To solve the problem of finding the volume of the frustum of a pyramid using calculus, we can consider the frustum as the difference between two pyramids: a larger pyramid and a smaller pyramid that has been removed from the top.

### Step-by-Step Solution

1. **Define the height function:**

   Consider a vertical axis along the height of the pyramid. Let the height of the entire larger pyramid (including the removed part) be \( H \). The height of the smaller pyramid that is removed is \( H - h \), where \( h \) is the height of the frustum.

   If we place the frustum along the \( z \)-axis with \( z = 0 \) at the base of the frustum and \( z = h \) at the top, the height \( z \) of any cross-section of the pyramid at height \( z \) is:

   \[
   \text{Base side length at height } z = s(z) = b - \left(\frac{b - a}{h}\right) z
   \]

   The area of this square cross-section is:

   \[
   A(z) = s(z)^2 = \left[b - \left(\frac{b - a}{h}\right) z\right]^2
   \]

2. **Set up the integral:**

   The volume of the frustum can be found by integrating the area of these cross-sections from \( z = 0 \) to \( z = h \):

   \[
   V = \int_0^h A(z) \, dz
   \]

   Substituting the expression for \( A(z) \):

   \[
   V = \int_0^h \left[b - \left(\frac{b - a}{h}\right) z\right]^2 dz
   \]

3. **Expand and integrate:**

   Expand the square:

   \[
   V = \int_0^h \left[ b^2 - 2b\left(\frac{b-a}{h}\right)z + \left(\frac{b-a}{h}\right)^2 z^2 \right] dz
   \]

   Now, integrate each term separately:

   \[
   V = \int_0^h b^2 \, dz - 2b\left(\frac{b-a}{h}\right) \int_0^h z \, dz + \left(\frac{b-a}{h}\right)^2 \int_0^h z^2 \, dz
   \]

   Evaluate each integral:

   - \( \int_0^h b^2 \, dz = b^2 h \)
   - \( \int_0^h z \, dz = \frac{h^2}{2} \)
   - \( \int_0^h z^2 \, dz = \frac{h^3}{3} \)

   So the volume becomes:

   \[
   V = b^2 h - 2b\left(\frac{b-a}{h}\right) \cdot \frac{h^2}{2} + \left(\frac{b-a}{h}\right)^2 \cdot \frac{h^3}{3}
   \]

4. **Simplify the expression:**

   Simplify each term:

   \[
   V = b^2 h - b(b-a)h + \frac{(b-a)^2 h}{3}
   \]

   Combine like terms:

   \[
   V = \frac{1}{3} \left[ 3b^2h - 3b(b-a)h + (b-a)^2h \right]
   \]

   Expand and simplify:

   \[
   V = \frac{h}{3} \left( b^2 + a^2 + ab \right)
   \]

### Final Answer

The volume of the frustum is:

\[
V = \frac{h}{3} \left( b^2 + a^2 + ab \right)
\]

This result is consistent with the geometric formula derived earlier.

Would you like further details or have any questions?

Here are five related questions you might explore:

1. How does the integral approach change if the frustum has a circular base and top (making it a truncated cone)?
2. Can you derive the surface area of the frustum using calculus by integrating the lateral surface area along the height?
3. How would the approach differ if the cross-sections were not squares but rectangles?
4. Can you derive the volume of a frustum using a double integral approach?
5. How does the calculus approach help in finding the centroid of the frustum?

**Tip:** When using calculus to find volumes, setting up the integral correctly based on the shape's cross-sectional area is key to solving the problem accurately.

Learn how to calculate the volume of a frustum of a pyramid using calculus with a detailed step-by-step solution. This problem covers integral calculus techniques applied to geometric solids. Suitable for advanced high school or college students.

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