After trying a couple of times, but failing to find a way to solve these problems, I decided I should perhaps ask the people on this forum for help.

Problem 1

Let $C$ be the curve $(x-1)^2+y^2=16$, $4x+3y+z=2$ oriented clockwise when viewed from high on the z-axis. Let $\vec{F} = \langle4z^2+y^2+sin(x^2), 2xy+3z, 2xz +yz\rangle$. Evaluate $\oint_C \vec{F} \cdot d\vec{r}$.

I tried solving this using Stoke's Theorem, but got stuck. I got $curl\vec{F}=\langle z-3, 6z, 0\rangle$ and the normal unit vector $\hat{N} = \langle 4, 3, 1 \rangle \cdot \frac{1}{\sqrt{26}}$. I got stuck on the boundaries while trying to evaluate the double integral, perhaps there is something obvious that I'm missing.

Problem 2

Calculate the flux of the curl of the field $\vec{F} = \langle 5z, x, y \rangle $ across the surface $S: \vec{r}(r, \theta) = \langle r cos\theta, rsin\theta, 16-r^2 \rangle$, $0 \leq r \leq 4$, $0 \leq 2\pi$ in the direction of the outward unit normal $\hat{N}$.

I also tried using Stoke's Theorem on this problem, but got stuck again. I tried converting $\vec{r}$ to cartesian coordinates in order to try to solve it, but got a whole load of gibberish.

So now I am stuck, I tried a couple of other things, but didn't really get to a concrete solution. Any help would be very much appreciated!


By Stokes theorem

$\iint \nabla \times F \ dS = \oint F\cdot dr$

And evaluate F around the circle in the xy plane.

$\int_0^{2\pi} (0, 4\cos\theta, 4\sin\theta )\cdot (-4\sin\theta, 4\cos\theta, 0)\ d\theta = \int_0^{2\pi} 16\cos^2\theta\ d\theta$

Tackling it directly

$\nabla \times F = (1,5,1)$

$dS = (\frac {\partial x}{\partial r},\frac {\partial y}{\partial r},\frac {\partial z}{\partial r})\times(\frac {\partial x}{\partial \theta},\frac {\partial y}{\partial \theta},\frac {\partial z}{\partial \theta})\\ (\cos\theta,\sin\theta, -2r)\times (-r\sin\theta,r\cos\theta, 0) = (2r^2\cos\theta,2r^2\sin\theta,r)$

$\int_0^{2\pi}\int_0^4 2r^2\cos\theta + 10r^2\sin\theta +r \ dr\ d\theta$

Evaluate by $\theta$ first, and the first two terms drop away.

$2\pi\int_0^4 r \ dr$

And while I am thinking about this. You could also evaluate this flux over the disk in the xy plane. Since it has the same contour as the hemisphere, and we are considering the curl of a field, you will get the same result.

  • $\begingroup$ Right, but I got something a bit different when calculating $dS$. I got that $dS=\langle cos\theta, sin\theta, 16-2r \rangle \times \langle -rsin\theta, rcos\theta, 0 \rangle$. After calculating the cross product, I got that $dS = \langle 16-2r-rcos\theta, 16-2r-rsin\theta, r \rangle$. I might have made a mistake, but I don't really understand how you got a $\sqrt{16-r^2}$ in the numerator $\endgroup$ – mrMoonpenguin Mar 31 at 13:39
  • $\begingroup$ I see my probelm, I used $z = \sqrt {16-r^2}$ however if $z = 16-r^2$, then $\frac {\partial z}{\partial r} = 2r$ $\endgroup$ – Doug M Mar 31 at 18:05
  • $\begingroup$ sorry, $\frac {\partial z}{\partial r} = -2r$ $\endgroup$ – Doug M Mar 31 at 18:12

Hint for the surface integral in (1): parametrize the surface using (almost) cylindrical coordinates: $$x = 1 + r\cos\theta,$$ $$y = r\sin\theta$$ $$z = \cdots$$ $$(r,\theta)\in\cdots$$

For (2), edit and post your calculations.


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