I want to calculate the area of this region that I'm showing in the picture.
First of all, I think that the area that the exercise refers to is the area I painted black on my 2nd pic.
I think it must be done with double integrals.
My first guess was to do the integrals with polar coordinates but I encountered the following problem (that I think that never happened to me before): I can't find an explicit equation in polar coordinates for that thing. And if I can't do that, how am I supposed to calculate that with polar coordinates??
My 2nd guess was solving it with cartesian coordinates. I've decided to do $2$ separate integrals (Only because I can't do it in a single one). I divided the region into $2$, $I_1$ and $I_2$ (shown in the 3rd pic).
$I_1$:
$$ 1- \sqrt(1- \frac{(x-2)^2}{4}) < y < x $$
$$ 0.4<x<2 $$
$I_2$:
$$0<y<2$$
$$ 2<x<2+\sqrt{4-4(y-1)^2} $$
In fact, the exercise doesn't ask to calculate the area, it asks the center of mass of that shape (with a uniform density), but my main problem is to do the set up of the area integral
$$y\le x, x^2 + 4y^2 – 4x – 8y +4 \le 0.$$
Best Answer
As you pointed out, the enclosed area $I$ consists of two parts
$$I=I_1+I_2$$
where $I_2=\pi$ represents the area of half ellipse and
$$I_1=\int_{2/5}^2 \left(x-1+\sqrt{1-\frac{(x-2)^2}4}\right)dx =\frac45+\sin^{-1}\frac45$$
Thus, the full area is
$$I=\pi+\frac45+\sin^{-1}\frac45$$