Answer
$$S =\frac {48 \pi} 5 $$
Work Step by Step
1. Write the formula for the area of the surface obtained by rotating a curve about the $x$-axis. $$S = \int_{\alpha}^{\beta} 2\pi y \sqrt {(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \space dt$$
2. Calculate: $\frac{dx}{dt}$ and $\frac{dy}{dt}$:
$\frac{dx}{dt} = \frac{d(3t - t^3)}{dt} = 3 - 3t^2$
$\frac{dy}{dt} = \frac{d(3t^2)}{dt} = 6t $
3. Substitute the given values into the formula:
$S = \int_{0}^{1} 2\pi (3t^2) \sqrt {(3-3t^2)^2 + (6t)^2} \space dt$
$S = \int_{0}^{1} 2\pi (3t^2) \sqrt {(9 - 18t^2 + 9t^4) + (36t^2)} \space dt$
$S = \int_{0}^{1} 2\pi (3t^2) \sqrt {(9 + 18t^2 + 9t^4) } \space dt$
$S = \int_{0}^{1} 2\pi (3t^2) \sqrt {(3 + 3t^2)^2 } \space dt$
$S = \int_{0}^{1} 2\pi (3t^2) (3 + 3t^2) \space dt$
$S = 2\pi \int_{0}^{1} (9t^2 + 9t^4) \space dt$
$S = 2\pi [3t^3 + \frac 9 5 t^5]^0_1$
$S = 2\pi (3(1)^3 + \frac 9 5 (1)^5) - 2\pi (3(0)^3 + \frac 9 5 (0)^5)$
$S = 2\pi (3 + \frac 9 5 ) $
$S = 2\pi (\frac {24} 5 ) $
$S =\frac {48 \pi} 5 $
