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.

\(\displaystyle{\left({x}-{1}\right)}^{{2}}+{y}^{{2}}={1}\)

\(\displaystyle{\left({r}{\cos{\theta}}-{1}\right)}^{{2}}+{r}^{{2}}{{\sin}^{{2}}\theta}={1}\)

\(\displaystyle{\left({r}^{{2}}{{\cos}^{{2}}\theta}-{2}{r}{\cos{\theta}}+{1}\right)}+{1}{r}^{{2}}{{\sin}^{{2}}\theta}={1}\)

\(\displaystyle{r}^{{2}}{{\cos}^{{2}}\theta}+{r}^{{2}}{{\sin}^{{2}}\theta}={2}{r}{\cos{\theta}}\)

\(\displaystyle{x}^{{2}}+{y}^{{2}}={2}{r}{\cos{\theta}}\)

\(\displaystyle{r}^{{2}}={2}{r}{\cos{\theta}}\)

\(\displaystyle{r}={2}\cos\theta\)

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.

\(\displaystyle{R}={\left({r},\theta\right)}{\mid}{1}\leq{r}\leq{2}{\cos{\theta}},-{\frac{{\pi}}{{{3}}}}\leq\theta\leq{\frac{{\pi}}{{{3}}}}\)

Step 5

Set up the integrals

\(\displaystyle{\int_{{-{\frac{{\pi}}{{{3}}}}}}^{{\frac{{\pi}}{{{3}}}}}}{\int_{{1}}^{{{2}{\cos{\theta}}}}}{r}{d}{r}{d}\theta\)

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

\(\displaystyle{\frac{{{1}}}{{{2}}}}{\sin{{2}}}\theta+{\frac{{{1}}}{{{2}}}}\theta{\mid}^{{\frac{{\pi}}{{{3}}}}}_{\left\lbrace-{\frac{{\pi}}{{{3}}}}\right\rbrace}\)

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: \(\displaystyle{\frac{{\pi}}{{{3}}}}+{\frac{{\sqrt{{3}}}}{{{2}}}}\)