The area of the region (x,y):y2≤4x , 4x2+4y2≤9

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 figureFrom 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−1⁡t313=2232+140+92sin−1⁡(1)−128+92sin−1⁡13=29π16−98sin−1⁡13+212