Plotting the area of the smaller region enclosed by the curves y2=8x+4 and x2+y2+43x−4=0 we get,

Now finding point of intersection of x2+y2+43x−4=0 and
y2=8x+4
We get, (0,2) and (0,−2)
Both are symmetric about x−axis
So, area will be =2∫02(16−y2−23)−(8y2−4)dy
=2[21(y16−y2+16sin−1(4y))−23y−24y3+21y]02
=31[8π+4−123]