Verify Stokes's theorem with the given surface and vector field

[jheld]

Homework Statement

verify Stokes's theorem for the given surface and vector field.
S is defined by x^2 + y^2 + z^2 = 4, z <= 4, oriented by downward normal;

F = (2y-z, x + y^2 - z, 4y - 3x)

Homework Equations

double integral over S of the curl F ds = integral over S' of F ds.

The Attempt at a Solution

I calculated the curl F= del operator cross-product F = 5i - 2j - k, feel free to check if you wish.

But in my vector calc book for a similar example, I noticed that they made an upward-pointing normal vector (as it was oriented upward) after del x F. I though del x F was a normal vector--am I wrong? The normal vector they came up with after the del x F (a separate calculation for my knowledge) was very different.

Any hints from where to go from there?

 

[nickmai123]

Homework Statement

verify Stokes's theorem for the given surface and vector field.
S is defined by x^2 + y^2 + z^2 = 4, z <= 4, oriented by downward normal;

F = (2y-z, x + y^2 - z, 4y - 3x)

Homework Equations

double integral over S of the curl F ds = integral over S' of F ds.

The Attempt at a Solution

I calculated the curl F= del operator cross-product F = 5i - 2j - k, feel free to check if you wish.

But in my vector calc book for a similar example, I noticed that they made an upward-pointing normal vector (as it was oriented upward) after del x F. I though del x F was a normal vector--am I wrong? The normal vector they came up with after the del x F (a separate calculation for my knowledge) was very different.

Any hints from where to go from there?

Check your curl again. Maple gave me a different answer by only a sign difference.

 

[jheld]

Yeah, that's been confusing me lately. I took elem. linear algebra, so I know how to calculate a determinant, so this is kind of ridiculous. I'll post my curl...

curl F = [d/dy(4y-3x)-d/dz(x+y^2-z)]i - [d/dx(4y-3x)-d/dz(2y-z)]j + [d/dx(x+y^2 -z)-d/dy(2y-z)]k. And from there I got = [4-(-1)]i -[-3-(-1)]j + [1-(2)]k = 5i+2j-k.
If you mean a sign change, do you mean 5i-2j-k?

 

[nickmai123]

Yeah, that's been confusing me lately. I took elem. linear algebra, so I know how to calculate a determinant, so this is kind of ridiculous. I'll post my curl...

curl F = [d/dy(4y-3x)-d/dz(x+y^2-z)]i - [d/dx(4y-3x)-d/dz(2y-z)]j + [d/dx(x+y^2 -z)-d/dy(2y-z)]k. And from there I got = [4-(-1)]i -[-3-(-1)]j + [1-(2)]k = 5i+2j-k.
If you mean a sign change, do you mean 5i-2j-k?

Well what you just showed me is correct. I think you just wrote the wrong thing up there then. B/c it is 5i + 2j -k.

 

[nickmai123]

S is defined by x^2 + y^2 + z^2 = 4, z <= 4, oriented by downward normal;

Are the bounds for S correct? Somehow I see that as the entire sphere itself, which I thought meant you can't use Stokes' Theorem.

 

[jheld]

Oh ok. Yeah, I probably didn't press the shift key. Do you know where I should go from here?

 

[nickmai123]

Did you get my private message?

 

[jheld]

just did. sry bounds are x^2 + y^2 + z^2 = 4 and z<= 0, just a typo.

 

[nickmai123]

just did. sry bounds are x^2 + y^2 + z^2 = 4 and z<= 0, just a typo.

that makes a lot more sense now, lol. Okay so ignore the PM haha.

So you want to take the flux integral instead of the line integral right? You could probably find it online, but here's the formula for a flux integral.

\int_{S}\vec{F}\cdot d\vec{A}=\int_{R}\vec{F}(x,y,f(x,y))\cdot (-f_{x}\vec{i} - f_{y}\vec{j} + \vec{k})dxdy

 

[nickmai123]

I reread your question, and I thought I'd mention that given what you're given, verify Stokes' Theorem here probably means taking the line integral instead of the curl flux integral. It looks like it would be a lot easier than the flux integral.

EDIT: Nevermind haha. That line integral would be pretty hairy.

 

[jheld]

alright, understandable. I saw that equation earlier, but didn't think I needed it.
In this case z = sqrt(4-x^2-y^2), but are x and y just as they are... for F(x,y,f(x,y))?
or should I parametrize x,y and z?

 

[jheld]

oh, ok. yeah, I've tried with the parametrization, and it doesn't seem that I can solve it by myself.

 

[nickmai123]

alright, understandable. I saw that equation earlier, but didn't think I needed it.
In this case z = sqrt(4-x^2-y^2), but are x and y just as they are... for F(x,y,f(x,y))?
or should I parametrize x,y and z?

They're just as they are. Plug in \sqrt(4-x^{2}-y^{2}) for z in the function. Do keep in mind that this is just one way to evaluate that flux integral. You *could* change to spherical coordinates and integrate. If you know how to parameterize that surface then you can always do that. Point is, see what each give you. Go back to the line integral and see if that's easier. Stokes' Theorem exists to simplify not complicate.

 

[nickmai123]

Also, any other method you use to evaluate that flux integral would require a "different formula." I put that in quotes b/c all of the formulas are related to each other.

 

[jheld]

Well, I plugged in everything and made the dot product...
Here is what I came up with (thus a little stuck again):
(2xy(4-x^2-y^2)^(-1/2)-1-2xy+xy(4-x^2-y^2)^(-1/2)-2y^3+y^3(4-x^2-y^2)^(-1/2)+3y-3x)dxdy

I was thinking of making u = 4-x^2-y^2, but that doesn't totally solve my problem.

 

[nickmai123]

Well, I plugged in everything and made the dot product...
Here is what I came up with (thus a little stuck again):
(2xy(4-x^2-y^2)^(-1/2)-1-2xy+xy(4-x^2-y^2)^(-1/2)-2y^3+y^3(4-x^2-y^2)^(-1/2)+3y-3x)dxdy

I was thinking of making u = 4-x^2-y^2, but that doesn't totally solve my problem.

Well that's not right. Remember that you're taking the flux of (\nabla \times \vec{F})\cdot d\vec{A}. That means you should just have:

\int\int (-5f_{x}-2f_{y}-1)dxdy

 

[jheld]

okay...but what about F(x,y,f(x,y))dot(-f_x,-f_y,k)?

 

[nickmai123]

However, that's still one very disgusting integral. I think what they meant for you to do was to take the line integral. The line itself would be a closed circle where z=0. Hence, your curve should reduce to 4=x^{2}+y^{2}.

 

[jheld]

yeah, which is what i did originally. from there i thought about reparametrizing the function in cylindrical coordinates, where z = 0. x = 2cos(t), y = 2sin(t), 0<=t<=2pi. but you're saying that if I do the aformented -5f_x, -2f_y, -1 da, that will work as well? seems rather simple :)

 

[nickmai123]

yeah, which is what i did originally. from there i thought about reparametrizing the function in cylindrical coordinates, where z = 0. x = 2cos(t), y = 2sin(t), 0<=t<=2pi. but you're saying that if I do the aformented -5f_x, -2f_y, -1 da, that will work as well? seems rather simple :)

Looks can be deceiving, lol. But yeah that integral's fairly simple. I mean it's still not exactly a nice problem, but there are no tricks involved as far as the rest go. Just pure integration.

I'm going to sleep, lol. I'll be up tomorrow if you need more help then.