The limits on y are sqrt (16 -x^2)=0, x = +/- 4 Take +4, The area
is int y dx with the limits +2 to -2. int y dx = int ( sqrt (16 -x^2)dx = Put x= 4sin a, dx= 4 cos a da
= 16int cos^a da= 8 int (cos 2a +1) da= -4 sin 2a +a Now put the limits
When x -4, sin a =-1, a =-pi/2, when x= 4, y= pi/2. In both the limits sin 2a becomes 0. We finally get the area= 8 pi. This is the answer given above. Please put brackets properly.