Stokes theorem!

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Homework Statement



Use stokes theorem to find double integral curlF.dS where S is the part of the sphere x2+y2+z2=5 that lies above plane z=1.
F(x,y,z)=x2yzi+yz2j+z3exyk

Homework Equations



stokes theorem says double integral of curlF.dS = [itex]\int[/itex]C F.dr

The Attempt at a Solution



boundary curve C is circle given by x2+y2=5, z=1.

Vector equation of c then is r(t)=[itex]\sqrt{5}[/itex]costi+[itex]\sqrt{5}[/itex]sintj+1k where 0<t<2pi

then r'(t)= -[itex]\sqrt{5}[/itex]sinti+[itex]\sqrt{5}[/itex]costj

[itex]\int[/itex]CF.dr = [itex]\int[/itex]F(r(t))dotr'(t).dt

F(r(t))=([itex]\sqrt{5}[/itex]cost)2([itex]\sqrt{5}[/itex]sint)i+([itex]\sqrt{5}[/itex]sint)j+e[itex]\sqrt{5}[/itex]cost[itex]\sqrt{5}[/itex]sint

then F(r(t))dotr'(t) = [([itex]\sqrt{5}[/itex]cost)2([itex]\sqrt{5}[/itex]sint)(-[itex]\sqrt{5}[/itex]sint)]i + [([itex]\sqrt{5}[/itex]cost)([itex]\sqrt{5}[/itex]sint)]j

simplifying down and i got to

5[itex]\int[/itex]sintcost-5cos2tsin2t.dt

how do i integrate that?
 
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Answers and Replies

  • #2
Dick
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Split it into two integrals and use simple substitution and double angle trig identities, like sin(2t)=2*sin(t)*cos(t) or sin(t)^2=(1-cos(2t))/2. You must have seen stuff like this before. Try it.
 
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  • #3
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Oh yeah! Just haunt got those identities in my table integrals so totally forgot. I'll see what answer I end up with.
 
  • #4
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okay so... i split it up to the two integrals.

[itex]\int[/itex]5costsint.dt - [itex]\int[/itex]25cos2tsin2t.dt

trig identities... 2costsint=sin2t and sin2tcos2t = (1-cos4t)/8

so

5/2[itex]\int[/itex]2costsint.dt - 25[itex]\int[/itex]cos2tsin2t.dt

= 5/2[itex]\int[/itex]sin2t.dt -25[itex]\int[/itex](1-cos4t)/8.dt

= 5/2[-cos2t/2] + 25/8[sin4t/4]

evaluating those between 0 and 2pi and blah blah blah and i ended up with 0!

did i go wrong? or is that the answer?
 
  • #5
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wait i think i found an error...

where i simplified -25[itex]\int[/itex](1-cos4t)/8 to +25/8[sin4t/4].... i took out the 1/8 then the negative..

but it should be -25[itex]\int[/itex](1-cos4t)/8 becomes -25/8[itex]\int[/itex](1-cos4t)

which = -25/8[t-(sin4t)/4]

so re-evaluating..

5/2[-cos2t/2] + 25/8[t-sin4t/4]

= 5/2[(-cos2(2pi)/2) - -cos2(0)/2] - 25/8[(2pi-sin4(2pi)/4) - (0-sin4(0)/4)]

= 5/2[-1/2+1/2] - 25/8[2pi-0-0+0]

= -25pi/8

feedback time!
 
  • #6
mege
I think you need to recalculate your boundry curve - when z=1, [itex]x^2+y^2 \neq 5[/itex]

Also, when using Stoke's Theorem, it should be an equation in 2 variables - since you're computing an area. It should be much simpler to compute using the curl of your vector field.

[itex]\oint \textbf{F} \cdot d\textbf{r} = \int\int \Delta X\textbf{F} \cdot \textbf{n} d\sigma[/itex]
 
  • #7
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So just use double integral of curlF? With bounds in xy plane and bounds on z axis. So since it's a sphere obviously bounds will be 0<theta<2pi and then with x and y = 0 then z^2=5 so 0<z<sqrt5 and curlF is curl of the Function of the sphere?
 
  • #8
HallsofIvy
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Homework Statement



Use stokes theorem to find double integral curlF.dS where S is the part of the sphere x2+y2+z2=5 that lies above plane z=1.
F(x,y,z)=x2yzi+yz2j+z3exyk

Homework Equations



stokes theorem says double integral of curlF.dS = [itex]\int[/itex]C F.dr

The Attempt at a Solution



boundary curve C is circle given by x2+y2=5, z=1.
This is incorrect. With z= 1, the equation of the sphere becomes [itex]x^2+ y^2+ 1= 5[/itex] or [itex]x^2+ y^2= 4[/itex], not [itex]x^2+ y^2= 5[/itex].

Vector equation of c then is r(t)=[itex]\sqrt{5}[/itex]costi+[itex]\sqrt{5}[/itex]sintj+1k where 0<t<2pi

then r'(t)= -[itex]\sqrt{5}[/itex]sinti+[itex]\sqrt{5}[/itex]costj

[itex]\int[/itex]CF.dr = [itex]\int[/itex]F(r(t))dotr'(t).dt

F(r(t))=([itex]\sqrt{5}[/itex]cost)2([itex]\sqrt{5}[/itex]sint)i+([itex]\sqrt{5}[/itex]sint)j+e[itex]\sqrt{5}[/itex]cost[itex]\sqrt{5}[/itex]sint

then F(r(t))dotr'(t) = [([itex]\sqrt{5}[/itex]cost)2([itex]\sqrt{5}[/itex]sint)(-[itex]\sqrt{5}[/itex]sint)]i + [([itex]\sqrt{5}[/itex]cost)([itex]\sqrt{5}[/itex]sint)]j

simplifying down and i got to

5[itex]\int[/itex]sintcost-5cos2tsin2t.dt

how do i integrate that?
 
  • #9
mege
So just use double integral of curlF? With bounds in xy plane and bounds on z axis. So since it's a sphere obviously bounds will be 0<theta<2pi and then with x and y = 0 then z^2=5 so 0<z<sqrt5 and curlF is curl of the Function of the sphere?

You'll be computing a surface integral over the 'dome' using your vector field to determine the circulation on the perimiter (I'm presuming). Stoke's Theorem states that the circulation on boundry is equivalent to the surface area bounded in a vector field (with conditions of being bounded, continuous, etc). Seeing as this is a part of a sphere, it will probably be much easier to convert coordinates to spherical, and you'll be integrating for surface over [itex]\phi[/itex] and [itex]\theta[/itex] since [itex]\rho[/itex] is constant. Does this sound familiar? You should still get the same answer, but it should be a bit quicker using Stoke's Theorem (And if the question is asking for a solution using that, a line integral on the boundry probably isn't good :p)
 
  • #10
HallsofIvy
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On the contrary, if the problem says "find the integral over the surface using Stoke's theorem" then an integration around the boundary is clearly intended!
 
  • #11
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the question says "use stokes theorem to evaluate [itex]\int[/itex][itex]\int[/itex]curlF.dS.."

then in my textbook it has under the definition of stokes theorem
"[itex]\int[/itex]CF.dr=[itex]\int[/itex][itex]\int[/itex]SCurlF.dS"

so i just assumed i had to use that exact forumla, and since im given the curl.F.dS in the question i thought i had to use the boundary curve one.

however, i noticed that error with the boundary curve last night.... and if i change that 5 to a 4 then that first integral breaks down to 2sintcost which is exactly the need for the identity to integrate so it ends up a lot neater, which seems more likely.

so which method should i use?
 
  • #12
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mege,

i have seen that before. end up with

r([itex]\phi[/itex],[itex]\theta[/itex])=asin[itex]\phi[/itex]cos[itex]\theta[/itex]i+asin[itex]\phi[/itex]sin[itex]\theta[/itex]j+acos[itex]\phi[/itex]k

then integrate with bounds 0<[itex]\phi[/itex]<[itex]\pi[/itex] and o<[itex]\theta[/itex]<2[itex]\pi[/itex]

just not sure how to account for the xyplane starting at plane z=1.

should the vector equation just become

r([itex]\phi[/itex],[itex]\theta[/itex])=sin[itex]\phi[/itex]cos[itex]\theta[/itex]i+asin[itex]\phi[/itex]sin[itex]\theta[/itex]j+1k??
 
  • #13
vela
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Integration around the boundary.

Or you could do it both ways and verify that Stoke's theorem works. :wink:
 
  • #14
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doing the boundary line one with the same method i used orignially outlined up there ^^^ but with

r(t)=2costi+2sintj+k

r'(t)=-2sinti+2costj

F(r(t))=(2cost)2(2sint)i+(2sint)j+e(2sint2cost)k

then [itex]\int[/itex]CF.dr=[itex]\int[/itex]F(r(t))dotr'(t)dt

= [itex]\int[/itex][(2cost)2(2sint)(-2sint)] + [(2sint)(2cost)].dt

=[itex]\int[/itex]4sintcost - 16cos2tsin2t.dt

=2[itex]\int[/itex]2sintcost.dt - 16[itex]\int[/itex]cos2tsin2t.dt

=2[itex]\int[/itex]sin(2t).dt - 16/8[itex]\int[/itex]1-cos4t.dt (see ideentities in first post)

=2[-cos(2t)/t] - 2[t-sin(4t)/t]

evaluating between 0 and 2pi...

=2[-1/2--1/2] - 2[2pi-0-0-0]

= -4pi

...
 
  • #15
vela
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Looks good (except for typos where you divided by t rather than constants).
 
  • #16
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oh whoops haha. that should be 2 and 4 respectively.

what does it mean when the answer is negative?
that just mean the orientation? the way i traversed around the curve?
 
  • #17
vela
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It means the vector field is generally "rotating" in the direction opposite to the direction of the integration path. This description of the curl in Wikipedia might help you visualize what that means.
 

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