1. ## Vector calculus question

I'm stuck on a question regarding work or perhaps better $\oint_{c} \vec{v} .d\vec{r}$
I know the vector $\vec{v} = -y\hat{i} + z\hat{j} + x\hat{k}$

I also know C is defined by ${z}^{2}+{x}^{2}=1$

Hence y = 0 and we're told the path starts at z = 0, x=1 then to x=0, z=1 back to the start (x=1, z=0)

I need to deduce the work or $\oint_{c} \vec{v} .d\vec{r}$ via stoke's law

- - - - - my attempt - - - -
Ok I keep coming out with zero for the answer, but im not certain it is correct as i'll explain in a minute:
$W = \iint_{s} (\nabla\times\vec{v}).d\vec{s}$
$d\vec{s} = (\vec{r}_x \times \vec{r}_y) dxdz$
However, for the matrix i keep obtaining zero.
i.e. here's the matrix code

$\left( \begin{array}{ccc} \frac{\partial x}{\partial x} & \frac{\partial y}{\partial x} & \frac{\partial z}{\partial x} \\ \frac{\partial x}{\partial z} & \frac{\partial y}{\partial z} & \frac{\partial z}{\partial z} \end{array} \right)$
Whereby there is no relationship between y with z and/or x. Resulting in N = 0 Hence W= 0

However my other method (which i need to know how to obtain the answer from stoke's rather than this) I obtained an answer of 2pi.
$W = \oint_{c} \vec{F}.d\vec{r} = \iint_{s} (\nabla \times \vec{v}).d\vec{s}$
Do dot product (as we're using f.dr here not stokes side) etc, notice y, dy = 0 xdz term left, use conversion of x = cos(u) z = sin(u)

Compute it all and i get w = pi.

- - - - - -- - -
I am not sure which one is correct, and why the other is wrong if one of them is correct, though my initial priority is to use stoke's theorem.

Any help is appreciated!

2. The curl is not a zero vector in this case. Have a look at (5) and (6) here: Curl -- from Wolfram MathWorld

3. The answer I'm getting for both the integrals is $\pm \pi$ by the way (depending on the orientation of the curve).

4. yes i agree with that answer but i am still having problems with the matrix, this is what i think:

i do not see how the j component is not zero surely here it is -[1x1 -(1)] I.e. the derivatives by chain rule equates to one

5. Can't get arrays to work with the TEX tags.

Using the second form we get $\text{Curl }\bold{v}=-\bold{i}-\bold{j}+\bold{k}$

6. Ok so $\vec{N} = \vec{r}_x \times \vec{r}_y = \vec{r} \times \vec{F} = -\hat{i}-\hat{j}+\hat{k}$ ?/

Though i thought
$w = \iint_{s} {\nabla \times \vec{F}).d\vec{s}$
So $\vec{r} \times \vec{F}$
Is as above, but then $d\vec{s}$ i though was:

$d\vec{s}= \vec{r}_x \times \vec{r}_y dxdz$

7. There must be a mistake here somewhere, but looking at my notes i can't really see it

8. $\text{Curl } \bold{v} = \bold{\nabla} \times \bold{v}$

Your mistake above is that the determinant isn't 3x3 to begin with. Also, both $\partial z / \partial x = 0$ and $\partial x / \partial z = 0$

9. I think I'm also wrong here. I don't think you can parametrize this surface using cartesian coordinates. Try switching directly to plane polar coordinates:

$\bold{r}=\bold{r}(r\cos\theta, 0, r\sin\theta)$

10. Okay I just worked through the problem using plane polar coordinates as the parametrization, and I get the right answer.

We have that $\text{Curl }\bold{v}=-\bold{i}-\bold{j}+\bold{k}$
and
$(\bold{r'}_r \times \bold{r'}_\theta)=-r\bold{j}$

11. Ok, so i get the following:
$w= \iint (\nabla \times \vec{v}) .d\vec{s}$
$\iint -1 dxdz = \iint -1 \frac{dx}{d\phi}\frac{dz}{dr} d\phi dr$
$x = rcos(\phi)$
$z = rsin(\phi)$

$\iint r{sin}^{2}(\phi) d\phi dr$
integration limits for r is 1 and 0, and 2pi and 0 for phi
though this gets pi/2 not pi

12. $\int_0^{2\pi}\int_0^1 (-1, -1, 1) \cdot (0, -r, 0) dr d\theta = \int_0^{2\pi}\int_0^1 r dr d\theta=\pi$