To find the surface area formed by rotating the parametric curve about the $x$-axis, we use the formula for surface area generated by rotating a parametric curve $(x(t), y(t))$ around the $x$-axis:

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

Given the parametric equations:

$x(t) = 3t^2, \quad y(t) = 2t, \quad t \in [0, 4]$

Step 1: Compute the derivatives

We need to find the derivatives of $x(t)$ and $y(t)$ with respect to $t$:

$\frac{dx}{dt} = \frac{d}{dt}(3t^2) = 6t$ $\frac{dy}{dt} = \frac{d}{dt}(2t) = 2$

Step 2: Plug into the surface area formula

Now, we plug these into the surface area formula:

$S = 2\pi \int_{0}^{4} y(t) \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt$

Substitute $y(t) = 2t$, $\frac{dx}{dt} = 6t$, and $\frac{dy}{dt} = 2$:

$S = 2\pi \int_{0}^{4} 2t \sqrt{(6t)^2 + (2)^2} \, dt$ $S = 2\pi \int_{0}^{4} 2t \sqrt{36t^2 + 4} \, dt$ $S = 4\pi \int_{0}^{4} t \sqrt{36t^2 + 4} \, dt$

Step 3: Solve the integral

We can perform a substitution to simplify the integral. Let:

$u = 36t^2 + 4$ Then, the differential of $u$ is:

$du = 72t \, dt$

Now substitute into the integral:

$S = 4\pi \int_{u(0)}^{u(4)} \frac{\sqrt{u}}{72} \, du$ $S = \frac{4\pi}{72} \int_{4}^{580} \sqrt{u} \, du$ $S = \frac{\pi}{18} \left[ \frac{2}{3} u^{3/2} \right]_{4}^{580}$ $S = \frac{\pi}{18} \left( \frac{2}{3} \left(580^{3/2} - 4^{3/2}\right) \right)$ $S = \frac{\pi}{18} \left( \frac{2}{3} \left(580\sqrt{580} - 8\right) \right)$

Step 4: Final calculation

Now, calculating the exact value of the expression:

$S = \frac{2\pi}{54} \left(580\sqrt{580} - 8\right)$

This represents the exact surface area formed by rotating the curve about the $x$-axis.

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Would you like more details on the simplifications or approximations?

Here are some related questions:

1. How do you find the surface area of a curve rotated about the $y$-axis?
2. What is the difference between parametric and polar coordinates in surface area calculations?
3. Can you find the volume generated by rotating the same curve about the $x$-axis?
4. How does the surface area formula change if the curve is rotated about the $y$-axis?
5. How would you compute the surface area if the curve was rotated about another arbitrary axis?

Tip: In parametric equations, always remember to differentiate correctly with respect to the parameter $t$, as errors in differentiation can affect the final outcome of integral calculations.