Bar, Ball, Cone

Since all the three have equal radius, let's set $\large r=1$

Volume of the cylindrical bar: $\large V_{cylinder}=\pi r^2h=\pi(1^2)(2)=2 \pi$

Volume of spherical ball: $\large V_{sphere}=\dfrac{4}{3} \pi r^3=\dfrac{4}{3} \pi (1^3)=\dfrac{4}{3} \pi$

Volume of cone: $\large V_{cone}=\dfrac{1}{3} \pi r^2h=\dfrac{1}{3} \pi (1^2)(2)=\dfrac{2}{3} \pi$

Compare:

$\large V_{cylinder} = V_{sphere} + V_{cone}$

$\large 2 \pi = \dfrac{4}{3} \pi + \dfrac{2}{3} \pi$

$\large 2 \pi = 2 \pi$

$\boxed{\color{#D61F06}\large \therefore Both~options~have~the~same~amount.}$

Let us compute the volume of the two options and then compare.

Option 1: the cylindrical bar

The standard formula for the volume of a cylinder is $\pi r^{2} h$ . In this case h=2r. Substituting this value in, the total volume for option 1 is $2\pi r^{3}$

Option 2: the ball and the cone

The standard formula for the volume of a ball (a sphere) is $\frac43 \pi r^{3}$ . The standard formula for the volume of a cone is $\frac13 \pi r^{2} h$ . h=2r so the volume of the cone is $\frac23 \pi r^{3}$ Thus the total volume for option 2 is $\frac43 \pi r^{3} + \frac23 \pi r^{3} = 2 \pi r^{3}$

Clearly the volume is the same for option 1 and option 2, thus it does not matter which option is chosen.

It's due to the Cavaliere's Principle. When it has same height and radius, the sum of the volume of a sphere and a cone is the same as the volume of a cilinder.

I did not want to work out the volumes of all the shapes, so I did this instead.

The volume of the cylinder is equal to the volume of the sphere plus two congruent cones of height r and base radius r. This is provable from a proof of the volume of a sphere.

All that remains to be shown is whether the volume of the two cones is greater than the single cone of double the height. It can be seen that $\frac{1}{3} \pi r^2 \cdot 2r=2(\frac{1}{3} \pi r^2\cdot r)$ and therefore the volume of the cylinder is equal to the sum of the volume of the cone and the sphere.

With h = 2 r,

Pi (r^2) h = (4/ 3) Pi (r^3) + (1/ 3) Pi (r^2) h

as 2 = 4/ 3 + 2/ 3

Same total volume.