Answer
$$S = \frac 6 5 a^2 \pi $$
Work Step by Step
1. Write the formula for the area of the surface obtained by rotating a curve about the $x$-axis. $$S = \int_{\alpha}^{\beta} 2\pi y \sqrt {(\frac{dx}{d\theta})^2 + (\frac{dy}{d\theta})^2} \space d\theta$$
2. Calculate: $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$:
$\frac{dx}{d\theta} = \frac{d(acos^3(\theta))}{d\theta} = -3acos^2(\theta)sin(\theta)$
$\frac{dy}{d\theta} = \frac{d(asin^3(\theta))}{d\theta} = 3asin^2(\theta)cos(\theta) $
3. Substitute the given values into the formula:
$S = \int_{0}^{\pi/2} 2\pi (asin^3(\theta)) \sqrt {(-3acos^2(\theta)sin(\theta))^2 + (3asin^2(\theta)cos(\theta) )} \space dt$
$S = \int_{0}^{\pi/2} 2\pi (asin^3(\theta)) \sqrt {9a^2cos^4(\theta)sin^2(\theta) + 9a^2sin^4(\theta)cos^2(\theta) } \space dt$
$S = \int_{0}^{\pi/2} 2\pi (asin^3(\theta)) \sqrt {9a^2cos^2(\theta)sin^2(\theta)( cos^2(\theta )+ sin^2(\theta)) } \space dt$
$S = \int_{0}^{\pi/2} 2\pi (asin^3(\theta))(3acos(\theta)sin(\theta)) \sqrt {( 1) } \space dt$
$S = \int_{0}^{\pi/2} 6\pi (a^2cos(\theta)sin^4(\theta)) \space dt$
$S = 6a^2 \pi \int_{0}^{\pi/2} cos(\theta)sin^4(\theta) \space dt$
** $u = sin(\theta) \longrightarrow \frac{du}{d\theta} = cos(\theta)$
** $u_1 = sin(0) =0$
** $u_2 = sin(\pi/2) = 1$
$S = 6a^2 \pi \int_{0}^{1} \frac{du}{d\theta} u^4 \space dt$
$S = 6a^2 \pi \int_{0}^{1} u^4 \space du$
$S = 6a^2 \pi [\frac{u^5}{5}]_0^1$
$S = 6a^2 \pi [\frac{(1)^5}{5}] - 6a^2 \pi [\frac{(0)^5}{5}]$
$S = 6a^2 \pi [\frac{(1}{5}] $
$S = \frac 6 5 a^2 \pi $
