This is a problem from Chennai Mathematical Institute, CMI Entrance 2014 based on area of a region. Try to solve it.
Problem: Area of a region
$latex \mathbf{ A= {(x, y), x^2 + y^2 \le 144 , \sin(2x+3y) le 0 } } $ . Find the area of A.
Discussion:
$latex \mathbf{ x^2 + y^2 \le 144 }$ is a disc of radius 12 with center (0, 0).
Now notice that $latex \mathbf{ \sin(2x + 3y) \le 0 \implies \sin ( 2(-x) + 3(-y) ) = \sin(-(2x+3y)) = -\sin(2x+3y) \ge 0 }$
Hence if a point (x, y) is in A then (-x, -y) is not in A.
Similarly if there is a point (x, y) not in A then we get a corresponding point (-x, -y) in A.
Therefore we have a bijection between points in A and not in A.
Thus area of A is exactly half the area of the disc = $latex \mathbf{ \frac{\pi (12)^2 }{2} = 72 \pi }$
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[…] . Find the area of A. Solution […]
This can in fact be generalised to any odd function f(x). f(ax+by) divides the area of the disc in half if f is odd.