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

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Expert Answer

tabuordg
Answered 2021-09-22 Author has 7361 answers

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

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