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.
Best Answer
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 $\oint=\frac{128}{5}$