Problem with the Stokes theorem I have the following problem:
Let $D:=\{(x,y)\in\mathbb{R}^2|(x-1)^2+y^2\leq1\}$ and $\omega=xy^2dx+2xydy$. Calculate $\int_{\partial D}\omega$ with the definition of the integral and with the Stokes theorem. My problem here is that I don't know how to calculate the integral because $\omega$ is a form. How do I apply that to the definition?
Also I don't know what to do with the Stokes theorem. We introduced that topic yesterday and had no examples. Could someone explain what I have to do?
Thanks in advance.
 A: This is for the path integral.
The boundary of the circle is $\{(x-1)^2+y^2=1\}$, which is a circle of radius $1$ around the point $(1,0)$. This circle can be parametrized by$$\gamma:[0,2\pi]\to\mathbb{R}^2,\quad\theta\mapsto(1,0)+(\cos\theta,\sin\theta).$$Using this parametrization we have$$x=1+\cos\theta,\quad y=\sin\theta,$$and so, $$dx=-\sin\theta d\theta,\quad dy=\cos\theta d\theta.$$Finally, $$\int_{\partial D}xy^2dx+2xydy=\int_0^{2\pi}\left((1+\cos\theta)\sin^2\theta\cdot(-\sin\theta)+2(1+\cos\theta)\sin\theta\cos\theta\right)d\theta,$$which is a usual integral on an interval.
A: $D$ is a disk in the $xy$-plane of radius $1$, centered around $(1,0)$. You are asked to calculate the integral of the function $\omega$ on the path surrounding $D$, where you can right $\omega = \vec{F} \cdot d\vec{r}$, $\vec{F}$ is a vector field. You can use Stokes' theorem to instead calculate the curl of $\vec{F}$ on the surface of the circle with the theorem which states that
$$ \int_{\mathcal{\partial D}} \vec{F} \cdot d\vec{r} = \iint_{\mathcal{D}} \left( \vec{\nabla} \times \vec{F} \right) \cdot d\vec{S} $$
with $d\vec{S}$ the vector normal to the disk.
I suggest you first find the vector field $\vec{F}$, then do the integral. I used this link to remember correctly, there are some examples as well.
I hope this helps you!
A: In two dimensions, Green's Theorem follows from Stoke's Theorem. 
We are trying to calculate,
$$\int_{\partial D} F \cdot dr$$
Where $F=\langle xy^2,2xy \rangle$ and $dr$ is shorthand for $\langle dx,dy \rangle$. Notice that: $(2xy)_x-(xy^2)_y=2y-2yx$.
We get,
$$\iint_{D} (2y-2xy) dA$$
Where $D$ is the disk of unit radius centered at $(1,0)$. To simplify the problem make the change of variables $u=x-1$ and $v=y$. Then $D$ becomes the unit disk centered at the origin in the $u-v$ plane, and the Jacobian corresponding to this transformation is $1$. We get,
$$\iint_{T(D)} (2v-2(u+1)(v)) dA$$
$$=-\iint_{T(D)} 2uv dA$$
$$=-\int_{0}^{2\pi} \int_{0}^{1} 2r\cos \theta r\sin \theta r dr d\theta$$
$$=-\frac{2}{4} \int_{0}^{2\pi} \frac{1}{2}\sin 2\theta d\theta$$
$$=0$$
