Calculate the volume of a frustum of a cone if the diameters of the top and bottom ends are 6 cm and 10 cm, respectively and the vertical distance between them is 4 cm.

The formula for the volume of a frustum is $V = \frac{1}{3}\pi h (R^2 + Rr + r^2)$.
Substitute the given values: $R = 25 \text{ cm}$, $r = 4 \text{ cm}$, $h = 42.5 \text{ cm}$.

$$V = \frac{1}{3} \times \pi \times 42.5 \times (25^2 + 25 \times 4 + 4^2)$$ $$V = \frac{1}{3} \times \pi \times 42.5 \times (625 + 100 + 16)$$ $$V = \frac{1}{3} \times \pi \times 42.5 \times 741$$ $$V = \pi \times 42.5 \times 247$$ $$V = 10497.5 \pi \text{ cm}^3$$

Using $\pi \approx 3.14159$:

$$V \approx 10497.5 \times 3.14159$$ $$V \approx 32979.6 \text{ cm}^3$$

The capacity of the bucket is approximately $\boxed{\text{32979.6 cm}^3}$.

This refers to the curved surface area of the frustum plus the area of the bottom circular base. The top is open.
First, we need to find the slant height $l$ of the frustum.
The formula for the slant height of a frustum is $l = \sqrt{h^2 + (R-r)^2}$.

$$l = \sqrt{42.5^2 + (25-4)^2}$$ $$l = \sqrt{42.5^2 + 21^2}$$ $$l = \sqrt{1806.25 + 441}$$ $$l = \sqrt{2247.25}$$ $$l \approx 47.405 \text{ cm}$$

The formula for the curved surface area of a frustum is $A_{curved} = \pi (R+r)l$.

$$A_{curved} = \pi (25+4) \times 47.405$$ $$A_{curved} = \pi \times 29 \times 47.405$$ $$A_{curved} = 1374.745 \pi \text{ cm}^2$$

Using $\pi \approx 3.14159$:

$$A_{curved} \approx 1374.745 \times 3.14159$$ $$A_{curved} \approx 4319.9 \text{ cm}^2$$

The formula for the area of a circle is $A_{base} = \pi r^2$.

$$A_{base} = \pi \times 4^2$$ $$A_{base} = 16 \pi \text{ cm}^2$$

Using $\pi \approx 3.14159$:

$$A_{base} \approx 16 \times 3.14159$$ $$A_{base} \approx 50.265 \text{ cm}^2$$

Total area $A_{total} = A_{curved} + A_{base}$.

$$A_{total} = 4319.9 + 50.265$$ $$A_{total} = 4370.165 \text{ cm}^2$$

The area of material required to make the bucket is approximately $\boxed{\text{4370.2 cm}^2}$.