# Homework Help: Verify Stokes' Theorem for F across a paraboloid

1. May 6, 2012

### jerzey101

1. The problem statement, all variables and given/known data
Verify Stokes' Theorem for F(x,y,z)=(3y,4z,-6x) where S is part of the paraboloid z=9-x2-y2 that lies above the xy-plane, oriented upward.

2. Relevant equations
Stokes' Theorem is ∫F*ds=∫∫scurl(F)*ds
Where curl(F)=∇*F

3. The attempt at a solution
I got curl(F)=(-4,6,-3)
then I'm not sure what to do. my notes say to parameterize s and then compute the normal vector n, then compute ∫∫curl(F)*n. I'm not sure how to parameterize z=9-x2-y2.

2. May 6, 2012

### sharks

Since $\hat n$ is the outward normal unit vector, directed upwards, above the plane z=0, the $\vec k$ vector component must be positive. Hence: $\phi (x,y,z)=x^2+y^2+z-9$

Now, to find $\hat n$, just use the formula:
$$\frac{∇\vec \phi}{|∇\vec \phi|}$$
You have found $∇\times \vec F=-4\hat i+6\hat j-3\hat k$
Now, all you have to do, is just project the surface S onto the x-y plane and find the area.

3. May 6, 2012

### jerzey101

So (2x,2y,1) / sqrt(4x2+4y2+1)

how do you go about simplifying this?

4. May 6, 2012

### sharks

You can't simplify the square root at this point. Just proceed by multiplying with Curl F and find the surface area.

5. May 6, 2012

### jerzey101

so -8x + 12y - 3 all over [itex]\sqrt{4x2+4y2+1}[\itex]
I'm sorry, I am so lost with this one.

And I am doing battle with the formatting. I don't get why it's not working

Last edited: May 6, 2012
6. May 6, 2012

### sharks

To close the itex tag, always use backslash \ instead of /

Your expression is correct. Now, project the surface to find the surface area.

7. May 6, 2012

### jerzey101

Meaning integrate it over the paraboloid?

8. May 6, 2012

### sharks

Read the last line of post #2.

9. May 6, 2012

### LCKurtz

When you are calculating a flux integral over a surface $S$$$\iint_S \vec F\cdot d\vec S=\iint_S\vec F \cdot \hat n dS$$you parameterize the surface in convenient variables $u$ and $v$ (which may be just $x$ and $y$)$$\vec R = \vec R(u,v)$$The normal to the surface is $\pm \vec R_u\times \vec R_v$, with the sign chosen to agree with the orientation. Then to get a unit normal you have$$\hat n =\pm \frac{\vec R_u\times \vec R_v}{|\vec R_u\times\vec R_v|}$$The scalar $dS\$is given by$$dS=|\vec R_u\times \vec R_v|dudv$$When you put these in the formula for the flux integral you get$$\iint_S \vec F\cdot d\vec S=\iint_S\vec F \cdot \hat n dS= \pm\iint_{(u,v)}\vec F\cdot \frac{\vec R_u\times \vec R_v}{|\vec R_u\times\vec R_v|} |\vec R_u\times \vec R_v|dudv =\pm \iint_{(u,v)}\vec F\cdot \vec R_u\times \vec R_vdudv$$where the limits are for the region in the $u-v$ plane. Note that you never have to calculate $|\vec R_u\times\vec R_v|$. In your problem what he means by parameterizing the surface is writing it as $\vec R(x,y) = \langle x,y,9-x^2-y^2\rangle$. If you calculate $\vec R_x\times \vec R_y$ you will get $\langle 2x,2y,1\rangle$, which happens to have $z$ positive so agrees with your orientation, and agrees with your gradient. You have calculated $\nabla \times \vec F = \langle -4,6,-3\rangle$, which is the $\vec F$ to put in the formula above:$$\iint_S \langle -4,6,-3\rangle\cdot d\vec S = +\iint_{(x,y)}\langle -4,6,-3\rangle\cdot \langle 2x,2y,1\rangle dydx=\iint_{(x,y)}-8x+12y-3\ dydx$$where the $x-y$ limits are over the region in the $xy\$plane. That is where $z=0$ and gives the region enclosed by $x^2+y^2=9$. At this point, you might want to work the integral in polar coordinates.

Then, of course, you still have to work the other side of the Stokes Theorem equation and see you get the same answer.

10. May 6, 2012

### jerzey101

Thanks LCKurtz, that is similar to what we were doing in my class. I was getting confused by what sharks was doing.

11. May 6, 2012

### sharks

If you had continued from what my suggestion, you would have ended up with the same double integral:
$$\iint_{(x,y)}-8x+12y-3\ dydx$$
But it's indeed easier for you to understand if you can relate directly to what you're doing in class.

12. May 7, 2012

### jerzey101

Ok I'm starting to get it. Thank sharks!