Given vector,
A=xy(ax−2xay)
⇒A=xyax−2x2yay
Now, x=acosθ
y=asinθ
z=z
B=acosθ×asinθ×ay−2×(acosθ)2×asinθ×ay
we know that a=ax=ay for the circle
=a3(cosθ.sinθ−2cos2θ.sinθ)
=a3(21sin2θ−cosθ.sin2θ)
=a3sin2θ(21−2cosθ)
=24a3sin2θcos2(2θ)
=2a3sin2θcos2(2θ)
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