Answer
$$S = \int_{0}^{\frac{\pi}{2}} 2\pi (sin(2t)) \sqrt {cos^2(t) + 4cos^2(2t)} \space dt \approx 8.0285$$
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 $\frac{dy}{dt}$:
$\frac{dx}{dt} = \frac{d(sin(t))}{dt} = cos(t)$
$\frac{dy}{dt} = \frac{d(sin(2t))}{dt} = 2cos(2t) $
3. Substitute the given values into the formula:
$S = \int_{0}^{\frac{\pi}{2}} 2\pi (sin(2t)) \sqrt {(cos(t))^2 + (2cos(2t))^2} \space dt$
$S = \int_{0}^{\frac{\pi}{2}} 2\pi (sin(2t)) \sqrt {cos^2(t) + 4cos^2(2t)} \space dt$
4. Write that integral on your calculator (you can use an online one).
The result is: $S \approx 8.0285$
