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A solid sphere with radius $r$ fits exactly inside a cylinder, touching the sides, the top and the bottom. What fraction of the cylinder is empty?

Area of circle: $\pi{r^2}$
Area of cylinder: $2\pi rh+2\pi r^2 $ (in this case, since the sides of the circle touch the edges of the cylinder, the height is $2r$) so: $6\pi r^2$.

I divided both and got $6$, however it's the wrong answer.

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The volume of the sphere is given by $\frac 43 \pi r^3$. From the condition in the problem we have that the height of the cylinder is $2r$, while the radius of the base is same as the radius of the ball. Hence:

$$\frac{V_{sph}}{V_{cyl}} = \frac{\frac 43 \pi r^3}{r^2 \pi \cdot 2r} = \frac{2}{3}$$

Hence a third of the cylinder is empty.

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