Let $\Gamma$ be the boundary of the closed circular region $A$ given by $x^{2}+y^{2} \leq 1$. Then $I=\int_{\Gamma}\left(3 x^{3}-9 x y^{2}\right) d s$ (where ds means integration along the bounding curve) is
(A) $\pi$
(B) $-\pi$
(C) 1
(D) 0

Question ID - 155902 :-

Let $\Gamma$ be the boundary of the closed circular region $A$ given by $x^{2}+y^{2} \leq 1$. Then $I=\int_{\Gamma}\left(3 x^{3}-9 x y^{2}\right) d s$ (where ds means integration along the bounding curve) is
(A) $\pi$
(B) $-\pi$
(C) 1
(D) 0

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