Stokes' Theorem parameterization

[LCKurtz]

The gradient of F divided by the magnitude of the gradient.
Is at located at z=2?
Sorry, I feel like I know nothing now...
In class, we discussed about how to find this normal vector. We did:
<a,b,c>⋅<cost,sint,2>=0 (since the normal vector should be orthogonal to the surface)
a=sint
b=-cost
c=0
Therefore, the normal vector = <sint,-cost,0>.

You are apparently talking about a normal vector to the curve, not the normal vector to the enclosed surface.

Yes, the surface lies in the plane z=2. You shouldn't need any calculus to tell me what a unit normal to that plane is. Have you drawn a picture? And it is easy to write down a standard area element in that plane. I think Vela's suggestion for you to read some examples is a good one, and I have to be gone the rest of the afternoon.

 

[vela]

Isn't it a unit circle centered at (0,0,2)? So the vector perpendicular to it will either point upwards or downwards directly from the surface... Am I even thinking correctly?

Yes, that's right! To choose between the two vectors depends on the way you go around the contour. Have you heard of the right hand rule?

For now, just take it on faith the correct vector is the one that points upward. Can you finish evaluating the integral?

 

[reminiscent]

Yes, that's right! To choose between the two vectors depends on the way you go around the contour. Have you heard of the right hand rule?

For now, just take it on faith the correct vector is the one that points upward. Can you finish evaluating the integral?

So the unit normal vector would be <0,0,2>?
All I have to do is take the curl and multiply it with the normal vector using the dot product, but I am confused about the bounds now. Do I also have to make the curl = <-cost, -sint, 6> (plugging in what they gave me for the circle)? Then are the bounds just from t=0 to t=2pi?

 

[vela]

So the unit normal vector would be <0,0,2>?

No. "Unit" means the length of the vector is equal to 1.

All I have to do is take the curl and multiply it with the normal vector using the dot product, but I am confused about the bounds now. Do I also have to make the curl = <-cost, -sint, 6> (plugging in what they gave me for the circle)? Then are the bounds just from t=0 to t=2pi?

It's a surface integral, so you have to integrate over the area bounded by the curve.

 

[reminiscent]

No. "Unit" means the length of the vector is equal to 1.It's a surface integral, so you have to integrate over the area bounded by the curve.

So would it be <0,0,sqrt(2)>?
Do I have to use polar coordinates?

 

[LCKurtz]

So would it be <0,0,sqrt(2)>?
Do I have to use polar coordinates?

This is getting a bit silly. Do you really have to ask whether <0,0,sqrt(2)> is a unit vector?

And before you talk about polar coordinates or rectangular coordinates, let's see what your integrand is. In post #15 we called the $\vec V$ the curl of $\vec F$ and the integrand is $\vec V\cdot d\vec S$. What does that work out to?