Find the area under the inequality $$\sin^2 \pi x + \sin^2 \pi y \le 1 \text{ for } x,y \in [-1,1]$$
I coudn't do this problem without using a graphing calculator:
It's easy to see now that in each quadrant, the area is $1/2$ unit, so the total area would be $2$ units.
How would one do this without access to a graphing calculator? Looking at the graph, It looks like there is a pattern I am missing out on. One thought would be to make the implicit inequality explicit, and obtain
$$|\sin \pi y \le \cos \pi x|$$
but I still couldn't couldn't graph this manually.
Best Answer
We have that
$$\sin^2 \pi x + \sin^2 \pi y = 1 \iff \sin^2 \pi x=\cos^2 \pi y \iff \sin \pi x=\pm\cos \pi y$$
and since by definition
$$ \begin{cases} \sin A= \cos B \iff A=\frac \pi 2\pm B+2k\pi \\\\ \sin A= -\cos B=\cos (-B) \iff A=-\frac \pi 2\pm B+2k\pi \end{cases} $$
we obtain
$$ \begin{cases} \pi x=\frac {3\pi} 2 \pm \pi y \iff x=\frac 32 \pm y\\\\ \pi x=\frac \pi 2 \pm \pi y \iff x=\frac 12 \pm y\\\\ \pi x=\frac \pi 2 \pm \pi y \iff x=\frac 12 \pm y\\\\ \pi x=-\frac {3\pi} 2 \pm \pi y \iff x=-\frac 32 \pm y \end{cases} $$
from here we can obtain the area under the inequality.