# Area inside $r = 1 + \cos \alpha$

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 $$S =\int_{\alpha_0}^{\alpha_1}\frac{r^2(\alpha)}{2}d\alpha~.$$

• Why don't you use the given formula ?? – Yves Daoust May 5 at 18:41
• I used the formula for the first equation but I don't know what to do next. – ScoobyDuh May 5 at 18:45
• This is not what you show. – 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 $$\dfrac{r^2}2\,d\alpha$$.
• I still don't understand what to fill in for $r$ when two equations are given. – ScoobyDuh May 5 at 19:04
• 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 – ScoobyDuh May 5 at 21:25
• @ScoobyDuh: well done! Now you solve $1+\cos\alpha\le1$. – Yves Daoust May 6 at 6:30
• Thank you I understand it now. And u mean greater or equal than one, right? Because we need the area outside $r=1$. – ScoobyDuh May 6 at 8:40