if c1 is a circle with radius 1 and origin of $\begin{bmatrix}1\\0\end{bmatrix}$ and c2 is a circle with radius 3 and origin of $\begin{bmatrix}0\\0\end{bmatrix}$ with counter counter clockwise direction how can I solve the integral of F on the circles (and F is a field function)
$ F= \frac{2y^2}{(x^2+ y^6+1)^2-(4x^2)}\begin{bmatrix}2xy\\3(y^6-x^2+1)\end{bmatrix} $
I know F is not defined in (1,0) and (-1,0) if for c1 , $\gamma $ = $\begin{bmatrix}cos(t)+1\\sin(t)\end{bmatrix}$ and $\gamma' $= $\begin{bmatrix}-sin(t)\\cos(t)\end{bmatrix}$
and for c2 , $\gamma $ = $\begin{bmatrix}3cos(t)\\3sin(t)\end{bmatrix}$ and $\gamma' $= $\begin{bmatrix}-3sin(t)\\3cos(t)\end{bmatrix}$
I tired writing :
$$\iint_{c2-c1} F$$
and then seperated the integrals for c1 and c2 and tried solving the integrals with the method :
$$\iint_cF(\gamma(t)).\gamma(t)'$$
but I can't solve the integrals. is there any easier way that the problem can be solved?
((also what if $ F= \frac{y^2}{(x+1)^2+(y^6)}\begin{bmatrix}-y\\3(x+1)\end{bmatrix} $ I also can't solve the integral with the aforementioned method but in this case we have the conditions of a conservative function and can write $$ \nabla g=f$$ and $ g = arctan(\frac{y^3}{(x+1)})$ however F is not defined everywhere. I would appreciate if anyone could help.))