JEE Main20202 Sep 2020Morning ShiftMathematicsArea Under CurvesActual
Area (in sq. units) of the region outside x 2 + y 3 = 1 and inside the ellipse x 2 4 + y 2 9 = 1 is
Options
- A6 π − 2
- B3 π - 2
- C3 4 - π
- D6 4 − π
Correct answer
A. 6 π − 2
Step-by-step solution
Area of ellipse = π a b = 6 π Area enclosed by | x | 2 + | y | 3 = 1 is = 4 × 1 2 2 3 = 12 So, required area = 6 π - 12 sq. units