Let $F(x,y)=<x,y>$. Let $C$ be the portion of the ellipse $(\frac{x^2}{a^2})+(\frac{y^2}{b^2})=1$ in quadrants 1 and 2. Show how to evaluate $\int_C F \cdot dr.$
this is the question i am given. i am not sure if what i am doing is right but this is what i have.
$$x=a \cos t \ \ \ \implies \ \ \ dx/dt=-a \sin t $$
$$y=b \sin t \ \ \ \implies \ \ \ dy/dt = b \cos t$$
$$0<t<\pi$$
$ds=\sqrt{(a^2sin^2t+b^2cos^2t)}dt$
(integral along the curve)$xdx$+(integral along the curve)$ydy$
$x = a\cos t\\y= b\sin t\\ dr = (\frac {dx}{dt},\frac {dy}{dt})dt$
That part is fine.
Now you substitute into $F(x,y)\to F(t) = (a\cos t, b\sin t)$
$\int_c F(t)\cdot dr = \int (a\cos t, b\sin t)\cdot (-a\sin t, b\cos t) dt = \int (b^2-a^2) \cos t\sin t\;dt$
Now $F(x,y) = \nabla (\frac 12 x^2 + \frac 12 y^2)$ Which means that F(x,y) is a conservative force and $\int_c F(x,y)\cdot dr$ depends only on the endpoints and not on the path taken. And $\oint F(x,y)\cdot dr = 0$ for any closed contour