I was attempting, for the exercise of it, to find the area of the a simple square with an infinite number of infinitesimal circle sectors. Let us say this square is $[5 x 5]$.
Alas, it's been proving as awkward to compute this as it sounds. First, I did the integral in rectangular coordinates, (much easier) $x=5$, so the area becomes $$\int_0^5 5 \space \mathrm{d}x = 25$$.
I then attempted to describe a rectangular area in polar coordinates. I did this using $r\cos(\theta)=5$, which simplifies in terms of $\theta$ to $r=\frac{5}{\sin{\theta}}$. Because this is a square, the angle the hypotenuse will make from the bottom left corner to the top right corner is going to be, surely, $45^{\circ}$. This produces the integral:
$$\frac{25}{2}\int_0^\frac{\pi}{4} \frac{1}{\sin^2 (\theta)} \mathrm{d}\theta$$.
This integral does not converge according to wolfram alpha. Would anyone be so kind as to tell me what I've done wrong?
Best Answer
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.