Let $D$ be the interior of the circle of $x^2+y^2=2x$. Find $$\int \int _D \sqrt{x^2+y^2} dA$$
I have a solution to this but it is not clear. It just says in the polar coordinates, the circle is $r^2 =2r\cosθ ⇒ r =0$ and $2\cosθ$. How did they straight away know this?
I just tried to let $x-1=r\cos\theta$ and $y=r\sin\theta$ but it turns out to be so hard to integrate.
The circle equation can be rewritten as:
$$(x-1)^2+y^2 = 1$$
Changing to polar one gets the circle must satisfy:
$$r^2-2r\cos\theta=0$$ That basically tells us the interior of the circle are the points with $r\in(0,2\cos\theta)$ for a given $\theta$, and the circle is centered in $(1,0)$ with radius $1$, so the integral is:
$$\int_{-\pi/2}^{\pi/2}d\theta\int_0^{2\cos\theta}dr r^2$$
The $r^2$ comes from your integrand function (thich gives one $r$) and the $r$ from the Jacobian.
I let the integral to you. Your doubt was how to know the circle was $r^2=2r\cos\theta$, right? I hope I made it clear, you just have to see that $x^2-2x=(x-1)^2-1$
EDITION: the second equations comes from the first one:
$$(x-1)^2+y^2 = 1$$
Changing to polar: $x=r\cos\theta$ and $y=r\sin\theta$. So the circle satisfies:
$$(r\cos\theta-1)^2+r^2\sin^2\theta=1$$ $$r^2\cos^2\theta-2r\cos\theta+1+r^2\sin^2\theta=1$$
The $1$ goes away, and $\sin^2+\cos^2 = 1$, so: $$r^2-2r\cos\theta = 0$$
We can also use Cavalieri's principle. With a bit of trigonometry, we have that the length of the arc in $D$ for which $x^2+y^2=d^2$, when $d\in[0,2]$, is given by $2d\arccos\frac{d}{2}$, hence: $$ I = \int_{0}^{2} d^2\arccos\frac{d}{2} = 8\int_{0}^{1}u^2\arccos u\,du=\frac{8}{3}\int_{0}^{1}\frac{u^3}{\sqrt{1-u^2}}\,du$$ (in the last step we applied integration by parts) and it follows that: $$ I = \frac{8}{3}\int_{0}^{\pi/2}\sin^3\theta\,d\theta = \color{red}{\frac{16}{9}}.$$
In the following, I try to answer just this part of the question: "How did they straightaway know this?"
In the usual definition of polar coordinates, as you know, the relationship between the Cartesian coordinates of a point, $(x,y)$, and the polar coordinates $(r,\theta)$ of the same point is
$$ x = r \cos \theta,$$ $$ y = r \sin \theta.$$
This means that if you have an equation that is satisfied by the Cartesian coordinates of a particular point, $(x,y)$, you can simply replace $x$ by $r\cos\theta$ and replace $y$ by $r \sin \theta$ in that equation, and you will have an equation satisfied by the polar coordinates of that point. This is true because in the descriptions of this point by its coordinates, $x$ and $r\cos\theta$ are the same number, likewise $y$ and $r \sin \theta$.
There is a particular curve in the plane, such that every point on that curve satisfies the given equation, $$x^2 + y^2 = 2x.$$
Replacing $x$ by $r\cos\theta$ and $y$ by $r \sin \theta$, we get another equation that must be true for every point on that curve: $$(r\cos\theta)^2 + (r\sin\theta)^2 = 2(r\cos\theta).$$
Since $\cos^2\theta + \sin^2\theta = 1$, the left side of this equation reduces to $r^2$. That reduction happens so often in coordinate geometry that most people who see $x^2 + y^2$ will immediately replace it by $r^2$. So you don't see the equation above written out in full, but instead simply write $r^2 = 2(r\cos\theta).$
One might well ask not just how "they" knew they could do this, but also why it might have occurred to them to try it. As with a lot of math problems, there are a couple of likely explanations: (1) you tried a bunch of things and this is the one that worked (and you threw out your notes on all the failed methods); (2) you were just lucky this was the first thing you tried; or (3) something in the problem reminded you of something you've seen before that works out nicely when you do this change of coordinates. By the time someone gets around to writing this in a textbook or class notes, it's probably due to the last reason, but someone seeing this for the first time only has the first two options.
I think you can legitimately construct a system of polar coordinates in which $r \cos \theta = x - 1$ instead of the usual $x$, but it's not the "obvious" coordinate conversion and apparently it didn't help much in this problem.
Partial answer regarding how $ x^2 + y^2 = 2 a x $ is a circle.
Substitute $ x = r \cos\theta , y = r \sin\theta, $ you get
$ r = 2 a \cos\theta .$ It is a circle of diameter $2 a $ touching y-axis at the origin as a tangent. It is easy to integrate it now. In this case $a=1.$
