The function is $F(x,y,z) = x^2y^2z$. Curve C is the intersection of the surfaces $z=2-x^2-y^2$ and $z=\sqrt{(x^2+y^2)}$. Calculate $\oint_C F ds$.
Attempt:
Firstly, to define the term inside the integral, I first find the gradient of the function, which in my case is: $\operatorname{grad}\vec F= 2xy^2z \vec i + 2x^2yz \vec j + x^2y^2 \vec k$
Next I find the intersection, for which I get the following: $2-x^2-y^2=\sqrt{x^2+y^2}$
I realize that I need use parametrization, but can't figure out the expression.
Edit#1
Ok, so I tried the following parametrization $r^2=x^2+y^2$ which gives the following: $$r^2+r-2=0$$ $$r_1=1$$ $$r_2=-2$$
It is obvious that $r_1=1$ is the correct solution. For $x, y, z$ I get: $$x=r\cos t$$ $$y=r\sin t$$ $$z=r$$ $$0≤t≤2\pi$$
Their respective derivates are: $$dx=-r\sin tdt$$ $$dy=r\cos tdt$$ $$dz=0dt$$
Now I add all these into integration formula:
$$\int 2xy^2zdx+2x^2yzdy+x^2y^2dz = \int_0^{2\pi} (-2r^5\sin^3t\cos t+2r^5\sin t\cos^3t)dt$$
The final solution is $0$, and according to the theorem, a closed line curve always gives $0$ as a solution. Can anyone confirm if this is the correct solution?