Dynamic Geometry: P3

The problem contained some inaccuracies. So I had to guess what was intended.

From the equation $y=\sqrt{a^2-x^2}$

it follows that if you fill in $x=a^2$ you get $P=(a^2;\sqrt{a^2-a^4})$ (Originally a square root was missing here in the problem - fixed now). In terms of x you get: $P=(x;\sqrt{x-x^2})$

This is the equation of the pink curve. We want the area $A=\int_0^1 \sqrt{x-x^2} dx$ Maybe the easiest way to do the integral is to do a substitution $u=x-\frac{1}{2}$ so that the integral becomes $\int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{\frac{1}{4}-u^2} du =\frac{1}{2} \int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{1-4u^2} du$ and do a second sub $v=2u$ to arrive at $\frac{1}{4}\int_{-1}^1 \sqrt{1-v^2}dv$ which corresponds to $\frac14$ of the area of half a unit circle.

Alternatively, this integral could be solved using a third sub $v=\sin(w)$ , via $\int_{-\frac{π}{2}}^{\frac{π}{2}} \cos(w)d\sin w=\int_{-\frac{π}{2}}^{\frac{π}{2}} \cos^2(w)dw=\int_{-\frac{π}{2}}^{\frac{π}{2}} (\frac12\cos(2w)+\frac12)dw=\frac12\cos π + \frac14 π -\frac12\cos (-π) - \frac14(-π))=\frac12π$ .

Either way $A=\frac{π}{8}$

This answer was originally given literally (without any $b$ in the expression!) so I guessed that $\frac{π}{b}$ was intended and submitted $\lceil π\rceil+b=4+8=\boxed{12}$

Edit 1: In the new version of the problem this was corrected, I renamed my $a$ accordingly to $b$ .

Edit 2: Explained the integration (in response to Jeff Giff's question)