Evaluate
$$\int_{A} (1+y-x)dxdy$$
Where $A=\left \{(x,y)\in\mathbb{R}^2 \ \text{s.t.} \ \arcsin x+ \arcsin x \leq \frac{\pi}{2}\right \}$.
My approach is the following: since $\arcsin$ is bounded from below by $-\pi/2$, we have that $-\pi \leq \arcsin x + \arcsin y$; so it is $-\pi-\arcsin x \leq \arcsin y \leq \frac{\pi}{2} – \arcsin x$.
Letting $x=\sin u$ and $y=\sin v$, we have that $\left|\det J(u,v)\right|=|\cos u \cos v|$ and the new integration set is $A'=\left \{(u,v)\in\mathbb{R}^2 \ \text{s.t.} \ -\pi-u \leq v \leq \frac{\pi}{2}-u\right \}$.
So I write $A'$ in this way: $A'=\left \{(u,v)\in\mathbb{R}^2 \ \text{s.t.} \ -\pi \leq u \leq \frac{\pi}{2}, \ -\pi-u \leq v \leq \frac{\pi}{2}-u\right \}$; and so
$$\iint_{A} (1+y-x)dxdy=\iint_{A'}(1-\sin v+\sin u)|\cos u \cos v|dudv$$
And now I'm struggling with $|\cos u \cos v|$, because I see that $\cos u \geq 0$ for $-\frac{\pi}{2} \leq u \leq \frac{\pi}{2}$ but I don't see how I can discuss $\cos v \geq0$ from $-\pi-u \leq v \leq \frac{\pi}{2}-u$; the only thing that comes into my mind is that I have to discuss $\text{min} \left \{\frac{\pi}{2}-u,\frac{\pi}{2} \right \}$ and $\text{max} \left \{-\pi-u,-\pi \right \}$ as $u$ varies in $\left[-\pi,\frac{\pi}{2}\right]$ and split the integral in a sum of something like 4 integrals.
Is my approach correct? I've some doubts especially about the substitution and the fact that $-\pi$ is a bound from below. Thanks.
Best Answer
Rewrite $\arcsin x + \arcsin y \leq \frac\pi 2$ as $$\arcsin y \leq \frac\pi 2 - \arcsin x.$$
If $x$ is negative, so is $\arcsin x$, and the inequality is automatically satisfied (because $\arcsin$ is bounded by $\pi/2$ from above).
If $x$ is positive, the left and the right hand side are inside $[-\frac\pi 2,\frac\pi 2]$ where $\sin$ is increasing, so apply $\sin$ to the inequality to get
$$y \leq \sin(\frac\pi 2 - \arcsin x) = \cos\arcsin x=\sqrt{1-x^2}. $$
So, $$\int_{A} (1+y-x)\,dxdy = \int_{-1}^0\int_{-1}^1(1+y-x)\,dydx + \int_{0}^1\int_{-1}^{\sqrt{1-x^2}}(1+y-x)\,dydx.$$
Regarding your approach, there is no problem. Just look at $\sin$ restricted to $[-\frac\pi 2,\frac\pi 2]$, so $u,v$ are in that segment. The domain of integration then becomes $u+v\leq \frac\pi 2$, with $u,v\in[-\frac\pi 2,\frac\pi 2]$. Also note that $\cos$ is nonnegative on $[-\frac\pi 2,\frac\pi 2]$, so the absolute values disappear. Your integral then becomes:
$$\int_{-\frac\pi 2}^0\int_{-\frac\pi 2}^{\frac\pi 2} (1+\sin v - \sin u)\cos u \cos v\,dvdu + \int_0^{\frac\pi 2}\int_{-\frac\pi 2}^{\frac\pi 2-u} (1+\sin v - \sin u)\cos u \cos v\,dvdu.$$