Area inside the astroid

Question

I need to find area inside the astroid:
$$ \left\{ \begin{array}{c} x=2\cos^3(\frac t 4) \\ y=2\sin^3(\frac t 4) \end{array} \right. $$ I've seen some formulas with integrals, but the first problem I run into is I've been trying to draw it with online services but (as far as I can understand) problem with $\frac t 4$ instead of simply $t$ what I see in all astroid equation.
So, for the beginning, is my $\frac t 4$ ok? Should I perceive it as regular argument?
How should I get my astroid draft? How to get points for drawing maybe?
Next is which formula or way should I use to calculate this area? Because on one site, I stumbled upon this formula:
$S=\frac {3\pi ab} {8}$ Is that correct? Can I use it?

Thanks to all in advance!

Answer 1

It doesn't matter what parameter you use. You can define $u=\frac t4$ and use $u$. You will get the same points. The formula give you a nice astroid, as plotted by Alpha

Answer 2

For calculating the area of your astroid, I would first consider transforming the eqations in one like this:

$x=2\cos^3(\frac{t}{4})\Rightarrow (\frac{x}{2})^{2/3}=\cos^2(\frac{t}{4})$

$y=2\sin^3(\frac{t}{4})\Rightarrow (\frac{y}{2})^{2/3}=\sin^2(\frac{t}{4})$

now you have $(\frac{x}{2})^{2/3}+(\frac{y}{2})^{2/3}=1$.

Because of the propreties from the trigonometric functions, you know that it's enough to calculate one forth of the astroid, bcs it's symetrical. We will calculate the part where $x>0$ and $y>0$, so: $y=(2^{2/3}-x^{2/3})^{3/2}$. Now we get that $$\int_0^2 (2^{2/3}-x^{2/3})^{3/2} = \frac{3\pi}{8}$$

So the area of the whole astroid is $\frac{3\pi}{2}$.