Evaluate $\int_{(x,y)\in \Bbb R^2,\ x^2+y^2=4}\ (3x-4x^2y)dx+(4x^2y+2y)dy$

Question

Let $C$ be the circle with centre at origin of radius $2$. Calculate line integral: $$\int_C(3x-4x^2y)dx+(4x^2y+2y)dy$$

I think the value is $32\pi$. But, the answer key given gives the value as $16\pi$.

Am I wrong somewhere?

I used the Green's theorem to evaluate the integral.

Any hints as to where the mistake lies? Thanks beforehand.

Answer 1

Using Green's theorem $$ \oint_C(3x−4x^2y)dx+(4x^2y+2y)dy=\int_S(8xy+4x^2)dxdy=\int_S(8r^2\sin(\theta)\cos(\theta)+4r^2\cos^2(\theta))rd\theta, $$ where polar coordinates were used and $S$ denotes the disc with radius 2. Computing the last integral we have $$ \int_S(8r^3\sin(\theta)\cos(\theta)+4r^3\cos^2(\theta))d\theta=\int_0^{2\pi} \Big(32\sin(\theta)\cos(\theta)+16\cos^2(\theta)\Big{)}d\theta=16\pi. $$