Q26 of 45 Page 1

Find the area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle x2 + y2 = 32.

The equations of the given lines are


y = x (i)


and x2 + y2 = 32 (ii)


from (i) and (ii), we get that the line and circle intersect each other at B(4,4) in the first quadrant.



Draw perpendicular BM to x-axis.


Therefore, the required area = area of the region OBMO + area of the region BMAB.


Now, the area of the region OBMO =


=


The area of the region BMAB =


=


=


=


=


=8π –(8+4π) =4π -8


Thus, the required area = area of the region OBMO + area of the region BMAB.


= 8 + 4π -8 = 4π sq units.


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