Áreas em Coordenadas Polares

Given a polar curve in the form $r = f(\theta)$, we typically want to calculate the area that it "sweeps out" as a point goes from $\theta_1$ to $\theta_2$. For example:

Recall that in polar coordinates, $r$ is the distance from the origin. and so the curve $r=4$ is the set of points that are $4$ units away from the origin, namely the circle centered at the origin with radius 4.

As $\theta$ goes from 0 to $\frac{\pi}{2}$, we get the part of this circle that's in the first quadrant, a quarter-circle.

So, the question asked us to compute the area of this quarter-circle, which is just $\frac{1}{4}(\pi\cdot 4^2) = 4\pi$.

How do we deal with more complicated things? In Cartesian coordinates the rectangle was the simplest shape, since when we made the interval small enough, $y = f(x)$ was almost constant, leading to an easy approximation which became an integral.

This, time, we'll be using circular sectors.

Now, to get an approximation, we would divide the given area into many pieces and then sum, just as we did for Cartesian areas. For a very small interval of $\theta$, $r = f(\theta)$ is almost constant, so the area swept out is almost a circular sector. Thus we see that a reasonable approximation is $$\sum_{i=0}^{N-1} \frac{1}{2}f(\theta_i)^2\Delta\theta$$ where we have divided the range $[\theta_i,\theta_f]$ into $N$ equal ranges, each with width $\Delta\theta$, the $i$th of which begins at $\theta_i$ and ends at $\theta_{i+1}$.

As you're hopefully expecting, we now take the limit as the number of pieces approaches $\infty$. Look at the form of this limit. It's exactly the same as a definite integral, just with $\frac{1}{2}f(\theta)^2$ as the integrand rather than just $f$. This gives us our formula:

$$A = \int_{\theta_i}^{\theta_f}\frac{1}{2}f(\theta)^2d\theta$$

which is also written $$A = \frac{1}{2}\int_{\theta_i}^{\theta_f}r^2d\theta$$