How to use Stokes theorem? Use Stokes theorem to evaluate $\int_{C} [ydx+y^{2}dy+(x+2z)dz]$ where $C$ is the curve of intersection of the sphere $x^{2}+y^{2}+z^{2}=a^{2}$ and the plane $y+z=a$ oriented counterclockwise as viewed from above.
I can't seem to understand the examples in the book. Can someone explain to me what this means and how to apply Stokes formula?
EDIT:
I still dont understand what exactly hes talking about i have figured out wha ti am supposed to take the curl of at least (now that i know what thing the curl is suppsoed to be taken on).
Curl F =  (0)dydz-(-1)dxdz+(-1)dxdy
from wha ti have been able to decifer i think n is supposed to be the vertor normal to the 2 sufaces  $x^{2}+y^{2}+z^{2}=a^{2}$ and $y+z=a$ in which case
$n=\begin{bmatrix} e_{1}&e_{2}&e_{3} \\ 0&1&1 \\ 2x&2y&2z\\ \end{bmatrix}$
$n=(2z-2y)i+(2x)j-(2z)k$
thus $\int\int_{S}2xdxdz+2xdxdy$
$=(x^{2}z+x^{2}y)||_{S}$ which again makes no sense
when i calculate K below i get 1-1=0 which also makes no sense 

We can rewrite out integral as
$$\oint_{C} P dx + Q dy + R dz$$
Where $P = y, Q = y^2, R = x + 2z$. We now compute
$$\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} = 0 - 0 = 0$$
$$\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} = 0 - 1 = -1$$
$$\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0 - 1 = -1$$
Stokes theorem tells us that 
$$\oint_{C} P dx + Q dy + R dz = \\\int\int_{\mbox{interior of $C$}} \left(\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}\right) dydz - \left(\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}\right) dxdz+ \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dxdy \\ = \int\int_{\mbox{interior of $C$}} dxdz - dxdy \\ = \int\int_{\mbox{interior of $C$}}(0,1,-1) \cdot d\vec{A}$$
What now remains is to compute $d\vec{A} = \vec{n}dA$, where $\vec{n}$ is the normal vector to the interior of $C$, chosen with correct orientation (right hand rule). You can find this by computing the normal of the plane $y+z = a$. Then take the dot product with $(0,1,-1)$ to get some constant $K$, and you will have 
$$ K\int\int_{\mbox{interior of $C$}} dA = K \times \mbox{Area of the interior of $C$, which is a disc}$$
Do you know how to proceed from here?
