Solving line integral without using Green’s theorem Good morning everyone, I am a new user here and I have the following problem that I am dealing with since yesterday but I cannot find the correct answer. Here is the problem
Calculate
$$\oint_C (x^2-2xy)dx+(x^2y+3)dy$$
where $C$ is the closed curve of the region bounded by $y^2=8x$ and $x=2$.
Okay, using Green’s theorem, I got
$$\oint_C (x^2-2xy)dx+(x^2y+3)dy=\int_{-4}^4\int_{y^2/8}^{2}(2xy+2x)dxdy=\frac{128}{5}$$
but without using the theorem, I got
The the region bounded by $y^2=8x$ and $x=2$ intersect at $(2,-4)$ and $(2,4)$, hence
The line integral along $x=2$ equals
$$\int_{x=2}^{2}(x^2-2xy)dx+(x^2y+3)dy=0\tag{*}$$
and the line integral along $y^2=8x$ equals
$$\begin{align}\int_{y=4}^{-4}\left(\left(\frac{y^2}{8}\right)^2-2\left(\frac{y^2}{8}\right)y\right)d\left(\frac{y^2}{8}\right)+\left(\left(\frac{y^2}{8}\right)^2y+3\right)dy&=\int_{y=4}^{-4}\left(\frac{5y^5}{256}-\frac{y^4}{16}+3\right)dy\\&=\frac{8}{5}\end{align}$$
So, where is my mistake? Could someone here help me out? Thanks for your comments and answers, appreciate it.

Use parametrisation. That is the trick to do line integral problems. 
Take $y=t$ and $x=\frac{t^2}{8}$ in Path- I and $x=2,dx=0$ in Path-II 
 
So your line integral $$\oint=\oint_{I}+\oint_{II}$$
Now $$\oint_1=\int_{4}^{-4}[(\frac{t^4}{64}-\frac{t^3}{4})\cdot \frac{t}{4}dt+(\frac{t^5}{64}+3)dt]$$
$$=\int_{4}^{-4}[(\frac{t^5}{256}-\frac{t^4}{16}+\frac{t^5}{64}+3)dt]$$
$$=[-\frac{t^5}{64}]_{4}^{-4}$$ since the other functions are odd functions, so their integrals are $0$
$$=\frac{128}{5}$$
And $$\oint_2=\int_{-4}^{4}(4y+3)dy$$
$$=0$$ since the function is odd.
So your answer is $\oint=\frac{128}{5}$
