Finding the Circulation of a Curve in a Solid. (Vector Calculus)

A solid can in spherical coordinates \begin{equation} x=\rho\sin\phi\cos\theta\\ y=\rho\sin\phi\sin\theta\\ z=\rho\cos\phi \end{equation}

be described by the following inequalities $$0<\rho<1-\cos\phi$$

Let a curve $C$ be the intersection of the boundary surface of the solid with the plane $y=0$ and equip $C$ with an anticlockwise orientation as seen from the positive $y-axis$. Find the circulation

$$\int_{C}(z+e^x)\:dx+e^{x^3+z^3}\:dy+(\sin y-x)\:dz$$

What i tried

The solid is clearly a sphere. Using Stokes theorem i took the curl of the following vector field and i got $$\nabla\times F=(\cos y-3z^2e^{x^3+z^3},2,3x^2e^{x^3+z^3})$$

The normal $n=(0,1,0)$ since the curve lies on the $xz$ plane as $y=0$

Then i took the dot product of $\nabla\times F$ and $n$ to get a value of $2$, which means to say the integral is $2$ times the area of $C$ when projected in the $xz$ plane. Im unsure about my workings though. Could anyone please help me with this. Thanks

Solution 1:

Parametrization of the surface: $$ \eqalign{ x&=\rho\sin\phi\cr y&=0\cr z&=\rho\cos\phi }$$ $$0\le\phi\le\pi,\qquad 0\le\rho\le 1-\cos\phi$$ union $$ \eqalign{ x&=-\rho\sin\phi\cr y&=0\cr z&=\rho\cos\phi }$$ $$0\le\phi\le\pi,\qquad 0\le\rho\le 1-\cos\phi$$

Normal vector in the first chunk: $$N=(\sin\phi,0,\cos\phi)\times(\rho\cos\phi,0,-\rho\sin\phi)=(0,\rho,0)$$

Normal vector in the second chunk: the same.

Parametrization of the curve: $$ \eqalign{ x&=(1-\cos\phi)\sin\phi\cr y&=0\cr z&=(1-\cos\phi)\cos\phi }$$ $$0\le\phi\le\pi$$ union $$ \eqalign{ x&=(\cos\phi-1)\sin\phi\cr y&=0\cr z&=(1-\cos\phi)\cos\phi }$$ $$0\le\phi\le\pi$$ Warning: with this parametrization the orientation of the curve is reversed in the first chunk.