I'm fascinated by how wonderful integrals are, and I wanted to see if I can use them to find the area of a square in the stupidest way possible!

My idea is: Define $f(\theta)$ as the length of the line you draw from the bottom-left corner to the edge of the square with angle $\theta$, integrate from $0$ to $\frac{\pi}{4}$, and double it. I had a slight feeling this wasn't going to work out quite right, but I didn't understand why.


Anyway, here's my work. We can find the length of the line with easy cosine, right? If the side length of the square is $s$, then $f(\theta)= \frac{s}{\cos{(\theta)}}$. Integrating and cheating with Wolfram Alpha, we get a perfect result of $\displaystyle 2\int_{0}^{\frac{\pi}{8}}f(\theta) \ \ d\theta={-2s}\sinh^{-1}{\Big{(}1-\sqrt{2}\Big{)}}$, which flawlessly resembles the area of a square?

I think that math complains when I try to integrate with respect to an angle. If this is the case, what do I do to fix this?

  • 4
    $\begingroup$ This will work if you're careful about the area you're integrating. Informally, if the line rotates from $\theta$ to $\theta+\mathrm d\theta$, it sweeps out a triangle of base $f(\theta)$ and altitude approximately $f(\theta)\mathrm d\theta$, covering an area $\mathrm dA \approx \frac12 f(\theta)^2 \mathrm d\theta$. Integrate that from $0$ to $\pi/4$ and you get the desired result. $\endgroup$ – Rahul Nov 10 '16 at 4:48

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