This is my first query on this site. I was asked to prove a statement in my maths textbook.
The figure shows a bowl in the shape of a hemisphere with a radius of r cm. It is filled with water to a depth of r/2 cm. Show that the volume of the water is more than 1/8 of the volume of the hemisphere.
I tried using both geometry and calculus for this to find the volume of the water
For geometry, I dissected the cross-section like this: Picture 1 . I tried using the ellipsoid volume formula ($V = \frac{4\pi(abc)}{3}$) to it, and it yielded $V = \frac{\pi r^3}{4}$.
For calculus, I used volumes of revolution in this picture: Picture 2 . The one on the left is the geometry working and the one on the right is using calculus. I obtained $V = \frac{5\pi r^3}{24}$
I have two questions:
Why are the results different?
How do you prove that without using calculus (volumes of revolution)?
Here's an idea for 2. In a hemisphere of radius $r$, consider the inscribed right circular cone whose base is the disk bounding the hemisphere, and partition both volumes by a plane at depth $r/2$. The volume of water inside the cone to depth $r/2$ is exactly one-eighth ($= (1/2)^{3}$) the volume of the cone, and it's geometrically clear that the corresponding volume of water in the hemisphere is a greater fraction of the hemisphere's volume.
This can be made quantitative using coordinates: The cross-sectional radius $\sqrt{r^{2} - (r - h)^{2}} = \sqrt{h(2r - h)}$ of the hemisphere at depth $h$ above the bottom divided by the cross-sectional radius $h$ of the cone at the same depth is a decreasing function of $h$: $$ \frac{\sqrt{h(2r - h)}}{h} = \sqrt{\frac{2r}{h} - 1}. $$