Find the area bounded by the hypocycloid? I have the answer. The hypobloid has parametrization = $x = acos^3(t)$ $y = asin^3(t)$ The explanation is you take a vector field $F(x,y) = (0, x) which has curl 1 than it says the area is equal to:
$$\int_{0}^{2\pi} (0,cos^3(t)) . (-3cos^2(t)sin(t),3asin^2(t)cos(t))$$
using Green's theorem you remake it.
What I don't understand is why create an artificial vector field to get the area? What's the logic begind that and how do I chose the right vector field?
 A: $\begin{align}x=a\cdot\cos^3t\\y=a\cdot\sin^3t\end{align}\quad=>\quad\bigg(\dfrac xa\bigg)^\tfrac23+\bigg(\dfrac ya\bigg)^\tfrac23=1$, which is a superellipse. A quarter of its area is 
given by $\displaystyle\int_0^1\sqrt[n]{1-x^n}~dx$, where $n=\dfrac23$ . At the same time, we know that $\displaystyle\int_0^1\sqrt[m]{1-x^n}~dx=$
$=\dfrac{\Big(\frac1m\Big)!\cdot\Big(\frac1n\Big)!}{\Big(\frac1m+\frac1n\Big)!}$ , see beta function for more details. Replacing, and using the fact that $\Gamma\bigg(\dfrac12\bigg)$ 
is $~\sqrt\pi,~$ we have $A=4\cdot I=4\cdot\dfrac{3\pi}{32}=\dfrac{3\pi}8$
A: Well, let's start with what we know.
We know that the area is given by: $$\int \int_A 1 dA. $$
This is the most general expression of the area that we are going to work with.
Look at Green's Theorem carefully. It states:
$$\int \int_A \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} dA$$
So the intuition as to why the vector field was chosen was because it could force either $\frac{\partial Q}{\partial x}$ or $\frac{\partial P}{\partial y}$ as $1$ or $-1$.
There are some $a$'s missing in your original expression.
So here we'll define the force field and parametrize the original area. Let's take it step by step.
$$F=P\vec i+Q\vec j$$
Choosing Q as x, P as 0, (alternatively, you can choose P is y and Q as 0)
$$F=x\vec j$$
As we have $x=a\cos^3t, y=a\sin^3t$, we can express the hypocycloid as:
$$C:\vec r=x\vec i+y\vec j=(a\cos^3t)\vec i+(a\sin^3t)\vec j$$
$$\int_c F.d\vec r=\int_c a\cos^3t dy$$
$$\frac{dy}{dt}=3 a \sin^2t \cos t$$
Now we sub dy into the expression:
$$\int_c a (\cos^3t) 3a (\sin^2t) cost dt=\int_0^{2\pi}(3a^2)(\cos^4t)(\sin^2t)dt=\frac{3a^2\pi}{8}$$
And remember that the initial expression you've started with $$\int_c F.d\vec r=\int \int_A 1 dxdy$$
Because you've chosen your vector field as such.
