Find the area of the region bounded by $x^2+y^2 \le 144$ and $\sin(2x+3y) \le 0$

Question

Find the area of the region bounded by $x^2+y^2 \le 144$ and $\sin(2x+3y) \le 0$

My try:

we have $$\sin(2x+3y) \le 0$$ when

$$-12 \le 2x+3y \le -3 \pi$$

$$ -2 \pi \le 2x+3y \le -\pi$$

$$-\pi \le 2x+3y \le 0$$

$$-\pi \le 2x+3y \le 0$$

$$ 3\pi \le 2x+3y \le 12$$

we get five strips of areas bounded by pair of parallel lines?

but the appraoch is very lengthy.

Any good appraoch

Answer 1

Note that the set of points within the circle where $\sin(2x+3y)=0$ is just the union of finitely many line segments, so has area $0$, hence can be ignored.

By symmetry, for the region where $\sin(2x+3y) < 0$, there is an opposite region, reflected through the origin, where $\sin(2x+3y) > 0$.

Since reflection preserves area, it follows that the required area is half that of the circular region.

Answer 2

Note that setting $u=2x+3y$ we have that $\sin u$ is an odd function with respect to $u=2x+3y=0$ and since the domain of the circle $x^2+y^2 \le 144$ is polar symmetric with respect to the origin, for symmetry the area bounded is an half of that circle with radius $R=12$ (see the following figure for the idea behind the symmetry argument).