The area of the region (x,y):y2≤4x , 4x2+4y2≤9
29π16−98sin−113+212
9π16−98sin−113+212
9π8−94sin−113+212
49π16−98sin−113+212
Given inequalities are
y2≤4x--------(1)
and 4x2+4y2≤9-----------(2)
Points satisfying (1) lies interior to parabola y2=4x and those of (2) lies inside circle 4x2+4y2=9
Solving the curves we get,
4x2+16x=9
or (2x−1)(2x+9)=0
or x=1/2 (as x=-92 not possible )
Therefore, the points of intersection of both curves are 12,2 and 12,−2 .
The graph of these two curves and the common region of the points satisfying both the inequalities is as shown in adjacent figure
From the figure, required area is given byA=2∫112 2xdx+2∫1232 129−4x2dx
Putting 2x=t , dx=dt2in the second integral, we get
∴ A=2∫012 2xdx+14∫13 (3)2−(t)2dt=2x3232012+14t29−t2+92sin−1t313
=2232+140+92sin−1(1)−128+92sin−113=29π16−98sin−113+212