find the measure of this set

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Let $\alpha>0$ and define $A_{\alpha}\subset\mathbb{R}^{2}$ as: $$A_{\alpha}=\{x>0, y>\alpha x, x^{2}+y^{2}<\alpha^{2}\}$$ The integreation is simple, but I can't find the mistake I keep making. I tried via:

$$m(A_{\alpha})=\int^{\frac{\alpha}{\sqrt{\alpha^{2}+1}}}_{0}\int^{\sqrt{\alpha^{2}-x^{2}}}_{0}dydx=\dots=\frac{1}{2}\alpha^{2}\arcsin\Bigg(\frac{\alpha}{\sqrt{\alpha^{2}+1}}\Bigg)\Bigg(1+\cos\arcsin\Bigg(\frac{\alpha}{\sqrt{\alpha^{2}+1}}\Bigg)\Bigg)$$ and through spherical coordinates, for which:

$$m(A_{\alpha})=\int^{\alpha}_{0}\int^{\frac{\pi}{2}}_{\arctan\alpha}\rho d\theta d\rho=\dots=\frac{\alpha^{2}}{2}\Big(\frac{\pi}{2}-\arctan\alpha\Big)$$

Although they may, I doubt these 2 results are the same but I can't understand which is the correct one, assuming I integrated properly.

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In the first method the integral w.r.t $y$ is from $\alpha x$ to $\sqrt {\alpha^{2}-x^{2}}$