## Find the centre of gravity of the sector of a circle subtending an angle 2α at the centre of the circle?

Ans: To find centre of gravity of the sector of a circle first draw the figure

AOB is the sector of the circle subtending an angle 2α at the centre. Ox is the bisector of angle AOB.

Equation of the curve is r=a, referred to O as pole and Ox as initial line, a being the radius of the circle.

Ox is the line of symmetry. Hence, the C.G. of the sector lies on Ox. Hence ȳ=0.

And x=2/3[( ʃ^{α}_{-α}r^{3}cosθdθ)/( ʃ^{α}_{-α}dθ)]

=2/3.a[( ʃ^{α}_{-α}cosθdθ)/( ʃ^{α}_{-α}dθ)]

=2a/3[( sinθ)^{α}_{-α}]/[(θ) ^{α}_{-α}]

= 2a/3[( sinα)/α]

So, r= a