AP EAMCET201825 Apr 2018Morning ShiftMathematicsArea Under CurvesActual
The area (in sq. units) of the region lying in the first quadrant and enclosed by the X -axis, the straight line x - 3 y = 0 and the circle x 2 + y 2 = 4 is
Options
- Aπ 3
- B2 π 3
- Cπ 2 3
- D2 π 3 2
Correct answer
A. π 3
Step-by-step solution
Required area = A r ∆ O A B + ∫ 3 2 4 - x 2 d x Now, A r ∆ O A B = 1 2 × 3 × 1 = 3 2     . . . i And, let I = ∫ 3 2 4 - x 2 d x = 1 2 x 4 - x 2 + 4 2 sin - 1 x 2 3 2 = 1 2 · 2 · 4 - 4 + 2 sin - 1 2 2 - 3 2 + 2 sin - 1 3 2 = 2 × π 2 - 3 2 + 2 π 3 = π 3 - 3 2     . . . i i By i   &   i i , we get Required area = π 3   sq . units .