How to calculate the area of this region (Double integrals - Ellipse intersected with a straight line)

Question

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

$$y\le x, x^2 + 4y^2 - 4x - 8y +4 \le 0.$$

Answer 1

Suppose that you do the change of coordinates $X=x-2$ and $Y=2y-2$. Then you ellipse becomes the ellipse $X^2+Y^2=2$ and your straight line becomes the straight line $Y=2X+2$, which intersects the ellipse at $(0,2)$ and at $\left(-\frac85,-\frac65\right)$. Can you take it from here?

Answer 2

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

$$I=\pi+\frac45+\sin^{-1}\frac45$$

Answer 3

Already $2$ good answers, but if we wanted to do the integral in a single shot, we could translate the origin to the top of the ellipse, $x=u+2$, $y=v+2$, so that $$\begin{align}\frac{(x-2)^2}{2^2}+\frac{(y-1)^2}{1^2}&=\frac14u^2+(v+1)^2\\ &=\frac14r^2\cos^2\theta+r^2\sin^2\theta+2r\sin\theta+1=1\end{align}$$ So along the ellipse, $$r\left(r\left(\frac14\cos^2\theta+\sin^2\theta\right)+2\sin\theta\right)=0$$ We also have $x=r\cos\theta+2$ and $y=r\sin\theta+2$. So we can get the area $$\begin{align}A&=\int_{-3\pi/4}^0\int_0^{\frac{-2\sin\theta}{\frac14\cos^2\theta+\sin^2\theta}}r\,dr\,d\theta=\int_{-3\pi/4}^0\frac12\cdot\frac{4\sin^2\theta}{\left(\frac14\cos^2\theta+\sin^2\theta\right)^2}d\theta\\ &=\int_{-\infty}^1\frac{2du}{\left(\frac14u^2+1\right)^2}=\int_{-\pi/2}^{\tan^{-1}\frac12}4\cos^2\phi d\phi=\int_{-\pi/2}^{\tan^{-1}\frac12}2(1+\cos2\phi)d\phi\\ &=\left[2\phi+\sin2\phi\right]_{-\pi/2}^{\tan^{-1}\frac12}=2\tan^{-1}\frac12+\frac45+\pi\end{align}$$ Having used the substitutions $u=\cot\theta$ and $\frac12\tan\phi=u$. This is the same area as everybody else has been getting. If I weren't out of time I could integrate with $r\cos\theta+2$ or $r\sin\theta+2$ in there and get $A\bar x$ or $A\bar y$, but...

EDIT: I'm back, and not just to fix typos! Let's do the integrals we threatened to do last time: $$\begin{align}A\bar x&=\int_{-3\pi/4}^0\int_0^{\frac{-2\sin\theta}{\frac14\cos^2\theta+\sin^2\theta}}(2+r\cos\theta)r\,dr\,d\theta\\ &=2A+\int_{-3\pi/4}^0\frac13\cdot\frac{-8\sin^3\theta\cos\theta}{\left(\frac14\cos^2\theta+\sin^2\theta\right)^3}d\theta\\ &=2A-\frac83\int_{-\infty}^1\frac{u\,du}{\left(\frac14u^2+1\right)^3}=2A+\left.\frac8{3\left(\frac14u^2+1\right)^2}\right|_{-\infty}^1\\ &=2A+\frac{128}{75}\end{align}$$ So we get $$\bar x=2+\frac{128}{75\left(2\tan^{-1}\frac12+\frac45+\pi\right)}$$ And $$\begin{align}A\bar y&=\int_{-3\pi/4}^0\int_0^{\frac{-2\sin\theta}{\frac14\cos^2\theta+\sin^2\theta}}(2+r\sin\theta)r\,dr\,d\theta\\ &=2A+\int_{-3\pi/4}^0\frac13\cdot\frac{-8\sin^4\theta}{\left(\frac14\cos^2\theta+\sin^2\theta\right)^3}d\theta\\ &=2A-\frac83\int_{-\infty}^1\frac{du}{\left(\frac14u^2+1\right)^3}=2A-\frac{16}3\int_{-\pi/2}^{\tan^{-1}\frac12}\cos^4\phi d\phi\\ &=2A-\frac23\int_{-\pi/2}^{\tan^{-1}\frac12}(3+4\cos2\phi+\cos4\phi)d\phi\\ &=2A-\frac23\left[3\phi+2\sin2\phi+\frac14\sin2\phi\right]_{-\pi/2}^{\tan^{-1}\frac12}\\ &=2A-2\tan^{-1}\frac12-\frac{92}{75}-\pi=A-\frac{32}{75}\end{align}$$ So $$\bar y=1-\frac{32}{75\left(2\tan^{-1}\frac12+\frac45+\pi\right)}$$ As a check, if we stretched the figure by a factor of $2$ in the $y$-direction, the centroid should lie on the line of symmetry that goes through the center of the circle and is perpendicular to the cut line: $$\frac{y-2}{x-2}=-\frac12$$ And in fact the stretched centroid $(x,y)=(\bar x,2\bar y)$ does pass this test.

Answer 4

So, first of all really thank you to those who helped me solving this question. I used José's hint to find the área of the elipse, and I got the correct answer. This part of the problem is solved.

But now, I'm facing other problem. As I said in the end of the intro post, the final goal of this exercise was to find the centre of mass of this shape.

I've already solved some problems about the centre of mass, but I've never solved this kind of exercise when I changed the system of coordinates.

What I don't fully understand is what formula I should use on this. It says that the density is uniform, so I chose to let density = k, with K being a constant.

I think that is ok, but I don't knowif I should calculate the center of mass in the new coordinate system (X,Y) and then just transform to the old, $ X = x - 1, Y = 2(y-1) $ or should I do with something else?

I'm asking this because I used what I described just now and the solution doesn't match with the solution provided (in the provided solution, x is approximately 2.60, and y is approximately 0,70).