I'm having some trouble with the following exercise:
Determine the area of the region with boundary $r=3\sin(2\theta)$ with $\theta\in [0,\pi/2]$ using Greens Theorem.
First I started by defining a curve $\alpha:[0,1] \to \mathbb R^2$ in cartesian coordinates with the same trace:
$\alpha(t) = (f\circ\beta)(t)$ where $\beta (\theta) = (3\sin(2\theta),\theta)$ and $f(r,\theta) = (r\cos(\theta), r \sin( \theta))$.
Let $C$ be the path traced by the curve $\alpha$. Then, according to Greens theorem: $$\iint_D\left( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}\right) dxdy = \int_C Fds$$
Where $D$ is the region with boundary $C$.
So if I find a vector field such that: $$\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} = 1$$
Then the area of $D$ will just be: $$Area(D) = \int _C F ds$$
For this I used the field $F(x,y) = (y,0)$. I tried to compute the line integral but got a very long and complicated integral with trig functions. Is this the best way to do this?
For ease of computation, I take the vector field $\vec F = (P, Q) = (-y, x)$ and then the area is half of the line integral as $Q_x - P_y = 2$.
The parametrized curve is $$r (\theta) = (3 \sin2\theta \cos\theta, 3 \sin2\theta \sin\theta), 0 \leq \theta \leq \pi/2$$ $$ r(\theta) = \frac 32 (\sin3\theta + \sin\theta, \cos\theta - \cos3\theta)$$
$$r'(\theta) = \frac 32 (3 \cos3\theta + \cos\theta, 3\sin3\theta - \sin\theta)$$
$$ \vec F(r(\theta)) = \frac 32 \left(\cos3\theta - \cos\theta, \sin3\theta + \sin\theta\right)$$
$$ \vec F(r(\theta))\cdot r'(\theta) = \frac 92 (1 + \sin 3\theta \sin\theta - \cos3\theta \cos\theta)$$ $$ = \frac 92 (1 - \cos 4\theta)$$
The line integral is,
$I = \displaystyle \int_0^{\pi/2} \vec F(r(\theta)) \cdot r'(\theta) ~ d\theta$
The integral of $\cos4\theta$ over $(0, \pi/2)$ is zero.
That leads to $A = \dfrac{I}{2} = \dfrac{9 \pi}{8}$