Answer
$$S = \int_{0}^{1} 2\pi (t + t^4) \sqrt {1 + 4t^2 -4t^3 + 9t^4 +
16t^6} \space dt \approx 12.7176$$
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 $\frac{dy}{dt}$:
$\frac{dx}{dt} = \frac{d(t^2 - t^3)}{dt} = 2t -3t^2$
$\frac{dy}{dt} = \frac{d(t + t^4)}{dt} = 1 + 4t^3 $
3. Substitute the given values into the formula:
$S = \int_{0}^{1} 2\pi (t + t^4) \sqrt {(2t - 3t^2)^2 + (1 + 4t^3)^2} \space dt$
$S = \int_{0}^{1} 2\pi (t + t^4) \sqrt {(4t^2 -12t^3 + 9t^4) + (1 +
8t^3 +16t^6)} \space dt$
$S = \int_{0}^{1} 2\pi (t + t^4) \sqrt {1 + 4t^2 -4t^3 + 9t^4 +
16t^6} \space dt$
4. Write that integral on your calculator (you can use an online one).
The result is: $S \approx 12.7176$
