I am having some trouble with the following question:
Solve for the circulation in five different ways for the velocity field $$\bar v=(e^{-x^2}-yz)\hat i+(e^{-y^2}-xz+2x)\hat j+(e^{-z^2})\hat k$$ around the curve C, the intersection of $x^2+y^2+z^2=2$ and $z=x^2+y^2$
My thoughts/attempts:
First of all, the intersection seems to be when $z=-2$ or $z=1$ but I am assuming we could consider the $z=1$ part above the xy plane to get the real component.
I thought, Circulation $= \int_{C}\bar v \bullet \hat T = \int_{C}\bar v \bullet \frac{dR}{dS}dS=\int_{C} \bar v \bullet dR$ also , stokes thereom saying it is equal to $\iint_{Rxy} \nabla \times F \bullet \hat k$
So we have a sphere of radius $\sqrt{2}$ and it intersects the paraboloid at $z=1$ ie when $x^2+y^2=1$ so my first thought would be that I need to parametrize this curve some how.
I also calculated, Curl $\bar F$$= (x,-y,2)$
I dont think it is correct but if I had to try to parametrize I would say $r(t)=(cost,sint,1)$, $0 \le t \le 2\pi$, and so $r'(t)=(-sint,cost,0)$
then I could plug into to solve $F(r(t))$ and evaluate $\int_{c} F(r(t)) \bullet r'(t) dt$, etc.
But I am just not confident in my work, or the other methods etc, I am looking for help.
Thanks all
I guess I can count to five. Your parametrization of the curve $C$ is correct, so just evaluate the line integral directly.
You can also apply Stokes's Theorem in various ways, taking different surfaces (being careful about orientation) with the same boundary curve $C$. Assuming we orient $C$ counterclockwise as viewed from above,
• You can fill $C$ in with the disk $x^2+y^2\le 1$, $z=1$.
• You can consider the spherical surface above $C$ (oriented with unit normal pointing outward).
• You can consider the spherical surface below $C$ (oriented with unit normal pointing inward).
• You can consider the surface of the paraboloid below $C$ (oriented with unit normal pointing upward).
Good luck with all these calculations! :)
EDIT: Since you've asked, I'll do one of these to illustrate. We wish to find the flux of $\text{curl}\,\vec F$ outward across that portion $S$ of the sphere $x^2+y^2+z^2=2$ with $z\ge 1$. We parametrize in spherical coordinates: $$\vec r(u,v)=\sqrt2(\sin u\cos v,\sin u\sin v,\cos u), \quad 0\le u\le \pi/4, 0\le v\le 2\pi.$$ Then you compute the fundamental vector cross product $$\frac{\partial\vec r}{\partial u}\times\frac{\partial\vec r}{\partial v} = 2\sin u(\sin u\cos v,\sin u\sin v,\cos u).$$ Thus, \begin{align*} \iint_S \text{curl}\,\vec F\cdot\vec n dS &= \int_0^{\pi/4}\int_0^{2\pi} \vec F(\vec r(u,v))\cdot \left(\frac{\partial\vec r}{\partial u}\times\frac{\partial\vec r}{\partial v}\right)du\,dv \\ &= \int_0^{\pi/4}\int_0^{2\pi} (\sqrt2\sin u\cos v,-\sqrt2\sin u\sin v, 2)\cdot 2\sin u(\sin u\cos v,\sin u\sin v,\cos u)\,du\,dv \\ &= 2\int_0^{\pi/4}\int_0^{2\pi} \big(\sqrt2(\sin^3u)(\cos^2v-\sin^2v) + 2\cos u\sin u\big)du\,dv \\ &= 4 (2\pi) \int_0^{\pi/4}\cos u\sin u\,du = 8\pi\big(\frac14\big)=2\pi. \end{align*} (Note that $\int_0^{2\pi} (\cos^2v-\sin^2v)\,dv = 0$.)