Calculating area between two polar curves $r = 1$ and $r = 1 + \cos{\theta}$

Question

I need to find the area of the shaded part shown below. Polar coordinates really confuse me for some reason - I tried switching to Cartesian coordinates but things got more complicated.
After browsing the internet for some time I came up with this integral, but not sure if it is ok $$ \int_{-1}^{0}{1^2 - (1 + \cos{\theta})^2 \, d\theta}. $$ Could anyone explain how to set the integral properly according to the picture? Thanks in advance.

Answer 1

Your formula should be :

$$\frac12 \int_{\pi/2}^{3\pi/2}\left( 1^2 - (1 + \cos{\theta})^2 \right) \, d\theta.$$

Indeed the area swept by the radius of the curve defined in polar form by $r=r(\theta)$ for $\theta = \theta_0 $ to $\theta_1$ is

$$\frac12 \int_{\theta_0}^{\theta_1} r(\theta)^2 \, d\theta.$$

Due to the symmetry of the issue with respect to $x$ axis, it suffices to calculate the half area (above $x$ axis) and double it:

$$\frac12 2 \int_{\pi/2}^{\pi}\left( 1 - (1 + \cos{\theta})^2 \right) \, d\theta=-\int_{\pi/2}^{\pi}\left(2\cos{\theta}+(\cos{\theta})^2 \right) \, d\theta$$

Can you take from here ?

Remark : some explanations for being more familar with polar equations : Let us consider some points $M$ on the cardioid (the name of the red curve),

$$\text{polar coord.} \ (r,\theta) \ \leftrightarrow \text{cartesian coord.} \ (x,y)$$

$$\pmatrix{(r=2, \theta=0)& \ \leftrightarrow \ &(x=2,y=0)\\ (r=3/2, \theta=\pi/3)& \ \leftrightarrow \ &(x=(3/2)\cos \pi/3,y=(3/2)\sin \pi/3)\\ (r=1, \theta=\pi/2)& \ \leftrightarrow \ &(x=0,y=1)\\ (r=1/2, \theta=2\pi/3)& \ \leftrightarrow \ &(x=(1/2)\cos 2\pi/3,y=(1/2)\sin 2\pi/3)}$$

etc.

Please note that there is no need to consider angles before reaching $\pi/2$ angle.