Evaluating Line Integrals! $3xy^2dx+2x^3dy$ 
where  is the boundary of the region between the circles $x^2+y^2=25$ and $x^2+y^2=64$ having positive orientation.
Not quite sure how to evaluate this...
 A: Ok, so let's assume what I asked in my comment is actually what you're looking for.
So the first step is to write your circle equations in parametric form:
First circle $C_1$: $x^2+y^2=25=5^2$
This means you have: $\cases{x=5\cos(t) \\ y=5\sin(t)}$
And: $\cases{dx=-5\sin(t)dt \\ dy=5\cos(t)dt}$
Second circle $C_2$: $x^2+y^2=64=8^2$
This means you have: $\cases{x=8\cos(t) \\ y=8\sin(t)}$
And: $\cases{dx=-8\sin(t)dt \\ dy=8\cos(t)dt}$
The area $A$ of the region between the two circles is equal to $A=A_2-A_1$ where $A_1$ and $A_2$ are respectively the areas of circles $C_1$ and $C_2$.
And you know that: $\displaystyle\cases{A_1=\frac{1}{2}\int_{C_1}xdy-ydx \\ A_2=\frac{1}{2}\int_{C_2}xdy-ydx}$
Now replace in both those equations the parametric forms of $x$,$y$,$dx$ and $dy$ and you'll be good to go (you'll see it simplifies a lot) ;)
Oh and don't forget, $t\in[0,2\pi]$ for your integrals !
A: One way of evaluating integral $\int\limits_{\partial{D}}{3xy^2 \, dx+2x^3\,dy}$ along the boundary  $\partial{D}$ of the domain $D=\{(x,\;y)\colon \;\;\; 25< x^2+y^2<64 \},$ which is an annulus between two circles $C_1=\{(x,\;y)\colon \;\;\;  x^2+y^2= 64 \}$ and $C_2=\{(x,\;y)\colon \;\;\;  x^2+y^2= 25 \},$ is parametrization, proposed by   @Dolma.
 These parametrizations are 
$$C_1 =\{(8\cos{t},\;8\sin{t}), \;\; {0}\leqslant{t}\leqslant{2\pi} \} \\
C_2 =\{(5\cos{t},\;5\sin{t}), \;\; {0}\leqslant{t}\leqslant{2\pi} \}.
$$
Then 
$$\int\limits_{\partial{D^{+}}}{3xy^2 \, dx+2x^3\,dy}=\int\limits_{{C_1^+}\cup{C_2^-}}{3xy^2 \, dx+2x^3\,dy}= \\
=\int\limits_{0}^{2\pi}{\left(3(8\cos{t})(8\sin{t})^2(-8\sin{t}) +2(8\cos{t})^3(8 \cos{t}) \right)\,dt} -\\
-\int\limits_{0}^{2\pi}{\left(3(5 \cos{t})(5 \sin{t})^2(-5\sin{t}) +2(5\cos{t})^3(5 \cos{t}) \right)\,dt} .$$
Another way  uses the Green's theorem
$$\int\limits_{\partial{D^{+}}}{3x^2 \, dx+2x^3\,dy}=\iint\limits_{D}{\left(\dfrac{\partial{}}{\partial{x}}(2x^3)-\dfrac{\partial{}}{\partial{y}}(3xy^2)\right)}\,dx\,dy= \iint\limits_{D}{\left(6x^2-6xy\right)}\,dx\,dy.$$
Last integral can be evaluated by using polar coordinates 
$$\begin{cases}x=\rho\cos{\theta} \\
 y=\rho\sin{\theta} \end{cases}$$
$$
\iint\limits_{D}{\left(6x^2-6xy\right)}\,dx\,dy=6\int\limits_{5}^{8}\int\limits_{0}^{2\pi}{\rho^2 \left(\cos^2{\theta}-\cos{\theta}\sin{\theta}\right)}\rho\,d{\theta}\,d{\rho}=\\
=6\int\limits_{5}^{8}{\rho^3 \,d{\rho} }\int\limits_{0}^{2\pi}{}\left(\cos^2{\theta}-\cos{\theta}\sin{\theta}\right)\,d{\theta}
$$
