# Evaluate $\int_{\Gamma}xy^2dx+xydy$ on $\Gamma={y=x^2}$

Evaluate $\int_{\Gamma}xy^2dx+xydy$ on $\Gamma=\{(x,y)\in\mathbb{R}^2:y=x^2,x\in[-1,1]\}$ with orientation clockwise using Green theorem

So $\Gamma$ is a parabola to use Green we have to close the curve, to do so we will add the line from $(1,1)$ to $(-1,1)$

Then

$\gamma_1(t)=(-t,1),t\in[-1,1]$

$\gamma_2(t)=(t,t^2),t\in[-1,1]$

$\int_{wanted}=\int_{\gamma_1(t)\cup \gamma_2(t)}-\int_{\gamma_1(t)}$

But we must have one parameterization of $2$ variables which is closed to use green?

maybe $\phi(r,\theta)=(\sin t\cos t,\sin ^2t-\sin t ),t\in [-\pi,-2\pi]$ is the closed curve?

$\int Mdx+Ndy = \iint (N_x-M_y)dxdy \\ M = xy^2, N = xy \\\int Mdx+Ndy = \int_{x=-1}^{x=1}\int_{y=0}^{y=x^2} (y-2xy) dydx \\ = \int_{x=-1}^1 (1-2x )x^4 \frac{1}{2} dx \\ = \frac{1}{5}$
• To be clear, this gives $-(\int_\Gamma + \int_{\Gamma'})$, where $\Gamma$ is the parabola traversed from left to right and $\Gamma'$ is the polyline from $(1, 1)$ to $(1, 0)$ to $(-1, 0)$ to $(-1, 1)$. – Maxim Aug 22 '18 at 16:12