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.
Calculating area between two polar curves $r = 1$ and $r = 1 + \cos{\theta}$
definite integralsintegrationpolar coordinates
Related Solutions
To generate the right-hand loop of the lemniscate $r^2=\cos(2\theta)$, you need to let $\theta$ vary from $-\frac{\pi}{4}$ to $\frac{\pi}{4}$; i.e. $\theta \in [-\frac{\pi}{4},\frac{\pi}{4}] $. To see this, we have
$$ r = \bar{+} \sqrt{\cos(2 \theta) } \implies \cos(2\theta)\geq 0 \implies -\frac{\pi}{2}\leq2\theta \leq \frac{\pi}{2}\implies -\frac{\pi}{4}\leq \theta \leq \frac{\pi}{4}. $$
Here is the plot
Here is the full graph
The area of a circular sector of radius $r$ and angle $d\theta$ is $\pi r^2 \frac{d\theta}{2\pi} = \frac{1}{2} r^2 d\theta$.
The right side of the square ($0 < \theta < \pi/4$) is the line $x = 5$, which in polar coordinates is $r = \frac{5}{\cos \theta}$ (not $\sin$).
Putting this together, the integrand should be $\frac{1}{2}(\frac{5}{\cos \theta})^2 d\theta = \frac{25}{2 \cos^2 \theta} d\theta$.
This works out to $12.5$. Since it only covers half of the square, double it to get the $25$ you expect.
Best Answer
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.