I said this incorrectly earlier, because of a lack of name for the counter-clockwise contour enclosing $R_{1}$ and the counter-clockwise contour enclosing $R_{2}$. Let's call those contours $C_{1}'$ and $C_{2}'$, respectively. I apologize for adding to the confusion.
When you add the path integral over $C_{1}'$ to the path integral over $C_{2}'$, the integrals over the extra lines (cuts) cancel because $C_{1}'$ contains right-to-left line integrals over the extra cuts, while $C_{2}'$ contains left-to-right line integrals over the extra cuts. So the sum of path integrals over $C_{1}'$ and $C_{2}'$ leaves you with the counter-clockwise integral over the outer contour $C_{1}$ plus the clockwise integral over the inner contour $C_{2}$. This is a standard, clever trick of line integrals.
$a)$ One thing you need to know by heart is this transformation: $x^2+y^2=r^2$, so the integrand becomes $\ln (1+r^2)$. We also have $dx\,dy=rdr\,dt$ (I will use $t$ instead of $\theta$ because it is easier for me) To compute the integral in the unit circle, you need to consider not $r=1$ but $r<1$ and $0<t<2\pi$ (they should have taught you this, It would be very difficult for me to explain this here using only text). So our integral is:
$$\int^{2\pi}_0\int^1_0\ln(1+r^2)r\,dr\,dt$$
Substitute $u=r^2$ $\frac{du}{2}=rdr$ and we will get:
$$\int^{2\pi}_0 \frac12 \int^1_0\ln(1+u)\,du\,dt$$
If you integrate by parts, you'll find that the antiderivative of $\ln(u+1)=(u+1) \ln (u+1)-u$, and the whole integral turns out to be:
$$2\pi\ln2-\pi$$
$b)$ I will assume we will find the surface of the Lemniscate. We can divide our function into 4 equal parts, only compute it for the first quadrant, and multiply by 4.
We start by inserting $x=r \cos t \, ,y=r \sin t$ into the equation of the Lemniscate. We will get:
$$r^4=4r^2(\cos ^2t-\sin ^2 t)=4r^2\cos (2t)$$
We want the positive $r$ value so we have $r=2\sqrt{\cos(2t)}$. Hence, $r$ goes from $0$ to $r=2\sqrt{\cos(2t)}$.
We also need the maximum $t$ value. We will need to find the angle of this tangent line:
To find it, we will use implicit derivative of the lemiscate evaluated at $(x,y)=(0,0)$. Which turns out to be $\pm1$. We are concerned with the first quadrant, so we take $+1$. The angle that makes the slope $+1$ is $\arctan(1)=\pi/4$
The surface integral is:
$$4\int^{\pi/4}_0\int^{2\sqrt{\cos(2t)}}_0 r\,dr\,dt$$
I can assume you can take it form here? The answer should come out to be $4$
Best Answer
The $xy$-integration is still easier than the polar coordinates for this problem.
Due to $y$-symmetry, the integration is between the two intersection points with the $x$-axis, which can be obtained by setting $y = 0$ in the curve equation (for $a=1$, $b=0$) $$ \frac{p}{(1-x)^2}+\frac{1-p}{x^2}=1 $$ or $$ x^4-2x^3+2(1-p)x-(1-p)=0 $$ Unfortunately, for general value of $p$, it has to be solved numerically. In your example of $p=0.2$, it has two real roots as expected. Let $x_1$ and $x_2$ be the two roots to be used as the integration bounds below.
The curve itself can be expressed as below by rearranging the original equation, $$ y(x)= \left[ -x(x-1)+\sqrt{x^2+(1-p)(1-2x)} \right]^{1/2} $$ The area can then be integrated with, $$ A = 2\int_{x_1}^{x_2}y(x)dx$$
It could not be done analytically, though. But, it is straightforward numerically. For the special case of $p=1$, the area integral reduces significantly and can be carried out by hand,
$$A_{p=1}=2\int_0^2 \sqrt{x(2-x)}=\pi$$ which is expected for a unit circle.