Question is:

Find the area of the circle 4x² + 4y² = 9 which is interior to the parabola x² = 4y

Solution:

The required area is represented by the shaded area OBCDO in the figure.

Solving the given equation of circle 4x² + 4y² = 9 and parabola x² = 4y, we get

B(√2, 1/2) and D(-√2, 1/2)

From the figure, it is observed that area is symmetrical about y-axis.

So, area OBCDO = 2 * Area OBCO

Draw BM perpendicular to OA. So, the coordinate of M is (√2, 0)

Hence, area OBCO = Area OMBCO - area OMBO

                           = ₀∫^√2 √(9 - 4x² )/4 dx - ₀∫^√2 (x² /4) dx

                          = (1/2) ₀∫^√2 √(9 - 4x² ) dx - (1/4) ₀∫^√2 x² dx

                          = (1/2)*[x√(9 - 4x² ) + (9/2)sin^-1 (2x/3) ₀]^√2  - (1/4)*[x² /3 ₀]^√2

                          = (1/2)*[√2 * √(9 - 8) + (9/2)sin^-1 (2√2/3)]  - (1/12)*(√2)³

                          = √2/4 + (9/8)sin^-1 (2√2/3) - √2/6  

                          = √2/12 + (9/8)sin^-1 (2√2/3)

                          = (1/2) * [√2/6 + (9/4)sin^-1 (2√2/3)]  

So, the required area OBCDO = 2 * (1/2) * [√2/6 + (9/4)sin^-1 (2√2/3)]

                                           = √2/6 + (9/4)sin^-1 (2√2/3) units