Another Obvious Offshoot

Let’s see if we can continue our Obvious Integral Spree. We’re going to calculate the area of a circle by integration.
Here we have a circle. It has radius r and can be described as f(x,y):x^2+y^2=r^2. From that we infer that y=\pm\sqrt{r^2-x^2}.

As all four sectors of the circle are equal, we just take the integral for the first quadrant, then multiply it by 4.
As area is positive we get


That won’t work, (if you must know) that leads us to


save for calculation errors; and we will end up with multiple divisions by zero and \ln(0) as an encore.

Let’s look at the circle again. There’s another way to describe y and x, because we know that x=r\cos\theta and y=r\sin\theta. We can rewrite r^2-x^2 as r^2-(r^2\cos^2\theta)=r^2(1-cos^2\theta). As we remember that \sin^2+\cos^2\equiv 1 we end up with r^2-x^2=r^2\sin^2\theta. Now we can take the square root and substitute back:

    \[4\int_0^r r\sin\theta\mathrm{d}x\]

Now we need to know how to integrate with respect to \theta:






There we are.
We also have to realize that evaluating the integral with x from 0 to r will now lead us to another interval.
As x=r\cos\theta evaluating from 0 to r now means evaluating from r\cos\theta=0 to r\cos\theta=r or \cos\theta=0 to \cos\theta=1 which is from \tfrac{\pi}{2} to 0.

So now we can rewrite the integral as follows:

    \[4\int_{\tfrac{\pi}{2}}^0 r\sin\theta\cdot -r\sin\theta\mathrm{d}\theta\]


    \[4\int_{\tfrac{\pi}{2}}^0 -r^2\sin^2\theta\mathrm{d}\theta\]

Factor out the constant,


Next problem is finding the primitive function of \sin^2\theta. But we remember (I hope so) that


and if h=i=\theta we get






leading to


Where were we? O, yes, let’s substitute:


Out with the constant

    \[-2r^2\int_{\tfrac{\pi}{2}}^0 1-\cos(2\theta)\mathrm{d}\theta\]

and we finally get to


    \[-2r^2(0-\tfrac{\pi}{2})=\pi r^2\]

And with that we’re done.