Let's break down the problem step by step.

You are asked to find the volume $V$ of a frustum of a right circular cone, which has height $h$, lower base radius $R$, and top radius $r$, using the disk method.

Part 1:

We are given that the integral setup for the volume is correct: $V = \pi \int_0^h \left( R - \frac{y(R - r)}{h} \right)^2 \, dy$

This integral represents the volume of the solid obtained by revolving the frustum around the y-axis.

Part 2: Finding the antiderivative

The next step is to compute the antiderivative of the expression $\left( R - \frac{y(R - r)}{h} \right)^2$. To do this, we expand the square term and then integrate with respect to $y$.

1. Expanding the square: $\left( R - \frac{y(R - r)}{h} \right)^2 = \left( R^2 - 2R \cdot \frac{y(R - r)}{h} + \left( \frac{y(R - r)}{h} \right)^2 \right)$ This expands to: $= R^2 - 2R \cdot \frac{y(R - r)}{h} + \frac{y^2(R - r)^2}{h^2}$

2. Integration: Now, integrate each term from $0$ to $h$:

   • For $R^2$, the integral is: $\int_0^h R^2 \, dy = R^2 h$

   • For $-2R \cdot \frac{y(R - r)}{h}$, the integral is: $\int_0^h -2R \cdot \frac{y(R - r)}{h} \, dy = -2R \cdot \frac{(R - r)}{h} \cdot \frac{h^2}{2} = -R \cdot h(R - r)$

   • For $\frac{y^2(R - r)^2}{h^2}$, the integral is: $\int_0^h \frac{y^2(R - r)^2}{h^2} \, dy = \frac{(R - r)^2}{h^2} \cdot \frac{h^3}{3} = \frac{h(R - r)^2}{3}$

3. Final antiderivative expression: Combining these, the antiderivative is: $V = \pi \left[ R^2 h - R h (R - r) + \frac{h (R - r)^2}{3} \right]$

Thus, this is the expression for the volume of the frustum of the cone.

Do you have any questions or need further details?

Here are 5 related questions to explore:

1. How do you apply the disk method to solids of revolution in general?
2. What is the geometric interpretation of each term in the expanded integral?
3. How would the volume change if $r = R$, making it a complete cone?
4. What is the relation between a frustum and a truncated cone?
5. How would the setup differ if the frustum was rotated around the x-axis instead?

Tip: When expanding squared terms inside integrals, make sure to simplify step-by-step before attempting integration.