Compute this line integral $\int_{C}(x+y){\mathbf{i}}+(x-y){\mathbf{j}}\ d\alpha $ Calculate the line integral of the function $$f(x,y)=(x+y){\mathbf{i}}+(x-y){\mathbf{j}}$$  around the ellipse $$b^2x^2+a^2y^2=a^2b^2$$ counter to clockwise. 
My approach is the following: The equation can be written in the form $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ then we can parametrize by taking $x=a\cos t$ and $y=b\sin t$ for $0\le t\le 2\pi$. So i have $\alpha(t)=(a\cos t)\mathbf{i}+(b\sin t)\mathbf{j}$. And to integrate i use the formula $$\int_{C}f\cdot d\alpha=\int_{a}^{b}f(\alpha(t))\cdot a'(t)dt$$ which is in this case $$\int_{0}^{2\pi}((a\cos t+b\sin t)\mathbf{i}+(a\cos t-b\sin t)\mathbf{j})\cdot ((-a\sin t)\mathbf{i}+(b\cos t)\mathbf{j})dt $$ which is equal to $$\int_{0}^{2\pi}-(a^2+b^2)\sin t\cos t-ab(\sin^2 t-\cos^2 t)dt$$ My problem is that this integral is not zero (which is the correct answer according to the book). Where is my error?
 A: Or use Stokes' Theorem:  if $G$ is a vector field, then
$\int_\Omega \nabla \times G dA = \int_\Gamma G \cdot d\alpha, \tag{1}$
which holds for any surface $\Omega$ bounded by the curve $\Gamma$, where $dA$ is the area element in $\Omega$ and $d\alpha$ is the line element along $\Gamma$.  If we take
$G(x, y) = f(x, y) = (x + y)\mathbf i + (x - y) \mathbf j, \tag{2}$
we can compute that
$\nabla \times f(x, y) = \nabla \times ((x + y)\mathbf i + (x - y) \mathbf j)$
$= -\partial / \partial z (x - y) \mathbf i + \partial / \partial z (x + y) \mathbf j + (\partial / \partial x (x - y) - \partial / \partial y (x + y)) \mathbf k$
$= 0 \mathbf i + 0 \mathbf j + (1 - 1)\mathbf k = 0; \tag{3}$
this shows the left hand side of (1) must be zero when $G(x, y) = f(x, y)$.  Thus
$\int_\Gamma f \cdot d\alpha = 0 \tag{4}$
as well.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
A: f(t) = (acost,bsint); f'(t) = (-asint,bcost) so that
$f \cdot f' = -a^2costsint  + b^2sintcost$
Both these terms are periodic in 2$\pi$ so when you integrate from 0 to $2\pi$ you get zero for each term.
