To solve the problem, we need to compute the surface area of the frustum generated by revolving the line segment joining the points $(0, 2)$ and $(6, 6)$ about the $x$-axis. The problem is given with a parameterization of the curve, which will be helpful in finding the surface area.

Step 1: Parameterization of the Line Segment

The parameterization of the line segment is given as: $x = \frac{3}{2}t, \quad y = t + 2, \quad 0 \leq t \leq 4$ We can check the parameterization at the endpoints:

• When $t = 0$, we get $x = 0$ and $y = 2$, which corresponds to the point $(0, 2)$.
• When $t = 4$, we get $x = \frac{3}{2}(4) = 6$ and $y = 4 + 2 = 6$, which corresponds to the point $(6, 6)$.

Step 2: Formula for Surface Area of Revolution

The surface area of a surface generated by revolving a curve $y = f(x)$ about the $x$-axis is given by: $A = 2\pi \int_{a}^{b} y \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx$ But here we use parameterization, so we need to use the formula for surface area in terms of a parameter $t$: $A = 2\pi \int_{t_1}^{t_2} y(t) \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$ In this case, the range for $t$ is from 0 to 4.

Step 3: Derivatives of $x(t)$ and $y(t)$

We first compute the derivatives: $\frac{dx}{dt} = \frac{3}{2}, \quad \frac{dy}{dt} = 1$

Step 4: Surface Area Integral

Now, substitute these into the surface area formula: $A = 2\pi \int_{0}^{4} (t + 2) \sqrt{\left(\frac{3}{2}\right)^2 + 1^2} \, dt$ Simplifying inside the square root: $A = 2\pi \int_{0}^{4} (t + 2) \sqrt{\frac{9}{4} + 1} \, dt = 2\pi \int_{0}^{4} (t + 2) \sqrt{\frac{13}{4}} \, dt$ $A = 2\pi \cdot \frac{\sqrt{13}}{2} \int_{0}^{4} (t + 2) \, dt$ The integral $\int_{0}^{4} (t + 2) \, dt$ is: $\int_{0}^{4} (t + 2) \, dt = \left[\frac{t^2}{2} + 2t\right]_{0}^{4} = \left(\frac{16}{2} + 8\right) - \left(0 + 0\right) = 8 + 8 = 16$

Thus, the surface area is: $A = 2\pi \cdot \frac{\sqrt{13}}{2} \cdot 16 = 16\pi \sqrt{13}$

Step 5: Checking with the Geometry Formula

The formula for the surface area of a frustum of a cone is: $A = \pi (r_1 + r_2) \cdot l$ Where:

• $r_1$ and $r_2$ are the radii of the frustum at the ends, which are the $y$-coordinates of the points: $r_1 = 2$, $r_2 = 6$.
• $l$ is the slant height, which is the distance between the two points $(0, 2)$ and $(6, 6)$: $l = \sqrt{(6 - 0)^2 + (6 - 2)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}$

Thus, the surface area using the geometry formula is: $A = \pi (2 + 6) \cdot 2\sqrt{13} = 8\pi \cdot 2\sqrt{13} = 16\pi \sqrt{13}$

Conclusion:

Both methods give the same surface area, confirming the result: $\boxed{16\pi \sqrt{13}}$

Do you need further details or have any questions?

Here are 5 related questions to expand on this topic:

1. How would you calculate the volume of the frustum generated in this problem?
2. What changes in the surface area formula if the curve was rotated around the $y$-axis instead?
3. Can you apply the same parameterization approach for different shapes, like a parabola?
4. How is the surface area affected if the length of the line segment changes?
5. What is the relationship between the slant height and the length of the line segment in such problems?

Tip: When calculating surface areas for solids of revolution, always verify the limits of integration and derivatives to ensure accuracy.