Green's theorem and intersection of plane with sphere

Question

Evaluate $$\int _{C}ydx+\left ( 2x-z \right )dy+\left ( z-x \right )dz$$ where C is the intersection of the sphere$$ x^{2}+y^{2}+z^{2}=4$$ and the plane $$z=1$$

Here's what I have thus far.

The intersection of the plane with the sphere produces $$x^{2}+y^{2}-3=0$$

Note that $$r\in[0,\sqrt{3}]$$

Observe that $M=y$ , $N=(2x-z)$ and $P=(z-x)$

How should I go further?

Answer 1

The intersection produces $x^2+y^2=3$ which is a circle on a plane $z=1$ with center at origin and radius $\sqrt{3}$ unit.

So applying Green's Theorem on a plane, and using $z=1 \Rightarrow z=$ constant $\Rightarrow dz = 0$
$$\oint_C ydx+(2x-z)dy+(z-x)dz$$ $$=\oint_C ydx+(2x-1)dy+(1-x)\cdot 0$$ $$=\oint_C ydx+(2x-1)dy$$ $$=\iint\left(\frac{\partial (2x-1)}{\partial x }-\frac{\partial y}{\partial y }\right) dx dy$$ $$=\iint (2-1) dx dy$$ $$=\iint dx dy$$ Using polar co-ordinates, we have $$=\int \int dr r d\theta$$ $$=\int_{0}^{\sqrt{3}} r dr \int_{0}^{2\pi} d\theta$$ $$=\frac{3}{2}\cdot 2\pi$$ $$=3\pi$$