3} (5r^2 - 2) \left( \frac{-r}{\sqrt{1-r^2}} \right)\\ &= \frac{\pi r}{3\sqrt{1-r^2}} \left( 10 (1-r^2) - (5r^2 - 2) \right)\\ &= \frac{\pi r (4-5r^2)}{\sqrt{1-r^2}}.\end{align*}$$ Thus $\tilde{V}$ has critical points at $r_1 = 0,$ $r_2 = \sqrt{\frac{4}{5}} = \frac{2\sqrt{5}}{5},$ and $r_3 = 1.$ Here since we have $\tilde{V}^\prime \gt 0$ when $r \lt \frac{2\sqrt{5}}{5}$ and $\tilde{V}^\prime \lt 0$ when $r \gt \frac{2\sqrt{5}}{5},$ we see that the maximum possible volume of our miraculously scooped out bread bowls with curved lower crusts is $$\tilde{V}^* = \frac{2\pi(5+\sqrt{5})}{15} \approx 3.03103706653\dots$$ cubic feet, which occurs when choosing a radius of $$\tilde{r}^* = \frac{2\sqrt{5}}{5} \approx 0.894427191\dots$$ feet.