# Double integral with domain $\arcsin x + \arcsin y \leq \frac{\pi}{2}$

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.

• is the domain not the same as the region inside the unit circle in the first quadrant? Commented Jun 27, 2020 at 13:17

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:

$$\int_{-\frac\pi 2}^0\int_{-\frac\pi 2}^{\frac\pi 2} (1+\sin v - \sin u)\cos u \cos v\,dvdu + \int_0^{\frac\pi 2}\int_{-\frac\pi 2}^{\frac\pi 2-u} (1+\sin v - \sin u)\cos u \cos v\,dvdu.$$

• Thanks, I suppose that $-1 \leq x \leq 1$ and $-1 \leq y \leq 1$ come from the existence condition of $\arcsin$. Can I ask you where are the flaws in my approach? Commented Jun 27, 2020 at 13:44
• @ZaWarudo, take a look at my edit. Commented Jun 27, 2020 at 14:01
• why not change to polars to make it all simpler? Commented Jun 27, 2020 at 14:06
• @David Quinn, I am not convinced that polar coordinates would simplify anything since the region $A$ is bounded by straight lines as well, but feel free to try. Commented Jun 27, 2020 at 14:13
• I'm just suggesting something you could try to make it simpler Commented Jun 27, 2020 at 14:16