Answer
$$S = \int_{0}^{\frac{\pi}{2}} 2 \pi (tcos(t)) \sqrt {1 + t^2} \space dt \approx 4.7394 $$
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 $\frac{dy}{dt}$:
$\frac{dx}{dt} = \frac{d(tsin(t))}{dt} = sin(t) + tcos(t)$
$\frac{dy}{dt} = \frac{d(tcos(t))}{dt} = cos(t) + t(-sin(t)) = cos(t) - tsin(t)$
3. Substitute the given values into the formula:
$S = \int_{0}^{\frac{\pi}{2}} 2\pi (tcos(t)) \sqrt {(sin(t) + tcos(t))^2 + (cos(t) - tsin(t))^2} \space dt$
$S = \int_{0}^{\frac{\pi}{2}} 2\pi (tcos(t)) \sqrt {(sin^2(t) + 2sin(t)tcos(t) + t^2cos^2(t)) + (cos^2(t) -2cos(t)tsin(t) + t^2sin^2(t))} \space dt$
$S = \int_{0}^{\frac{\pi}{2}} 2\pi (tcos(t)) \sqrt {sin^2(t) + t^2cos^2(t) + cos^2(t) + t^2sin^2(t)} \space dt$
$S = \int_{0}^{\frac{\pi}{2}} 2\pi (tcos(t)) \sqrt {1 + t^2(sin^2(t) + cos^2(t))} \space dt$
$S = \int_{0}^{\frac{\pi}{2}} 2 \pi (tcos(t)) \sqrt {1 + t^2} \space dt$
4. Write that integral on your calculator (you can use an online one).
The result is: $S \approx 4.7394$
