Line integral over a non-central ellipse. I have to find the integral $$ \int_C ydx + x^2dy$$ over the curve $C$ given as a intersection of plane $z=0$ and surface $\frac{x^2}{a^2} + \frac{y^2}{b^2}=\frac{x}{a}+\frac{y}{b} $, curve $C$ is positively oriented $(a\geq b>0)$.
This is what i have this far:
given plane is $xy$ plane, it's intersection with this surface (whatever it is in the three-dimensional space) should be some sort of ellipse (non-origin ellipse) obviously, and the limits of integration should be from $0$ to $2\pi$. Now, after little bit of algebraic manipulation of the surface equation i got the following equation: $$\frac{(x-\frac{a}{2})^2}{\frac{a^2(a^2+b^2)}{4}} + \frac{(x-\frac{b}{2})^2}{\frac{b^2(b^2+a^2)}{4}}=1$$
Which is indeed an ellipse eqation, now , i suppose i should introduce polar coordinates here, in order to get the parametric equations for this curve, for this ellipse they should look something like this:
$x=\frac{a}{2}+\frac{a^2(a^2+b^2)}{4}\cos t \\y=\frac{b}{2}+\frac{b^2(a^2+b^2)}{4}\sin t \\ dx=-\frac{a^2(a^2+b^2)}{4}\sin t \\dy=\frac{b^2(a^2+b^2)}{4}\cos t $
Now, all i should do is to insert this into the given integral, but, i am not quite sure is this legitimate approach. Any suggestions or comments if this is incorrect is appreciated or if it is correct, any advice on how to do this more easily is appreciated too.
 A: By Green's theorem,
$$I:=\int_C ydx + x^2dy=\iint_E (2x-1)dxdy=|E|(2\bar{x}-1)$$
where $E$ is the domain bounded by $C$ in the $xy$-plane, $|E|$ is its area and $\bar{x}$ is the $x$-coordinate of the centroid of $E$. 
Since $C$ is the ellipse given by (check your algebraic manipulations!!)
$$\frac{(x-\frac{a}{2})^2}{\frac{a^2}{2}} + \frac{(y-\frac{b}{2})^2}{\frac{b^2}{2}}=1$$
then
$$|E|=\pi\frac{a}{\sqrt{2}}\cdot\frac{b}{\sqrt{2}}=
\frac{\pi ab}{2}\quad\text{and}\quad\bar{x}=\frac{a}{2}.$$
Therefore
$$I=\frac{\pi ab(a-1)}{2}.$$
A: Seems your computation is not correct. A slightly easier way to rearrange that ellipse is
$$\left(\frac{x}{a}-\frac{1}{2}\right)^2+\left(\frac{y}{b}-\frac{1}{2}\right)^2=1.$$
You can then let
$$x=a\left(\frac{1}{\sqrt{2}}\cos\theta+\frac{1}{2}\right)\\
y=b\left(\frac{1}{\sqrt{2}}\sin\theta+\frac{1}{2}\right).$$
The integral will be easier to solve using this transformation.
A: First observe that the correct equation for this ellipse is
$$\frac{x^{2}}{a^{2}}-\frac{x}{a}+\frac{y^{2}}{b^{2}}-\frac{y}{b}=0$$
$$\left(\frac{x}{a}-\frac{1}{2}\right)^{2}+\left(\frac{y}{b}-\frac{1}{2}\right)^{2}=\frac{1}{2}$$
$$\frac{\left(x-\frac{a}{2}\right)^{2}}{\frac{a^{2}}{2}}+\frac{\left(y-\frac{b}{2}\right)^{2}}{\frac{b^{2}}{2}}=1$$
As for the integral, it turns out to be very simple if you use Green's theorem
$$I=\int_{\rm \partial S}y{\rm d}x+x^{2}{\rm dy}=\iint_{\rm S}\left(\frac{\partial}{\partial x}x^2-\frac{\partial}{\partial y}y\right){\rm d}x{\rm d}y=\iint_{\rm S}\left(2x-1\right){\rm d}x{\rm d}y$$
Now note that
$$\iint_{\rm S}x{\rm d}x{\rm d}y=\frac{a}{2}{\rm Area}\left({\rm S}\right)$$
is the average of $x$ over the ellipse and
$$\iint_{\rm S}{\rm d}x{\rm d}y={\rm Area}\left({\rm S}\right)$$
is just the area. Using ${\rm Area}\left({\rm S}\right)=\pi\frac{a}{\sqrt{2}}\cdot\frac{b}{\sqrt{2}}=\frac{\pi ab}{2}$, you get
$$I=\frac{\pi ab\left(a-1\right)}{2}$$
