Martina has a cylinder with radius $10 \mathrm{~cm}$ and height $70 \mathrm{~cm}$ that is closed at the bottom. Martina puts some solid spheres of radius $10 \mathrm{~cm}$ inside the cylinder and then closes the top end of the cylinder. If she puts the largest possible number of spheres in the cylinder, what is the volume, in $\mathrm{cm}^{3}$, of the cylinder that is not taken up by the spheres?
The volume of a cylinder with radius $r$ and height $h$ is $\pi r^{2} h$.

The volume of a sphere with radius $r$ is $\frac{4}{3} \pi r^{3}$.

The given cylinder has radius $10 \mathrm{~cm}$ and height $70 \mathrm{~cm}$ and so has volume $\pi(10 \mathrm{~cm})^{2}(70 \mathrm{~cm})=$ $7000 \pi \mathrm{cm}^{3}$.

Since the radius of the spheres and the radius of the cylinder are equal, then we can view each of the spheres as having a "height" of $2 \cdot 10 \mathrm{~cm}=20 \mathrm{~cm}$. (In other words, the spheres can only stack perfectly on top of each other.)

Since the height of the cylinder is $70 \mathrm{~cm}$ and the height of each sphere is $20 \mathrm{~cm}$, then a maximum of 3 spheres will fit inside the closed cylinder.

Therefore, the volume of the cylinder not taken up by the spheres is $7000 \pi \mathrm{cm}^{3}-3 \cdot \frac{4}{3} \pi(10 \mathrm{~cm})^{3}$ or $7000 \pi \mathrm{cm}^{3}-4000 \pi \mathrm{cm}^{3}$ or $3000 \pi \mathrm{cm}^{3}$.

ANSWER: $3000 \pi \mathrm{cm}^{3}$