 # Use a double integral to find the area of the sodni3 2021-09-21 Answered
Use a double integral to find the area of the region. The region inside the circle ${\left(x-1\right)}^{2}+{y}^{2}=1$ and outside the circle ${x}^{2}+{y}^{2}=1$
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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.
${\left(x-1\right)}^{2}+{y}^{2}=1$
${\left(r\mathrm{cos}\theta -1\right)}^{2}+{r}^{2}{\mathrm{sin}}^{2}\theta =1$
$\left({r}^{2}{\mathrm{cos}}^{2}\theta -2r\mathrm{cos}\theta +1\right)+1{r}^{2}{\mathrm{sin}}^{2}\theta =1$
${r}^{2}{\mathrm{cos}}^{2}\theta +{r}^{2}{\mathrm{sin}}^{2}\theta =2r\mathrm{cos}\theta$
${x}^{2}+{y}^{2}=2r\mathrm{cos}\theta$
${r}^{2}=2r\mathrm{cos}\theta$
$r=2\mathrm{cos}\theta$
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.
${x}^{2}+{y}^{2}=1$
${r}^{2}=1$
$r=1$
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.
$1=2\mathrm{cos}\theta$
$\mathrm{cos}\theta =\frac{1}{2}$
$\theta =-\frac{\pi }{3},\frac{\pi }{3}$
Step 4
Our Region can now be defined.
$R=\left(r,\theta \right)\mid 1\le r\le 2\mathrm{cos}\theta ,-\frac{\pi }{3}\le \theta \le \frac{\pi }{3}$
Step 5
Set up the integrals
${\int }_{-\frac{\pi }{3}}^{\frac{\pi }{3}}{\int }_{1}^{2\mathrm{cos}\theta }rdrd\theta$
Step 6
Integrate with respect to r and run
${\int }_{-\frac{\pi }{3}}^{\frac{\pi }{3}}\frac{1}{2}{r}^{2}{\mid }_{1}^{2\mathrm{cos}\theta }d\theta$
Step 7
At this point the power reduction formula is necessary for further integration.
${\int }_{-\frac{\pi }{3}}^{\frac{\pi }{3}}2{\mathrm{cos}}^{2}\theta -\frac{1}{2}d\theta$
Step 8
Based on symmetry you can double the integral and run from 0 to pi/3 but I will show it without.
${\int }_{-\frac{\pi }{3}}^{\frac{\pi }{3}}2\mathrm{cos}2\theta +\frac{1}{2}d\theta$
Step 9
Use trig to find the values of the inputs
$\frac{1}{2}\mathrm{sin}2\theta +\frac{1}{2}\theta {\mid }_{\left\{-\frac{\pi }{3}\right\}}^{\frac{\pi }{3}}$
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.
$\frac{\pi }{3}+\frac{\sqrt{3}}{2}$
Result: $\frac{\pi }{3}+\frac{\sqrt{3}}{2}$