To find the exact area of the surface obtained by rotating the given curve about the $x$-axis, we use the formula for the surface area of revolution:

$S = 2\pi \int_{a}^{b} y \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$

Step 1: Compute $\frac{dx}{dt}$ and $\frac{dy}{dt}$

The given parametric equations are: $x = 9t - 3t^3, \quad y = 9t^2$ Differentiating $x$ and $y$ with respect to $t$: $\frac{dx}{dt} = 9 - 9t^2, \quad \frac{dy}{dt} = 18t$

Step 2: Find $\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}$

Substitute the derivatives: $\left(\frac{dx}{dt}\right)^2 = (9 - 9t^2)^2 = 81 - 162t^2 + 81t^4$ $\left(\frac{dy}{dt}\right)^2 = (18t)^2 = 324t^2$ Add them: $\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = 81 - 162t^2 + 81t^4 + 324t^2$ Simplify: $\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = 81 + 162t^2 + 81t^4$

Thus: $\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} = \sqrt{81 + 162t^2 + 81t^4}$

Step 3: Set up the integral

The surface area formula becomes: $S = 2\pi \int_{0}^{2} y \sqrt{81 + 162t^2 + 81t^4} \, dt$ Substitute $y = 9t^2$: $S = 2\pi \int_{0}^{2} (9t^2) \sqrt{81 + 162t^2 + 81t^4} \, dt$ Factor out $81$ from the square root: $S = 2\pi \int_{0}^{2} 9t^2 \sqrt{81(1 + 2t^2 + t^4)} \, dt$ $S = 2\pi \int_{0}^{2} 9t^2 \cdot 9\sqrt{1 + 2t^2 + t^4} \, dt$ $S = 162\pi \int_{0}^{2} t^2 \sqrt{1 + 2t^2 + t^4} \, dt$

Step 4: Simplify and solve the integral

The integrand is simplified as: $\sqrt{1 + 2t^2 + t^4} = (1 + t^2)$ Thus: $S = 162\pi \int_{0}^{2} t^2 (1 + t^2) \, dt$

Expand the product: $S = 162\pi \int_{0}^{2} (t^2 + t^4) \, dt$ Split into two integrals: $S = 162\pi \left( \int_{0}^{2} t^2 \, dt + \int_{0}^{2} t^4 \, dt \right)$

Compute the integrals:

1. For $\int_{0}^{2} t^2 \, dt$: $\int t^2 \, dt = \frac{t^3}{3}, \quad \text{evaluate from } 0 \text{ to } 2:$ $\int_{0}^{2} t^2 \, dt = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3}$

2. For $\int_{0}^{2} t^4 \, dt$: $\int t^4 \, dt = \frac{t^5}{5}, \quad \text{evaluate from } 0 \text{ to } 2:$ $\int_{0}^{2} t^4 \, dt = \frac{2^5}{5} - \frac{0^5}{5} = \frac{32}{5}$

Substitute back into $S$: $S = 162\pi \left( \frac{8}{3} + \frac{32}{5} \right)$

Step 5: Combine terms

Find a common denominator for $\frac{8}{3} + \frac{32}{5}$: $\frac{8}{3} = \frac{40}{15}, \quad \frac{32}{5} = \frac{96}{15}$ $\frac{8}{3} + \frac{32}{5} = \frac{40}{15} + \frac{96}{15} = \frac{136}{15}$

Thus: $S = 162\pi \cdot \frac{136}{15}$ Simplify: $S = \frac{162 \cdot 136 \cdot \pi}{15} = \frac{22032\pi}{15} = 1468.8\pi$

The exact surface area is: $S = \frac{22032\pi}{15}$

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Related Questions

1. What is the significance of parametric equations in calculating surface areas of revolution?
2. How does the choice of axis affect the formula for surface area?
3. Can the integral for surface area always be solved explicitly, or are numerical methods sometimes necessary?
4. How would the problem change if the curve were revolved about the $y$-axis instead?
5. How does factoring out common terms simplify the computation of integrals?

Tip: Always check if the square root in the integral can be simplified, as it can save significant computational effort!