Physics Help Forum Vector Calculus: Divergence theorem and Stokes' theorem

 Apr 3rd 2010, 06:05 AM #1 Junior Member   Join Date: Mar 2010 Posts: 10 Vector Calculus: Divergence theorem and Stokes' theorem Hi =) I've been working on some exercises, but unfortunately we are not given the answers, so have no idea if I am going wrong or not. I attemted part a) which reads: Consider the vector field V= 3x^2y i - 2xy^2 j - xyz k Determine the divergence of this vector field and evaluate the integral of this quantity over the interior of a cube of side a in z> or equal to 0, whose base has the vertices (0,0,0), (a,0,0), (0,a,0), (a,a,0). Evaluate the flux of V over the surface fo the cube and thereby verify the divergence theorem. So for this I got both sides to be 1/4 a^5 Its the next part that has got me completely confused! (its long, here goes..! if anyone can teach me how to use the integral symbols and stuff, i will edit this post to make more sense!) [B]b) Given V = (3x-2y) i + x^2z j + (1-2z) k Find del.V and del x V and evaluate:[B] I found them to be (3x-2y) i + x^2y j + (1-2z)k and -x^2 i + (2xz-2) k respectively. [B]i) double integral of V.dS over the circular region in the xy-plane bounded by x^2 + y^2 = a^2. Regard the area element as positive in the positive z direction.[B] What is dS supposed to be?! ii) Evaluate the double integral of (del x V) . dS over the same region. iii) Evaluate the closed loop integral of V.dr clockwise looking in the positive z direction around the circle x^2 + y^2 = a^2 in the xy-plane. iv) Evaluate the triple integral of (del x V)dV over the volume of the hemisphere bounded by the spherical surface x^2 + y^2 + z^2= a^2 for z>0 and the xy-plane. Which two answers provide a verification of Stokes' Theorem? Thank you SO much Last edited by maple_tree; Apr 4th 2010 at 05:19 AM.
 Apr 8th 2010, 09:48 PM #2 Physics Team   Join Date: Feb 2009 Posts: 1,425 Craig where are you? Your reply is eagerly awaited.
Apr 9th 2010, 10:29 AM   #3
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 Originally Posted by physicsquest Craig where are you? Your reply is eagerly awaited.
I concur!!

Apr 12th 2010, 01:40 PM   #4
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 [b]b) Given V = (3x-2y) i + x^2z j + (1-2z) k Find del.V and del x V and evaluate:[b] I found them to be (3x-2y) i + x^2y j + (1-2z)k and -x^2 i + (2xz-2) k respectively.
Just curious, what do you mean by "del.V"? Do you mean the divergence of the V field?
If so, check again your work. (Or show us how you reached that div V=V)
I'm guessing that "del x V" is the rotor of V. If so, I reach (2+2xy)k. How did you find your value?

 [b]i) double integral of V.dS over the circular region in the xy-plane bounded by x^2 + y^2 = a^2. Regard the area element as positive in the positive z direction.[b] What is dS supposed to be?!
dS is supposed to be the differential surface element. As they deal with a circular region, it is convenient to use polar coordinates to express dS. Using a theorem (involving a Jacobian determinant), one gets that dA=r dr dtheta where dA is the differential area element of the region. I don't remember how dA and dS are related. If I'm not wrong, n dA=dS where n is the normal unit vector to dA.
In this question they are asking you what is the net flux through the circular surface.
Now maybe it's convenient to use Gauss's theorem (although one can get the result without using the theorem).
I too will wait for a more formal answer. Good luck!
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 Apr 14th 2010, 01:58 PM #5 Senior Member     Join Date: Apr 2008 Posts: 815 b]i) And they tell you that n =(0,0,1). So that the double integral becomes int _S (-xyz) dA. But z=0 in all S... I'm confused. We need help. __________________ Isaac If the problem is too hard just let the Universe solve it.
Apr 14th 2010, 03:29 PM   #6
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 Originally Posted by maple_tree Hi =) I've been working on some exercises, but unfortunately we are not given the answers, so have no idea if I am going wrong or not. I attemted part a) which reads: Consider the vector field V= 3x^2y i - 2xy^2 j - xyz k Determine the divergence of this vector field and evaluate the integral of this quantity over the interior of a cube of side a in z> or equal to 0, whose base has the vertices (0,0,0), (a,0,0), (0,a,0), (a,a,0). Evaluate the flux of V over the surface fo the cube and thereby verify the divergence theorem. So for this I got both sides to be 1/4 a^5 Its the next part that has got me completely confused! (its long, here goes..! if anyone can teach me how to use the integral symbols and stuff, i will edit this post to make more sense!) [b]b) Given V = (3x-2y) i + x^2z j + (1-2z) k Find del.V and del x V and evaluate:[b] I found them to be (3x-2y) i + x^2y j + (1-2z)k and -x^2 i + (2xz-2) k respectively. [b]i) double integral of V.dS over the circular region in the xy-plane bounded by x^2 + y^2 = a^2. Regard the area element as positive in the positive z direction.[b] What is dS supposed to be?! ii) Evaluate the double integral of (del x V) . dS over the same region. iii) Evaluate the closed loop integral of V.dr clockwise looking in the positive z direction around the circle x^2 + y^2 = a^2 in the xy-plane. iv) Evaluate the triple integral of (del x V)dV over the volume of the hemisphere bounded by the spherical surface x^2 + y^2 + z^2= a^2 for z>0 and the xy-plane. Which two answers provide a verification of Stokes' Theorem? Thank you SO much
Ok so I've attemted the problem again and here are my answers:

a) Divergence= xy
The integral over the cube = 1/4 a^5
Veryfying using the divergence theorem: the only contributions to the flux come from the sides where x=y=z=a, for the rest it is zero. Adding up the three contrbutions gives 1/4 a^5

b) del.V = 3+0-2 = 1
del x V = =x^2 i + (2xz-2)k

i figured dS out to be = rdrd(phi) k
giving the overall result pi a^2

Next using dr= acoscita i + a sincitaj dcita to be 3pia^2

anyone agree/disagree?

Apr 14th 2010, 03:44 PM   #7
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 Originally Posted by maple_tree Ok so I've attemted the problem again and here are my answers: b) del.V = 3+0-2 = 1 del x V = =x^2 i + (2xz-2)k i figured dS out to be = rdrd(phi) k giving the overall result pi a^2 anyone agree/disagree?
I agree for del.V.
Just curious, have you read the post #4? I was asking you some clarification and how did you reach your results.
I do not agree with your curl (V). Read post #4. I used the determinant of a matrix to find curl (V) which was confirmed by this program: Wolfram|Alpha and the result is different from yours. So once again let me ask you what method did you use and please show your work so that we can see any error.
Your dS is the same than mine (I used theta instead of phi) and as I stated in my last post, n=(0,0,1) thus the unit vector k. Good.
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Apr 15th 2010, 08:40 AM   #8
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 Originally Posted by arbolis I agree for del.V. Just curious, have you read the post #4? I was asking you some clarification and how did you reach your results. I do not agree with your curl (V). Read post #4. I used the determinant of a matrix to find curl (V) which was confirmed by this program: Wolfram|Alpha and the result is different from yours. So once again let me ask you what method did you use and please show your work so that we can see any error. Your dS is the same than mine (I used theta instead of phi) and as I stated in my last post, n=(0,0,1) thus the unit vector k. Good.

For the curl= del x V i also used the determinant of the matrix to find:
[the partial derivative wrt y of (1-2z)] - [the partial derivative wrt z of ((x^2)z) = -x^2 i
[the partial derivative wrt x of (1-2z)] - [the partial derivative wrt z of (3x-2y) = 0j
[the partial derivative wrt x of ((x^2)z)] - [the partial derivative wrt y of (3x-2y) = (2xz-2)k

is that the explanation you were refering to? =)

Apr 15th 2010, 04:09 PM   #9
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 Originally Posted by maple_tree For the curl= del x V i also used the determinant of the matrix to find: [the partial derivative wrt y of (1-2z)] - [the partial derivative wrt z of ((x^2)z) = -x^2 i [the partial derivative wrt x of (1-2z)] - [the partial derivative wrt z of (3x-2y) = 0j [the partial derivative wrt x of ((x^2)z)] - [the partial derivative wrt y of (3x-2y) = (2xz-2)k is that the explanation you were refering to? =)
Yes. You're right on this, it seems I copied a wrong expression and I put the same wrong expression in wolframalpha which returned the result I had.
You should double check your work with wolframalpha for the curl of the field in part a).
By the way have you done the double integral? I don't really know how to treat the "z" in the integrand while converting to polar coordinates.
The integral is int _S (1-2z) dA= int_{0}^{2 pi} int_{0}^{a} [(1-2z)] r dr dtheta . Notice that this expression is incorrect since I'd need to convert 1-2z into polar coordinates I believe, but I think it's impossible, thus I'm making a mistake. If you overcame this, please tell me how you did. Thanks.
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Apr 16th 2010, 03:04 AM   #10
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 Originally Posted by arbolis Yes. You're right on this, it seems I copied a wrong expression and I put the same wrong expression in wolframalpha which returned the result I had. You should double check your work with wolframalpha for the curl of the field in part a). By the way have you done the double integral? I don't really know how to treat the "z" in the integrand while converting to polar coordinates. The integral is int _S (1-2z) dA= int_{0}^{2 pi} int_{0}^{a} [(1-2z)] r dr dtheta . Notice that this expression is incorrect since I'd need to convert 1-2z into polar coordinates I believe, but I think it's impossible, thus I'm making a mistake. If you overcame this, please tell me how you did. Thanks.
Hi arbolis!
Indeed, you are making a mistake by including the z in your expression to be integrated. We are looking at a surface on the xy plane, and so z=0, diving you the double integral of = -2 rdrdphi which eventually gives -2pia^2, do you get it?

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