Problem: Assume that $C$ is the positively-oriented elliptic curve $x^2 +4y^2 = 4$. What is the value of $\oint_C (x^3 + 2y)dx + (4x-3y^2)dy$?
Attempt: I say we do this the Green way; that is, using Green's theorem: First we put the ellipse in standard form $x^2/4 + y^2 = 1$, then $$\oint_C (x^3 + 2y)dx + (4x-3y^2)dy=\iint \bigg(\frac{\partial}{\partial x}(4x- 3y^2)-\frac{\partial}{\partial y}(x^3 + 2y)\bigg)dxdy=\iint_C (4 - 2)dxdy = 2(\text{ellipse area})= 4\pi.$$ However, as usual there is a solution manual that disagrees with me, saying that $\pi$ is the value of the integral not $4\pi$. What do you think?
Much obliged!