Help me with Calculus question

[JohnCU]

It's driving me crazy because I can't figure out how to get the right answer. I know the right answer: it is -2*pi (I asked the teacher after the exam). I really want to know how to get it, though.

Verify Stoke's Theorem where F(x,y,z) = <2y,-z,3> and S is the surface z = 4 - x^2 - y^2 that is in the cylinder x^2 + y^2 = 1, where S has upward orientation.

Okay, Stoke's Theorem says: The line integral of F dotted with dr over C should be the same as the double integral over S of curl F dotted with dS.

Using the line integral, I got -2*pi.
Using the double integral with the curl I keep getting -pi. Here's how I'm setting it up:

Parametrize S by the following: x = x, y = y, z = 4 - x^2 - y^2
r(x,y) = <x, y, 4-x^2-y^2>

Taking partials:
partial of r with respect to x: <1, 0, -2x>
partial of r with respect to y: <0, 1, -2y>

The cross product dr/dx x dr/dy = <2x, 2y, 1>
The curl of F is <1, 0, -1>

So, to evaluate (I'll use S for integral notation) the integral SS curl F dot dS you take the double integral over the region D and re-write the integral as SS curl F dot [(dr/dx)x(dr/dy)] dA which is SS <1, 0, -1> dot <2x, 2y, 1> dA which comes out to SS 2x-1 dA, where you switch to polar coordinates and get S(from 0 to 2*pi),S(0 to 1) (2*r*cosy-r)drdy, which evaluates to -pi.

Help me. Please.

:disgust:

 

[rayray2]

/head explodes

 

[Legendary]

Man last semester I would be able to help you, but Calc III got wiped from my mind - sorry dude.

 

[TuffGirl]

Originally posted by: rayray2
/head explodes

lol

 

[chuckywang]

Space reserved for solution.....trying to solve it now....

I think you did the curl wrong. The curl that I got is <1, 0, -2>. From this, n = <0,0,1> for the bottom of the surface. The sides of the surface cancels out in the integral due to symmetry. Therefore, integral over S for (curl)*n dS is just -2*Area(S) = -2*pi.

 

[BRObedoza]

you got the curl wrong

curl F = <1, 0 , -2>

dot this with <2x,2y, 1> and you get 2x-2

this becomes 2rcosy-2r in polar
integrate this over the double integral and i got -2pi


edit - what he said *point up*

 

[JohnCU]

ahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

all that agony over the CURL! the simplest operation ever! *shoots self in the foot*

thank you for the help.

 

[JustAnAverageGuy]

Must think how relates to real world. Can't.

Originally posted by: rayray2
/head explodes

 

[JohnCU]

Originally posted by: JustAnAverageGuy
Must think how relates to real world. Can't.

Originally posted by: rayray2
/head explodes

Without Stoke's theorem you wouldn't have electricity.

 

[chuckywang]

.whoops

 

[chuckywang]

Originally posted by: chuckywang

Originally posted by: JohnCU

all that agony over the CURL! the simplest operation ever! *shoots self in the foot*

thank you for the help.

You'll forgive me if I don't quote your "ahhhhh..."

Actually, what sleepmachine did was more accurate because the problem asked you to verify Stokes's Theorem. What I did was apply Stokes's theorem to solve an easier surface integral.

http://omega.albany.edu:8008/c...theorem-dir/lec10.html

i.e. I did method #3 from that page, and sleepmachine did method #1. The problem asked you to do method #1.

I do have one question though...does the surface include the bottom circle? I don't think it does...

 

[JohnCU]

Originally posted by: chuckywang

Originally posted by: JohnCU

all that agony over the CURL! the simplest operation ever! *shoots self in the foot*

thank you for the help.

You'll forgive me if I don't quote your "ahhhhh..."

Actually, what sleepmachine did was more accurate because the problem asked you to verify Stokes's Theorem. What I did was apply Stokes's theorem to solve an easier surface integral.

http://omega.albany.edu:8008/c...theorem-dir/lec10.html

i.e. I did #3 from that page, and sleepmachine did #1.

I do have one question though...does the surface include the bottom circle? I don't think it does...

It can be any surface that has the boundary C, which is the beauty of Stoke's theorem.

 

[chuckywang]

Originally posted by: JohnCU

Originally posted by: chuckywang

Originally posted by: JohnCU

all that agony over the CURL! the simplest operation ever! *shoots self in the foot*

thank you for the help.

You'll forgive me if I don't quote your "ahhhhh..."

Actually, what sleepmachine did was more accurate because the problem asked you to verify Stokes's Theorem. What I did was apply Stokes's theorem to solve an easier surface integral.

http://omega.albany.edu:8008/c...theorem-dir/lec10.html

i.e. I did #3 from that page, and sleepmachine did #1.

I do have one question though...does the surface include the bottom circle? I don't think it does...

It can be any surface that has the boundary C, which is the beauty of Stoke's theorem.

What about a closed surface? Then there is no boundary C. Can you not apply Stoke's Theorem in that case, or would the integrals just be 0?

 

[JohnCU]

Originally posted by: chuckywang
What about a closed surface? Then there is no boundary C. Can you not apply Stoke's Theorem in that case, or would the integrals just be 0?

If you're talking about a solid region, you'd use the Divergence Theorem and set up a triple integral I think.

 

[Yossarian]

test's over. you should be working on forgetting that stuff as soon as possible, not studying more.

 

[chuckywang]

Originally posted by: JohnCU

Originally posted by: chuckywang
What about a closed surface? Then there is no boundary C. Can you not apply Stoke's Theorem in that case, or would the integrals just be 0?

If you're talking about a solid region, you'd use the Divergence Theorem and set up a triple integral I think.

Ah, it's all coming back to me now.

If you really want to get to know the beauty of Stokes's Theorem, then try to learn the Generalized Stokes's Theorem, which gets into differential forms.

Stokes's Theorem - 2d Surface bounded by 1d Curve
Divergence Theorem - 3d Solid bounded by 2d Surface
Generalized Stokes's Theorem - N-d Solid bounded by (N-1)-d Solid


The fundamental Theorem of Calculus has never been more generally expressed!

 

[JohnCU]

Originally posted by: Yossarian
test's over. you should be working on forgetting that stuff as soon as possible, not studying more.

I'm an electrical engineering major. I'll use this stuff later for electric flux or something.

 

[BRObedoza]

it'll probably come back on finals as well. i'm pretty sure you won't be getting the curl wrong on the final now haha

 

[JohnCU]

Originally posted by: chuckywang

Originally posted by: JohnCU

Originally posted by: chuckywang
What about a closed surface? Then there is no boundary C. Can you not apply Stoke's Theorem in that case, or would the integrals just be 0?

If you're talking about a solid region, you'd use the Divergence Theorem and set up a triple integral I think.

Ah, it's all coming back to me now.

If you really want to get to know the beauty of Stokes's Theorem, then try to learn the Generalized Stokes's Theorem, which gets into differential forms.

Stokes's Theorem - 2d Surface bounded by 1d Curve
Divergence Theorem - 3d Solid bounded by 2d Surface
Generalized Stokes's Theorem - N-d Solid bounded by (N-1)-d Solid


The fundamental Theorem of Calculus has never been more generally expressed!

Yeah, it's really very interesting. Thanks, I'ma check it out!
<-- Geek. :thumbsup:

 

[chuckywang]

Originally posted by: JohnCU

Originally posted by: chuckywang

Originally posted by: JohnCU

Originally posted by: chuckywang
What about a closed surface? Then there is no boundary C. Can you not apply Stoke's Theorem in that case, or would the integrals just be 0?

If you're talking about a solid region, you'd use the Divergence Theorem and set up a triple integral I think.

Ah, it's all coming back to me now.

If you really want to get to know the beauty of Stokes's Theorem, then try to learn the Generalized Stokes's Theorem, which gets into differential forms.

Stokes's Theorem - 2d Surface bounded by 1d Curve
Divergence Theorem - 3d Solid bounded by 2d Surface
Generalized Stokes's Theorem - N-d Solid bounded by (N-1)-d Solid


The fundamental Theorem of Calculus has never been more generally expressed!

Yeah, it's really very interesting. Thanks, I'ma check it out!
<-- Geek. :thumbsup:

<--uber-geek Linky!

 

[TuxDave]

If you asked me 5 years ago, I probably could've whipped out an answer in no time.... college has made me retarded.

 

[chuckywang]

Don't worry...I think ATOT has solved this problem, and then some!

 

[chuckywang]

Originally posted by: sleepmachine
you got the curl wrong

curl F = <1, 0 , -2>

dot this with <2x,2y, 1> and you get 2x-2

this becomes 2rcosy-2r in polar
integrate this over the double integral and i got -2pi


edit - what he said *point up*

beating a dead horse here, but isn't that 2r^2*cos(theta)-2r?

 

[JohnCU]

Originally posted by: chuckywang

Originally posted by: sleepmachine
you got the curl wrong

curl F = <1, 0 , -2>

dot this with <2x,2y, 1> and you get 2x-2

this becomes 2rcosy-2r in polar
integrate this over the double integral and i got -2pi


edit - what he said *point up*

beating a dead horse here, but isn't that 2r^2*cos(theta)-2r?

Yup, the Jacobian is r when you replace dA with drd(theta), so you have to multiply by r, and x is replaced with rcos(theta), so two r's make an r^2

 

[BRObedoza]

Originally posted by: JohnCU

Originally posted by: chuckywang

Originally posted by: sleepmachine
you got the curl wrong

curl F = <1, 0 , -2>

dot this with <2x,2y, 1> and you get 2x-2

this becomes 2rcosy-2r in polar
integrate this over the double integral and i got -2pi


edit - what he said *point up*

beating a dead horse here, but isn't that 2r^2*cos(theta)-2r?

Yup, the Jacobian is r when you replace dA with drd(theta), so you have to multiply by r, and x is replaced with rcos(theta), so two r's make an r^2

you guys are right. my mistake. i used y instead of theta since that's what the OP used. missed the squared but the answer's still the same. that probably would have snuck by the grader