Surface Area of the top half of an Astroid

Question

How would I go about beginning this question? I have applied the standard surface area integral formula but it becomes complicated quickly.

Answer 1

If we use the parametrization $x=a\cos^3\theta, y=a\sin^3\theta, \;0\le\theta\le\pi$ and symmetry, we have

$\displaystyle S=2\int_0^{\frac{\pi}{2}}2\pi(a\sin^3\theta)\sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2}\;d\theta=4\pi\int_0^{\frac{\pi}{2}}(a\sin^3\theta)(3a\sin\theta\cos\theta) d\theta$

$=\displaystyle 12\pi a^2\int_0^{\frac{\pi}{2}}\sin^4\theta\cos\theta\;d\theta=12\pi a^2\int_0^1u^4\;du=12\pi a^2\left[\frac{1}{5}u^5\right]_0^1=\frac{12\pi a^2}{5}.$