Tricky

[BigGlenntheHeavy]

\(\displaystyle \int_{0}^{\pi/3} \int_{0}^{sin(y)}\frac{1}{\sqrt(1-x^{2})}dx \ dy \ = \ \frac{\pi^{2}}{18}\)

\(\displaystyle Now. \ reverse \ the \ order \ of \ integration \ and \ solve.\)

\(\displaystyle \int_{?}^{?}\int_{?}^{?}\frac{1}{\sqrt(1-x^{2})}dy \ dx \ = \ \frac{\pi^{2}}{18}\)

 

[galactus]

\(\displaystyle \int_{0}^{\frac{\sqrt{3}}{2}}\int_{sin^{-1}(x)}^{\frac{\pi}{3}}\frac{1}{\sqrt{1-x^{2}}}dydx=\frac{{\pi}^{2}}{18}\)

 

[BigGlenntheHeavy]

Right on galactus, where I slipped up , I was looking at y=sin(x) instead of x =sin(y). See my graph.

[attachment=0:2cpoqlk8]new.jpg[/attachment:2cpoqlk8]