The area (in sq. units) of the region lying in the first quadrant and enclosed by the X-axis, the straight line x-√3y=0 and the circle x^2+y^2=4 is

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MARKS Content Team · Accepted Answer

Correct answer: (A) (π)/3. Required area=Ar∆OAB+∫√32√(4-x^2)dx Now, Ar∆OAB=1/2×√3×1=√32 ...i And, let I=∫√32√(4-x^2)dx =1/2x√(4-x^2)+4/2sin^(-1)x/2√32 =1/2·2·√(4-4)+2sin^(-1)2/2-√32+2sin^(-1)√32 =2×(π)/2-√32+2(π)/3 =(π)/3-√32 ...ii By i & ii, we get Required area=(π)/3 sq.units.

The area (in sq. units) of the region lying in the first quadrant and enclosed by the $X$ -axis, the straight line $x - \sqrt{3} y = 0$ and the circle $x^{2} + y^{2} = 4$ is

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Relevant Questions

The area (in sq. units) of the region lying in the first quadrant and enclosed by the X-axis, the straight line x-3y=0 and the circle x2+y2=4 is

Solution

Relevant Questions

The area (in sq. units) of the region lying in the first quadrant and enclosed by the $X$ -axis, the straight line $x - \sqrt{3} y = 0$ and the circle $x^{2} + y^{2} = 4$ is