The value of the line integral $\oint_C ( (y - 3x)dx + (2x + y^2)dy )$ around the circle $C: x^2 + y^2 = 4$ traversed counter-clockwise is:

(A) $4\pi$

(B) $8\pi$

(C) $2\pi$

(D) $16\pi$

Answer Key:

A

### Solution: By Green's theorem: $\oint_C (P dx + Q dy) = \iint_R \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA$. Here $P = y - 3x \implies \frac{\partial P}{\partial y} = 1$. $Q = 2x + y^2 \implies \frac{\partial Q}{\partial x} = 2$. $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2 - 1 = 1$. Integral $= \iint_R 1 dA = \text{Area of circle of radius } 2 = \pi (2^2) = 4\pi$. Correct: A.

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