• Support PF! Buy your school textbooks, materials and every day products Here!

Stokes' Theorem problem

  • Thread starter mathman44
  • Start date
  • #1
206
0

Homework Statement



Use stokes' theorem to find I = [tex]\int\int (\nabla x F) n dS[/tex] where D is the part of the sphere [tex]x^2 + y^2 + (z-2)^2 = 8[/tex] that lies above the xy plane, and

[tex]F=ycos(3xz^2)i + x^3e^[-yz]j - e^[zsinxy]k[/tex]


Attempt at solution:
I want to use the line integral [tex]\int F dr[/tex] to solve this.

I parametrize the boundary [tex]r(t) = (2cos\theta)i + (2sin\theta)j + 0k)[/tex] (the sphere on the xy plane)

Then F dotted with r'[t] is [tex]\int -4(sin\theta)^2 + (2cos\theta)^4 [/tex]

and 0 < theta < 2pi

And the answer that I get is 8pi. Is this correct? More specifically is my boundary curve legit, considering that the sphere "bulges" out above this boundary? :S
 
Last edited:

Answers and Replies

  • #2
206
0
I just realized that on the xy plane the radius would be 2, not sqrt(8). Nonetheless, is this otherwise okay?
 
  • #3
ideasrule
Homework Helper
2,266
0

Homework Statement



Use stokes' theorem to find I = [tex]\int\int (\nabla x F) n dS[/tex] where D is the part of the sphere [tex]x^2 + y^2 + (z-2)^2 = 8[/tex] that lies above the xy plane, and

[tex]F=ycos(3xz^2)i + x^3e^[-yz]j - e^[zsinxy]k[/tex]


Attempt at solution:
I want to use the line integral [tex]\int F dr[/tex] to solve this.

I parametrize the boundary [tex]r(t) = (\sqrt(8)cos\theta)i + (\sqrt(8)sin\theta)j + 0k)[/tex] (the sphere on the xy plane)

Then F dotted with r'[t] is [tex]64(cos\theta)^4 - 8sin\theta[/tex]
This is not what I got. Did you get dr/d(theta)=-2sin(theta)i + 2cos(theta)j?

More specifically is my boundary curve legit, considering that the sphere "bulges" out above this boundary? :S
That's not a problem. The boundary can curve in or out as much as it wants, but Stokes' theorem always holds true.
 
  • #4
407
0
Is the answer you got 0?
 
  • #5
ideasrule
Homework Helper
2,266
0
Is the answer you got 0?
Me? Yes, I got 0.
 
  • #6
699
5
Since stoke's theorem is an alt of green's, wouldn't it be easier to just compare the 6 partial derivatives? Because the integral will be 0 if it is conservative which occurs if they are all equal, correct?
 
  • #7
206
0
I changed the r[t] to [tex]2cos\theta,2sin\theta,0[/tex]. For the F dot r'[t] I got [tex]\int -4(sin\theta)^2 + (2cos\theta)^4[/tex] with theta from 0 to 2pi. Why is this not giving me zero :S ?
 
  • #8
ideasrule
Homework Helper
2,266
0
Since stoke's theorem is an alt of green's, wouldn't it be easier to just compare the 6 partial derivatives? Because the integral will be 0 if it is conservative which occurs if they are all equal, correct?
Try it. I don't think the field is conservative.
 
  • #9
ideasrule
Homework Helper
2,266
0
I changed the r[t] to [tex]2cos\theta,2sin\theta,0[/tex]. For the F dot r'[t] I got [tex]\int -4(sin\theta)^2 + (2cos\theta)^4[/tex] with theta from 0 to 2pi. Why is this not giving me zero :S ?
You've done everything right so far. If you don't get 0, it's just a mistake in your integration somewhere.
 
  • #10
206
0
I double checked on maple and 8pi seems correct, but I get zero when I directly compute the double integral without using Stokes'...
 
Last edited:
  • #11
gabbagabbahey
Homework Helper
Gold Member
5,002
6
I double checked on maple and 8pi seems correct, but I get zero when I directly compute the double integral without using Stokes'...
I get [itex]8\pi[/itex] using both methods, so I'm pretty sure you are just doing something wrong when you calculate the double integral directly.
 
  • #12
206
0
Can anyone confirm this? We have two homework helpers with two different answers!
 
  • #13
206
0
Shameless bump
 
  • #14
gabbagabbahey
Homework Helper
Gold Member
5,002
6
Can anyone confirm this? We have two homework helpers with two different answers!
I've asked in the homework helpers forum for another HH to take a look when they get time, so hopefully someone will reply.
 
  • #15
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,623
1,257
I get [itex]8\pi[/itex] for the line integral. Is there a trick to evaluating the double integral? I get a mess for curl F.
 
  • #16
407
0
When I found the curl I took z to equal 0, and so the curl ends up being 3x2-1.
I then did [tex]\int\int3x^{2}-1 dA = \int_{0}^{2\pi}\int_{0}^{2}(r^{3}cos^2 {\phi} - r) drd{\phi} [/tex]
this gives me zero. There is a similar problem in my book and this seems to be how it's done...
 
  • #17
ideasrule
Homework Helper
2,266
0
Are we all integrating [tex]
\int -4(sin\theta)^2 + (2cos\theta)^4
[/tex] from 0 to 2pi?
 
  • #18
407
0
I did my integral a different way.
 
  • #19
ideasrule
Homework Helper
2,266
0
Which way?
 
  • #20
407
0
When I found the curl I took z to equal 0, and so the curl ends up being 3x2-1.
I then did [tex]\int\int3x^{2}-1 dA = \int_{0}^{2\pi}\int_{0}^{2}(r^{3}cos^2 {\phi} - r) drd{\phi} [/tex]
this gives me zero. There is a similar problem in my book and this seems to be how it's done...
I took the curl on the cirlce in the xy plane, so z=0.
 
  • #21
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,623
1,257
That's not right. You have to integrate over the surface of the sphere, so z won't be zero. Also, you apparently set z=0 before you took the curl of F, which you shouldn't do either.
 
  • #22
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,623
1,257
Are we all integrating [tex]
\int -4(sin\theta)^2 + (2cos\theta)^4
[/tex] from 0 to 2pi?
Yeah, gabba, mmmboh, and I all get 8 pi for that integral.
 
  • #23
gabbagabbahey
Homework Helper
Gold Member
5,002
6
When I found the curl I took z to equal 0, and so the curl ends up being 3x2-1.
I then did [tex]\int\int3x^{2}-1 dA = \int_{0}^{2\pi}\int_{0}^{2}(r^{3}cos^2 {\phi} - r) drd{\phi} [/tex]
this gives me zero. There is a similar problem in my book and this seems to be how it's done...
Two things:

(1) You seem to be integrating over the flat disk that is bounded by the same circle that the OPs original surface is bounded by. This should give the same result, but is not really a direct computation of the original double integral

(2) What happened to the factor of 3? ....[itex](3x^2-1)dA=(3r^{3}cos^2 {\phi} - r) dr d\phi\neq (r^{3}cos^2 {\phi} - r) dr d\phi[/tex]
 
  • #24
gabbagabbahey
Homework Helper
Gold Member
5,002
6
Also, you apparently set z=0 before you took the curl of F, which you shouldn't do either.
By an odd coincidence, you actually get the same result either way.
 
  • #25
407
0
Ah you're right, I actually do have the factor 3 written down, but I lost it somewhere...I fixed my work and I get [tex]8\pi[/tex]
 

Related Threads on Stokes' Theorem problem

  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
5K
  • Last Post
Replies
3
Views
475
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
7
Views
928
  • Last Post
Replies
1
Views
718
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
3
Views
633
  • Last Post
Replies
4
Views
1K
Top