A $5 \mathrm{~m} \times 5 \mathrm{~m}$ flat square roof receives $6 \mathrm{~mm}$ of rainfall. All of this water (and no other water) drains into an empty rain barrel. The rain barrel is in the shape of a cylinder with a diameter of $0.5 \mathrm{~m}$ and a height of $1 \mathrm{~m}$. Rounded to the nearest tenth of a percent, what percentage of the barrel will be full of water?
Since the roof measures $5 \mathrm{~m}$ by $5 \mathrm{~m}$, then its area is $5^{2}=25 \mathrm{~m}^{2}$.

When this roof receives $6 \mathrm{~mm}$ (or $0.006 \mathrm{~m}$ ) of rain, the total volume of rain that the roof receives is $25 \cdot 0.006=0.15 \mathrm{~m}^{3}$. (We can imagine that the rain that falls forms a rectangular prism with base that is $5 \mathrm{~m}$ by $5 \mathrm{~m}$ and height $6 \mathrm{~mm}$.)

The rain barrel has diameter $0.5 \mathrm{~m}$ (and so radius $0.25 \mathrm{~m}$ ) and a height of $1 \mathrm{~m}$, and so has volume $\pi \cdot 0.25^{2} \cdot 1=0.0625 \pi \mathrm{m}^{3}$.

Thus, the percentage of the barrel that is full is

$$
\frac{0.15}{0.0625 \pi} \times 100 \% \approx 76.39 \%
$$

To the nearest tenth of a percent, the rain barrel will be $76.4 \%$ full of water.