For a given vector field and curve C, evaluate $\int_C \mathbf{F.dr}$
$\mathbf{F}(x,y)= -y \mathbf{i} +x \mathbf{j}$, C:Ellipse $\frac{x^2}{4}+\frac{y^2}{9}$=1
First we parametrise
$x=2\cos{t}$; $dx=-2\sin{t}$
$y=3\sin{t}$; $dy=3\cos{t}$
Then we write a proper integrand
The limits are found by replacing $x=2$ and $x=-2$ in $x=2\cos{t}$
$\int_{\pi}^{0}{6\sin^2{t}+6\cos^2{t}}$= $\int_{\pi}^{0}{6}=-6\pi$
But the answer is $12\pi$
Parametrize the ellipse :
$$\gamma(t)=(x(t),\,y(t))\;,\;\;\begin{cases}x=2\cos t\\{}\\y=3\sin t\end{cases}\;,\;\;0\le t\le2\pi\;\;\implies$$$${}$$
$$\int_CF\cdot dr=\int_0^{2\pi}F(\gamma(t))\cdot\gamma'(t)\;dt=\int_0^{2\pi}\left(-3\sin t,\,2\cos t\right)\cdot\left(-2\sin t,\,3\cos t\right)dt=$$
$$=\int_0^{2\pi}6(\cos^2t+\sin^2t)\;dt=6\int_0^{2\pi}dt=12\pi$$