  Question

# Consider a region M which contains all the points (x,y) such that x2+y2≤100 and sin(x+y)≥0 where x,y∈R. If the area of region M is mπ sq units, then the value of m is

Solution

## x2+y2≤100 represents set of all points (x,y) which lies on or inside the circle x2+y2=100. Let C denote the disk (x,y) with x2+y2≤100. Now, sin(x+y)=0 if and only if x+y=kπ for k∈Z. So, disk C has been cut by parallel lines x+y=kπ 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. 50π sq units. Alternate Solution:  x2+y2≤100 represents set of all points (x,y) which lies on or inside the circle x2+y2=100. Also, sin(x+y)≥0 ⇒0≤x+y≤πor2π≤x+y≤3πor4π≤x+y≤5πand so on ⋯ ⇒x∈[0,π]∪[2π,3π]∪[4π,5π]∪⋯ Clearly from figure, Required Area = Shaded Area Also, Shaded Area + Unshaded Area = Area of circle By symmetry, Shaded Area = Unshaded Area  ∴ Required Area=12(Area of circle)                               =12×π(102)                               =50π sq units  Suggest corrections   