We are studying polar equations.

Calculate the surface inside $r = 1 + \cos \alpha $ and outside $r = 1$.

I know the area inside $r = 1 + \cos \alpha $ being $\frac{3 \pi}2$ because I calculated $\int_0^{2\pi} 1 + \cos(\alpha) \,\mathrm{d}\alpha$

The given solution is $2+\frac{\pi}4$

How can I visualize this integral? Which formula I should use to obtain the answer and why?

A formula we are given is $$S =\int_{\alpha_0}^{\alpha_1}\frac{r^2(\alpha)}{2}d\alpha~.$$

  • $\begingroup$ Why don't you use the given formula ?? $\endgroup$ – Yves Daoust May 5 at 18:41
  • $\begingroup$ I used the formula for the first equation but I don't know what to do next. $\endgroup$ – ScoobyDuh May 5 at 18:45
  • $\begingroup$ This is not what you show. $\endgroup$ – Yves Daoust May 5 at 18:46


The area of a rectangle of height $y$ and basis $dx$ is $y\,dx$.

The area of a circular sector of radius $r$ and aperture $d\alpha$ is $\dfrac{r^2}2\,d\alpha$.

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  • $\begingroup$ I still don't understand what to fill in for $r$ when two equations are given. $\endgroup$ – ScoobyDuh May 5 at 19:04
  • 1
    $\begingroup$ @ScoobyDuh Study the intersection of the curves. $\endgroup$ – Yves Daoust May 5 at 19:09
  • $\begingroup$ I see it now I didn't use the correct formula in my post but I did in the Symbolab link. Anyways I managed to find the correct solution by first calculating $$\int_\frac{-π}2^\frac{π}2 \frac{(1+cos(x))^2}2 dα$$ this gave me $$\frac{6π+16}8$$ Next I calculated the area of the unit circle $$\int_\frac{-π}2^\frac{π}2 \frac{1^2}2 dα$$ which gave me $$\frac{π}2$$ $$\frac{6π+16}8 - \frac{π}2$$ gives the correct solution. Now I did this all graphical. I'm wondering how I can find the right bounderies of the integral by calculation. I can imagine the graphical way isn't always an option $\endgroup$ – ScoobyDuh May 5 at 21:25
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    $\begingroup$ @ScoobyDuh: well done! Now you solve $1+\cos\alpha\le1$. $\endgroup$ – Yves Daoust May 6 at 6:30
  • $\begingroup$ Thank you I understand it now. And u mean greater or equal than one, right? Because we need the area outside $r=1$. $\endgroup$ – ScoobyDuh May 6 at 8:40

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