Stokes' theorem question

**Poison** · June 1st 2010, 09:06 PM

I have the following homework problem:

In part, we compute both integrals from Stokes' theorem for the helicoid ${\Psi(r,\theta) = (r\cos \theta, r\sin \theta, \theta)}$ where ${(r,\theta)}$ lies in the rectangle ${[0,1] \times [0,\pi/2]}$ , and ${\mathbf{F}}$ is the vector field ${\mathbf{F} = (5 z, 3 x, 7 y)}$ .

First, compute the surface integral:
${\iint_M (\nabla \times \mathbf{F})\cdot d\mathbf{S}= \int_a^b\int_c^d f(r, \theta) dr\,d\theta}$ , where
${a = }$ _, ${b = }$ _, ${c = }$ _, ${d =}$ _, and
${f(r, \theta) = }$ ____ (use "t" for theta).
Finally, the value of the surface integral is ____.

Next compute the line integral on that part of the boundary from ${(1,0,0)}$ to ${(0,1,\pi/2)}$ .
${\int_C \mathbf{F}\cdot d\mathbf{s} = \int_a^b g(\theta)\,d\theta}$ , where
${a =}$ _, ${b =}$ _,
and ${g(\theta) =}$ ____ (use "t" for theta).

----

So far I have a=0, b=pi/2, c=0, d=1, and those are right. But, I'm stuck at ${f(r, \theta) = }$ ____. I tried finding:

${\Psi_r = (\cos \theta, \sin \theta, 0)}$
${\Psi_\theta = (-r\sin \theta, r\cos \theta, 1)}$
${\Psi_r \times \Psi_\theta = (\sin \theta, - \cos \theta, r)}$

Then I plugged ${\Psi}$ into ${\mathbf{F} = (5 z, 3 x, 7 y)}$ to get:
$(5 \theta, 3r \cos \theta, 7r \sin \theta)$

I dotted that with the cross product and found:
$5 \theta \sin \theta - 3r \cos^2 \theta + 7 r^2 \sin \theta$

Which is what I thought ${f(r, \theta) = }$ should be.. but my webwork says that it is incorrect. Can anyone find where I am going wrong?

**HallsofIvy** · June 2nd 2010, 02:56 AM

[QUOTE=Poison;522315]I have the following homework problem:

In part, we compute both integrals from Stokes' theorem for the helicoid ${\Psi(r,\theta) = (r\cos \theta, r\sin \theta, \theta)}$ where ${(r,\theta)}$ lies in the rectangle ${[0,1] \times [0,\pi/2]}$ , and ${\mathbf{F}}$ is the vector field ${\mathbf{F} = (5 z, 3 x, 7 y)}$ .

First, compute the surface integral:
${\iint_M (\nabla \times \mathbf{F})\cdot d\mathbf{S}= \int_a^b\int_c^d f(r, \theta) dr\,d\theta}$ , where
${a = }$ _, ${b = }$ _, ${c = }$ _, ${d =}$ _, and
${f(r, \theta) = }$ ____ (use "t" for theta).
Finally, the value of the surface integral is ____.

Next compute the line integral on that part of the boundary from ${(1,0,0)}$ to ${(0,1,\pi/2)}$ .
${\int_C \mathbf{F}\cdot d\mathbf{s} = \int_a^b g(\theta)\,d\theta}$ , where
${a =}$ _, ${b =}$ _,
and ${g(\theta) =}$ ____ (use "t" for theta).

----

So far I have a=0, b=pi/2, c=0, d=1, and those are right. But, I'm stuck at ${f(r, \theta) = }$ ____. I tried finding:

${\Psi_r = (\cos \theta, \sin \theta, 0)}$
${\Psi_\theta = (-r\sin \theta, r\cos \theta, 1)}$
${\Psi_r \times \Psi_\theta = (\sin \theta, - \cos \theta, r)}$ [quote]
This is wrong. $\Psi_r \times \Psi_\theta = \left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ \cos(\theta) & \sin(\theta) & 1 \\ -sin(\theta) & cos(\theta) & 0 \end{array}\right|= -cos(\theta)\vec{i}- sin(\theta)\vec{j}+ r\vec{k}$

Then I plugged ${\Psi}$ into ${\mathbf{F} = (5 z, 3 x, 7 y)}$ to get:
$(5 \theta, 3r \cos \theta, 7r \sin \theta)$

I dotted that with the cross product and found:
$5 \theta \sin \theta - 3r \cos^2 \theta + 7 r^2 \sin \theta$

Which is what I thought ${f(r, \theta) = }$ should be.. but my webwork says that it is incorrect. Can anyone find where I am going wrong?

**Poison** · June 2nd 2010, 04:56 AM

Your Latex code isn't showing up correctly for me, but I'm assuming that you crossed:
$(cos \theta, sin\theta, 1) \times (-sin \theta, cos\theta, 0)$

To get:
$= -cos(\theta)\vec{i}- sin(\theta)\vec{j}+ r\vec{k}$

But im not sure where the two vectors you crossed came from, or why neither of them has r but your answer does. I think it may be a problem with me trying to read the latex code.

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