Centroid of a conical frustrum

The height of the centroid of the frustrum is $\frac{\text{Weighted Mean}}{\text{Volume}}$ $=$ $\frac{\pi \int_{0}^{H}x(B + (A - B)\frac{x}{H})^2 dx}{\pi \int_{0}^{H}(B + (A - B)\frac{x}{H})^2 dx}$ $=$ $\frac{\frac{1}{12}\pi H^2(3A^2 + 2AB + B^2)}{\frac{1}{3}\pi H(A^2 + AB + B^2)}$ $=$ $\frac{H(3A^2 + 2AB + B^2)}{4(A^2 + AB + B^2)}$ .

When $A = 20 \text{cm}$ , $B = 40 \text{cm}$ , and $H = 28 \text{cm}$ , the height of the centroid is $\frac{28(3(20)^2 + 2 \cdot 20 \cdot 40 + 40^2)}{4(20^2 + 20 \cdot 40 + 40^2)} = \boxed{11 \text{cm}}$ .