Area of Two Inequalities in the $XY$ Plane

Question

This was question B1 on the 2014 CMI (Chennai Mathematical Institute) Entrance Examination for Undergraduate Courses. We were asked to find the area of the region in the $XY$ plane consisting of all points in the set $S=\{(x,y):x^2+y^2\leq144 \text{ and} \sin(2x+3y)\leq0\}.$ While this question is interesting, I do not know how to even begin. To me, it isn't readily obvious why this set has to have an area at all. What if only finitely many values of $x$ and $y$ satisfy the above conditions? I mean, given that the question is asking for the area, I can guess that there are infinitely many such pairs $(x,y),$ but apart from that, I don't know what to do. I suppose we could begin by writing $S=S_1\cap S_2,$ where $S_1=\{(x,y):x^2+y^2\leq144\},$ and $S_2=\{(x,y):\sin(2x+3y)\leq0\}.$ What if we look for a more general answer by putting $S_1=\{(x,y):x^2+y^2=r^2,r\in\mathbb{R}\setminus\{0\}\},$ and $S_2=\{(x,y):\sin(ax+by),a,b\in\mathbb{R}\setminus\{0\}\}?$

Answer 1

As mentioned in the comments, $\sin(2x+3y)\le 0$ means $$(2k-1)\pi\le 2x+3y\le 2k\pi$$ Let's take a look at $k=0$. This is a diameter of the circle of radius $12$, centered in origin, $x^2+y^2=(12)^2$. All the lines $2x+3y=k\pi$ are parallel to this one. You can calculate the distance between lines, but it's not necessary. All you need is to notice that all the distances are the same. For that, draw a straight line (say $x=0$) that intersects the set of parallel lines. Notice that the $y$ intersections are equidistant. Now notice that if you select an area between $k\pi\le 2x+3y\le (k+1)\pi$, if the value of $\sin$ is positive, then the same expression for $-k$ will be negative. And vice versa. So for example, you are keeping the area between $k=-1$ and $k=0$, and you are throwing away the area between $k=0$ and $k=1$. But, due to symmetry, these areas are the same. And this continues for all $k$ values, until $|k|\pi>12$. Then you disregard the rest.