The bucket described in the problem is in the shape of a frustum of a cone. The volume of a frustum of a cone can be calculated using the formula \(V = \frac{1}{3}πh(r_1^2 + r_2^2 + r_1r_2)\), where h is the height of the frustum, \(r_1\) is the radius of the top circle, and \(r_2\) is the radius of the bottom circle.

In this case, the height of the frustum is equal to the depth of the bucket, which is 4 cm. The radius of the top circle is half the diameter of the top, which is \(\frac{1}{2} \cdot 12 cm = 6 cm\). The radius of the bottom circle is half the diameter of the bottom, which is \(\frac{1}{2} \cdot 8 cm = 4 cm\).

Substituting these values into the formula for the volume of a frustum, we get:

\(V = \frac{1}{3}π \cdot 4 \cdot (6^2 + 4^2 + 6 \cdot 4)\)

\(= \frac{4}{3}π \cdot (36 + 16 + 24)\)

\(= \frac{4}{3}π \cdot 76\)

\(= 304π/3 cm^3\)

So, the volume of the bucket is 304π/3 cm^3.