Use a double integral to find the area of the

Step1
A basic understanding of equations tells us that this equation is a circle shifted one unit to the right. I noticed that visualizing this in polar can be difficult so I laid out ever step for those that like to hand-jam.
${\displaystyle {(x-1)}^2+y^2=1}$
${\displaystyle {(r\cos \theta -1)}^2+r^2{\sin }^2\theta =1}$
${\displaystyle (r^2{\cos }^2\theta -2r\cos \theta +1)+1r^2{\sin }^2\theta =1}$
${\displaystyle r^2{\cos }^2\theta +r^2{\sin }^2\theta =2r\cos \theta }$
${\displaystyle x^2+y^2=2r\cos \theta }$
${\displaystyle r^2=2r\cos \theta }$
${\displaystyle r=2\cos \theta }$
Step2
Again, an understanding that this is the equation of a circle with radius 1 is useful, but if that cannot be seen you can convert to polar.
${\displaystyle x^2+y^2=1}$
${\displaystyle r^2=1}$
${\displaystyle r=1}$
Step3
Graphing this is simple and recommended to better understand the limits of integration. To find your limits of integration for r make the appropriate substitution and solve for theta.
${\displaystyle 1=2\cos \theta }$
${\displaystyle \cos \theta =\frac{1}{2}}$
${\displaystyle \theta =-\frac{\pi }{3},\frac{\pi }{3}}$
Step 4
Our Region can now be defined.
${\displaystyle R=(r,\theta )\mid 1\le r\le 2\cos \theta ,-\frac{\pi }{3}\le \theta \le \frac{\pi }{3}}$
Step 5
Set up the integrals
${\displaystyle {\int }_{-\frac{\pi }{3}}^{\frac{\pi }{3}}{\int }_1^{2\cos \theta }rdrd\theta }$
Step 6
Integrate with respect to r and run
${\displaystyle {\int }_{-\frac{\pi }{3}}^{\frac{\pi }{3}}\frac{1}{2}{r}^2{\mid }_1^{2\cos \theta }d\theta }$
Step 7
At this point the power reduction formula is necessary for further integration.
${\displaystyle {\int }_{-\frac{\pi }{3}}^{\frac{\pi }{3}}2{\cos }^2\theta -\frac{1}{2}d\theta }$
Step 8
Based on symmetry you can double the integral and run from 0 to pi/3 but I will show it without.
${\displaystyle {\int }_{-\frac{\pi }{3}}^{\frac{\pi }{3}}2\cos 2\theta +\frac{1}{2}d\theta }$
Step 9
Use trig to find the values of the inputs
${\displaystyle \frac{1}{2}\sin 2\theta +\frac{1}{2}\theta {\mid }_{\{-\frac{\pi }{3}\}}^{\frac{\pi }{3}}}$
Step10
I went more in detail with the set up of the problem than I did with the integration because I believe that's where the most confusion comes from. Please let me know if there are any problems with this solution.
${\displaystyle \frac{\pi }{3}+\frac{\sqrt{3}}{2}}$
Result: ${\displaystyle \frac{\pi }{3}+\frac{\sqrt{3}}{2}}$