Lets say we have a figure described as follows:
$r=2\cdot\sqrt{\cos(2\theta)}$
Click here to see a plot.
Now lets say that we want to calculate the area of this rotated eight.
I'd like to understand what has to be done in order to integrate in polar coordinates.
So far I've noticed, that the following integral is wrong:
$4 \int_0^\frac{\pi}{2} r \ dr$
And I've found that the following integral would lead to the correct result:
$4 \int_0^\frac{\pi}{2} \frac{1}{2} r^2 \ dr$
However, I don't know how this formula for the correct integral can be derived. What is the process of getting to the correct formula? Does this relate to the functional-determinant, volume-element of coordinate transforms or multidimensional integrals in some way?
Notice, the curve: $\ r=2\sqrt{\cos(2\theta)}$ is symmetrical about x-axis & y-axis
area bounded by the curve $\ r=2\sqrt{\cos(2\theta)}$ is $$2\int_{-\pi/4}^{\pi/4}\frac{1}{2}r^2d\theta $$ $$=\int_{-\pi/4}^{\pi/4}r^2d\theta $$ $$=2\int_{0}^{\pi/4}4(\cos2\theta)d\theta $$ $$=8\int_{0}^{\pi/4}\cos2\theta d\theta $$ $$=8\left[\frac{\sin 2\theta}{2}\right]_{0}^{\pi/4}$$ $$=8\left[\frac{\sin \frac{\pi}{2}}{2}-0\right]$$ $$=8\left[\frac{1}{2}\right]=4$$