The area of the circle 4x2 + 4y2 = 9 which is interior to the parabola x2 = 4y is
A. \(\frac{\sqrt{2}}{6}+\frac{9}{4} \sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)\) sq. units
B. \(\frac{\sqrt{2}}{6}-\frac{1}{4} \sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)\) sq. units
C. \(\frac{3}{2}\) sq. units
D. \(\frac{7}{2}\) sq. units
Answer:A
The area of the circle 4x2 + 4y2 = 9 which is interior to the parabola x2 = 4y is
A. \(\frac{\sqrt{2}}{6}+\frac{9}{4} \sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)\) sq. units B. \(\frac{\sqrt{2}}{6}-\frac{1}{4} \sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)\) sq. units C. \(\frac{3}{2}\) sq. units D. \(\frac{7}{2}\) sq. units