# Arc Length and Surface Area

Hi! I am currently working on some online calc2 homework problems on arc length and surface area. I understand the formula for the arc length of y=f(x) over a given interval, but this question isn't set up quite like that and I am rather confused on how to solve it. If someone could help me answer this problem I would greatly appreciate it.

You can parametrize the astroid by taking advantage of the trig identity $\cos^2(\theta)+\sin^2(\theta) = 1$; namely, take $x = 2^{3/2}\cos^3(t)$ and $y = 2^{3/2}\sin^3(t)$.

By symmetry, we can simply find the arclength of $1/4$th the astroid and multiply by $4$ at the end, so let $0 \leq t \leq \pi/2$.

Now simply use the formula for arclength:

$$\int_0^{\pi/2} \sqrt{\frac{dx}{dt}^2 + \frac{dy}{dt}^2} dt$$

See the wikipedia article for a general idea of what an Astroid looks like:

http://en.wikipedia.org/wiki/Astroid

If you solve the equation for $y$:

$y = (2 - x^{2/3})^{3/2}$

$y' = -x^{-1/3}(2 - x^{2/3})^{1/2}$

And if $x=0$, $y^{2/3} = 2$, and $y = \pm \sqrt{8}$. So integrating from 0 to $2\sqrt{2}$ will give 1/4 of the length of the curve, by symmetry. So we have:

$$L = 4 \int_0^{2\sqrt{2}} \sqrt{1 + (-x^{-1/3}(2 - x^{2/3})^{1/2})^2} \, dx$$

I'll let you take it from there.