The figure below shows the graphs of the circles and The graphs intersect at the point and . Let be the shaded region in the first quadrant bounded by the two circles and the -axis.
(a) Set up, but do not evaluate, an expression involving one or more integral with respect to x that represent the area of .
(b) Set up, but do not evaluate, an expression involving one or more integral with respect to y that represent the area of .
(c) The polar equation of the circles are and , respectively. Set up, but do not evaluate, an expression involving one or more integrals with respect to the polar that represent area of .
For part (a), we express both circles as functions of , giving , , where we take the positive square root since we are only interested in . Then clearly, over , we have , and over , we have . Thus the area of is given piecewise as ,
For part (b), we express both circles as functions of , giving , , where we have carefully chosen the negative root for the second function because this represents the part of the circle that lies to the left of . Thus for , we have , and the area is simply
For part (c), we use the normal area formula of polar curve:
So we get
As for the actual value of , one can use either integral described above to find , but a simpler method utilizes basic geometry. We first observe that the radius joining the point of intersection to the origin forms a 45-degree angle with the positive -axis. Also, the radius joining the same point of intersection to the center of the smaller circle at forms a 90-degree angle with the -axis. This suggests that we draw the right , and consider the three regions into which this triangle partitions . Then the area of is immediately seen to be the sum of an eighth of the larger circle’s area plus a quarter of the smaller’s, minus the area of : as claimed.