Question

Consider a region $M$ which contains all the points $(x,y)$ such that $x^2+y^2\le 100$ and $\sin (x+y)\ge 0$ where $x,y\in R$. If the area of region $M$ is $m\pi$ sq units, then the value of $m$ is

Answer 1

$x^2+y^2\le 100$ represents set of all points $(x,y)$ which lies on or inside the circle $x^2+y^2=100.$
Let $C$ denote the disk $(x,y)$ with $x^2+y^2\le 100$.
Now, $\sin (x+y)=0$ if and only if $x+y=k\pi$ for $k\in Z.$
So, disk $C$ has been cut by parallel lines $x+y=k\pi$ and in between those lines there are regions containing points $(x,y)$ with either $\sin (x+y)>0$ or $\sin (x+y)<0$.

Since, $\sin (-x-y)=-\sin (x+y)$, the regions containing points $(x,y)$ with $\sin (x+y)>0$ are symmetric with respect to the origin to the regions containing points $(x,y)$ with $\sin (x+y)<0$.


Thus, from the given figure, the area of region $M$ is half the area of disk $C$, i.e. $50\pi$ sq units.


Alternate Solution:
$x^2+y^2\le 100$ represents set of all points $(x,y)$ which lies on or inside the circle $x^2+y^2=100.$
Also, $\sin (x+y)\ge 0$
$\begin{array}{c}\Rightarrow 0\le x+y\le \pi \\ \text{or}\\ 2\pi \le x+y\le 3\pi \\ \text{or}\\ 4\pi \le x+y\le 5\pi \\ \text{and so on}\cdots \end{array}$
$\Rightarrow x\in [0,\pi ]\cup [2\pi ,3\pi ]\cup [4\pi ,5\pi ]\cup \cdots$

Clearly from figure,
Required Area = Shaded Area
Also, Shaded Area + Unshaded Area = Area of circle
By symmetry,
Shaded Area = Unshaded Area

$\therefore \text{Required Area}=\frac{1}{2}\left(\text{Area of circle}\right)$
$=\frac{1}{2}\times \pi \left({10}^2\right)$
$=50\pi \text{sq units}$