<s>[INST] What is the area of this closed curve $\alpha(t) = (\sin(2t), 2\sin t)$ on the plane? The closed curve $\alpha(t) = (\sin(2t), 2\sin t)$ with $0 \leq t \leq \pi$.
Find the area of the $A$ enclosed by the $\alpha$ on the $\mathbb{R}$.

I took the formula $\int_{0}^{\pi} \frac{1}{2}{f(\theta)}^2 d\theta$ for the $r = f(\theta)$.
In my thought, $\alpha(t) =(x(t), y(t)) = (\sin(2t), 2\sin t)$. So the $r^2(t) =   x^2(t) + y^2(t)  = \sin^2(2t) + 4\sin^2t$
My answer is $\int_{0}^{\pi} \frac{1}{2}{f(t)}^2 dt = \int_{0}^{\pi} \frac{1}{2} r^2 dt = \int_{0}^{\pi} \frac{1}{2} (\sin^2(2t) + 4\sin^2t) dt  = \frac{5\pi}{4}$
But the answer was not my thing. Actually the answer was $\frac{8}{3}$. So my question is What the point did I have a mistake? Why is the formula $\int_{0}^{\pi} \frac{1}{2}{f(\theta)}^2 d\theta$ not hold for this case?
 [/INST] In polar coordinates, $x = r \cos\theta, y = r\sin\theta$ where $\theta$ is the angle with the positive x coordinate axis. That is not the case here and hence the formula you write cannot be applied as is. You can check it yourself. For the $r$ you obtained, does $r \cos\theta ~$ give you $x$?
$y = 2\sin t$
$x = \sin2t \implies x^2 = 4 \sin^2 t (1 - \sin^2 t)$
$ ~ \displaystyle x^2 = y^2 - \frac{y^4}{4} ~ $ is the equation of the curve in cartesian coordinates.
That leads to $\displaystyle - \sqrt{y^2 - \frac{y^4}{4}} \leq x \leq \sqrt{y^2 - \frac{y^4}{4}}$
So the integral to find area can be written as,
$ 2 \displaystyle \int_0^2 \sqrt{y^2 - \frac{y^4}{4}} ~ dy$
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