How to find the area of region $R$? Region $R$ contains all the points $(x,y)$ such that $x^2+y^2\leq100\;$  and $\:\sin(x+y)\geq0$. Find the area of region $R$.

$\:\:\:\:\:\:\:\:\:$$\sin(x+y)\geq0$
$\:\:\:\:\:\:\:\:\:$$\implies2n\pi\leq x+y\leq(2n+1)\pi$

Thus graph would be probably look like this:

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$

Now how to find the area?
 A: For the points in the circle $(a,b)$ if $$\sin(a+b)\leq0$$  then for $(-a,-b)$ $$\sin((-a)+(-b))\geq0$$ 
Thus if our condition does not hold true for a single point we can take a symmetrically opposite point to make our conditions hold. Thus half the are of our circle will satisfy our condition 
(note there will be lines that have $\sin(x+y)=0$ but as they are lines their contribution to the area is negligible, not affecting the probability, in reality it should be slightly greater than 1/2) 
A: try to divide it to regains, start with the top right quarter, find the equations for the start and end of the region that $\sin(x+y)\geq0$ is true(where it is equal $0$)
 at them and then integrate them with each other, the axises and the circle of $r=10$. after that do the top left one. now if you want you can prove symmetry and just multiply by $2$ or keep going and integrate the bottom left and bottom right(i think keep integrating is better for practice)
A: Due to innate symmetry, we can say that area of the common region is half of the area of circle. So required area is 
$$A = \frac{1}{2} 100 \pi = 50 \pi$$
Explanation
If $P(x,y)$ satisfies $x^2 + y^2 \le 100$, then $P'(-x,-y)$ also satisfies. Since $P$ and $P'$ are diametrically opposite and have same modulus, there are as many $P$ as there are $P'$.
Now if $\sin(x+y) <0$, then $-\sin(x+y) > 0$ That means $\sin(-x-y) >0$.
So if $P$ lies in required region, $P'$ does not lie. Since they are equal in number, required area is half of the area of circle.
