5. A mouse is sitting in a toy car on a negligibly small turntable. The car cannot turn on its own, but the mouse can control when the car is launched and when the car stops (the car has brakes). When the mouse chooses to launch, the car will immediately leave the turntable on a straight trajectory at 1 meter per second.
Suddenly someone turns on the turntable; it spins at $30 \mathrm{rpm}$. Consider the set $S$ of points the mouse can reach in his car within 1 second after the turntable is set in motion. (For example, the arrows in the figure below represent two possible paths the mouse can take.) What is the area of $S$, in square meters?
Solution: $\pi / 6$
The mouse can wait while the table rotates through some angle $\theta$ and then spend the remainder of the time moving along that ray at $1 \mathrm{~m} / \mathrm{s}$. He can reach any point between the starting point and the furthest reachable point along the ray, $(1-\theta / \pi)$ meters out. So the area is given by the polar integral
$$
\int_{0}^{\pi} \frac{(1-\theta / \pi)^{2}}{2} d \theta=\frac{1}{2} \cdot \frac{1}{\pi^{2}} \int_{0}^{\pi} \phi^{2} d \phi=\pi / 6
$$
(where we have used the change of variables $\phi=\pi-\theta$ ).