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Vector field

  1. May 23, 2012 #1

    sharks

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    1. The problem statement, all variables and given/known data
    Vector field ##\vec F= 4x \hat i+4y \hat j +3 \hat k##
    Let S be the open surface above the xy-plane defined by ##z=4-x^2-y^2##

    a. Evaluate normal outward flux of F through S.

    b. Use Stokes' theorem to evaluate the normal outward flux of ##∇ \times \vec F## through S.

    c. Use Gauss' theorem to evaluate the volume integral $$\iiint_V (∇\times \vec F)\,. dxdydz$$ where V is the volume enclosed by S and the xy-plane.


    2. Relevant equations
    vector field, curl, div, flux formula, etc.


    3. The attempt at a solution

    Part (a):
    First, i found the unit normal vector, ##\hat n## and then i used the formula:
    $$\iint_S \vec F. \hat n .dS$$
    Is that correct? I calculated the integrand and then projected on the xy-plane for the surface area.

    Part (b): I used the following formula:
    $$\iint_S ∇\times F. \hat n .dS$$
    Is that correct? If yes, then for this part i got the curl F = 0, so i didn't calculate the rest as the double integral of 0 is 0.

    Part (c): I used the formula:
    $$\iiint_V ∇. F .dxdydz$$
    I found div F, then used cylindrical coordinates to find the volume. I'm quite sure that i got this part right.
     
  2. jcsd
  3. May 23, 2012 #2

    HallsofIvy

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    Yes. What did you get for dS? It is, of course, impossible to say whether you have done this correctly since you don't show what you did. (What you show is not a formula, it is an "expression". A formula says that two expressions are equal. I assume you simply mean that you did the surface integral directly.)

    Again, that's not a formula. If you mean that you calculate the integral directly, that is not "using Stokes theorem". Do you know what "Stokes theorem" is?

    One more time, that is NOT a "formula". Also there is NO formula that says the triple integral of a grad is equal to the triple integral of a curl, which is what you want to find. The problems says "use Gauss's theorem". Do you know what that is?
     
    Last edited: May 23, 2012
  4. May 23, 2012 #3

    sharks

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    From what i understand, Gauss divergence theorem is used for calculating the flux in a solid volume.
     
  5. May 23, 2012 #4

    HallsofIvy

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    Are you saying that you do not know what Gauss's theorem is?
     
  6. May 23, 2012 #5

    sharks

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    OK, i think it's simpler if i just show my work.

    (a) ##z=4-x^2-y^2## gives ##\phi (x,y,z)=z-4+x^2+y^2## as the surface is oriented upwards, so z has to be positive.
    ##∇\phi = 2x\hat i +2y\hat j + \hat k## and ##|∇\phi| = \sqrt{4x^2 +4y^2 + 1}##
    ##\hat n = 1/ \sqrt{4x^2 +4y^2 + 1}.(2x\hat i +2y\hat j + \hat k)##
    ##\vec F= 4x \hat i+4y \hat j +3 \hat k##
    $$\iint_S \vec F.\hat n\,.dS=\iint_S (4x \hat i+4y \hat j +3 \hat k).\frac{1}{ \sqrt{4x^2 i +4y^2 + 1}}.(2x\hat i +2y\hat j + \hat k) \,.dS$$
    Projecting the surface S onto the xy-plane:
    ##z_x=2x## and ##z_y=2y##
    $$\iint_S \vec F.\hat n \,.dS= \iint_R (4x \hat i+4y \hat j +3 \hat k).\frac{1}{ \sqrt{4x^2 i +4y^2 + 1}}.(2x\hat i +2y\hat j + \hat k). \sqrt{4x^2+4y^2+1}\,.dxdy
    \\=\iint_R (8x^2+8y^2+3)\,.dxdy$$Parametrization of the region R:$$
    \int^{2\pi}_0 \int^2_0 (8r^2+3)\,.rdrd\theta=76\pi$$
    (b) Using Stoke's theorem: $$\iint_S ∇\times \vec F. \hat n .dS$$ but first calculating the curl of F. I get the answer 0. This is a bit surprising as it voids any other calculations involving the product with ##\hat n## and subsequently calculating the surface area, so here is my work:
    $$∇\times \vec F=\begin{vmatrix}\hat i & \hat j & \hat k \\ \partial /\partial x & \partial /\partial y & \partial /\partial z \\ 4x & 4y & 3\end{vmatrix}=0$$
    Therefore, $$\iint_S ∇\times \vec F. \hat n .dS=0$$
    (c)$$\iiint_V div\, \vec F .dV$$
    ##div\, \vec F=4+4=8##
    Describing the volume in terms of cylindrical coordinates:
    For ##r## and ##θ## fixed, ##z## varies from ##z=0## to ##z=4-r^2##
    For ##\theta## fixed, r varies from r=0 to r=2
    θ varies from 0 to ##2\pi##
    Therefore, $$\iiint_V div\, \vec F .dV=\int^{2\pi}_0 \int^2_0 \int^{4-r^2}_0 8\,.rdzdrd\theta=64\pi$$
    Is my work correct?
     
    Last edited: May 23, 2012
  7. May 23, 2012 #6

    HallsofIvy

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    Well, that's what you should do from the start!:approve:

    There should be no "i" inside that square root, of course.

    Okay, all of that's correct but I really hate that way of representing the differential of surface area! Did you notice that you had to calculate the length of [itex]\nabla \phi[/itex] twice and then they canceled?

    Simpler, in my opinion, is this. If a surface is given by the "position vector" [itex]\vec{r}(s, t)= x(s,t)\vec{i}+ y(s,t)\vec{j}+ z(s,t)\vec{k}[/itex], with parameters s and t, then the partial derifvatives, [itex]\vec{r}_s= x_s\vec{i}+ y_s\vec{j}+ z_s\vec{k}[/itex] and [itex]\vec{r}_t= x_t\vec{i}+ y_t\vec{j}+ z_s\vec{k}[/itex] are vectors in the tangent plane to the surface at every point. Their cross product, [itex]\vec{r}_s\times\vec{r}_t[/itex], is normal to the surface and gives the "vector differential of surface area": [itex]\vec{n} dS= (\vec{r}_s\times\vec{r}_t)dsdt[/itex]. (And, by the say, the "differential of surface area is given by the length of that vector: [itex]dS= |\vec{r}_s\times\vec{r}_t| dsdt[/itex].

    In this case, [itex]z= 4- x^2- y^2[/itex] so the "position vector" of any point on the surface, using x and y as parameters, is [itex]\vec{r}= x\vec{i}+ y\vec{j}+ (4- x^2- y^2)\vec{k}[/itex]. So [itex]\vec{r}_x= \vec{i}- 2x\vec{k}[/itex] and [itex]\vec{r}_t= \vec{j}- 2y\vec{k}[/itex]. Their cross product is [itex]2x\vec{i}+ 2y\vec{j}+ \vec{k}[/itex] and so [itex]d\vec{S}= (2x\vec{i}+ 2y\vec{j}+ \vec{k})dxdy[/itex] which gives exactly what you have- but is much easier don't you think?

    That's true, of course, but it does NOT use "Stoke's theorem". That is a direct calculation (which happens to be remarkably easy).

    Stoke's theorem says that
    [tex]\int \nabla\times \vec{F}\cdot d\vec{S}= \oint \vec{F}\cdot d\vec{\sigma}[/tex]
    where the integral on the right is the the integral around the boundary of the surface, here, the circle [itex]x^2+ y^2= 4[/itex], z= 0.
    We can take as parameterization [itex]x= 2cos(\theta)[/itex], [itex]y= 2 sin(\theta)[/itex], so that [itex]d\vec{\sigma}= dx\vec{i}+ dy\vec{j}+ 0\vec{k}= (-2sin(\theta)\vec{i}+ 2cos(\theta)\vec{j})d\theta[/itex] and the integral becomes
    [tex]\int_0^{2\pi} (8 cos(\theta)\vec{i}+ 8sin(\theta)\vec{j}+ 3\vec{k})\cdot (-2sin(\theta)\vec{i}+ 2 cos(\theta)\vec{j})d\theta[/tex]

    What is that?

     
    Last edited: May 23, 2012
  8. May 23, 2012 #7

    sharks

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    Yes, it was a typo. I fixed it.
    Honestly, i have practiced so much using that other method which i'm now very familiar with, that i'll need to re-write and re-wire the ROMs in my brain. Your method was first suggested to me by LCKurtz some days ago, and i went to dig more on that by myself in some books. It looks easy but by reflex, i always revert to the first method that i learnt, especially during exams when the stress and doubt of a newly learned method is often pushed out to favor the old/trusted, even if longer, method.

    So, for part (b) what i did in post #5, is wrong? I actually learned Stokes' theorem in this way:
    $$\oint \vec{F}\cdot d\vec{\sigma}=\int \nabla\times \vec{F}\cdot d\vec{S}$$
    In all the problems that i've worked through, i've never had to use the left integral without first converting to the right integral, so i'm frankly quite lost here.
    Evaluating your integral i get ##32\pi##

    I suppose you're asking about the curl F which i calculated to be 0.

    Part (c) deals with the Gauss divergence theorem. I hope that my answer is correct.
     
    Last edited: May 23, 2012
  9. May 23, 2012 #8
    For part C, you need to use Gauss' Law, which states that the electric flux of an electric field [itex]\vec{E}(x,y,z)[/itex] through a closed surface S is proportional to the total charge of the solid E enclosed by the surface S:

    [itex]\int\int_{S}\vec{E}\bullet d\vec{S}=\frac{1}{ε_{0}}\int\int\int_{E}ρdV[/itex]

    where [itex]ρ(x,y,z)[/itex] is the charge density and [itex]ε_{0}[/itex] is the electric constant.

    I can't really understand what you've written though. From what I can tell, you wrote [itex]\int\int\int_{E}(∇\times\vec{F})\bullet dV[/itex], where E is the solid enclosed by the surface S and the xy-plane. This doesn't make much sense to me, since [itex]dV[/itex] is not a vector, and so it doesn't make sense to take a dot product. Unless it's actually [itex]d\vec{V}[/itex] and not [itex]dV[/itex]. But what could [itex]d\vec{V}[/itex] possibly represent? A "volume normal"? And even if you didn't mean to include the dot product, it still doesn't make any sense, since then we'd be integrating a vector field.
     
    Last edited: May 23, 2012
  10. May 24, 2012 #9

    sharks

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    This is part (c) and div F is the divergence of F, which means ##\nabla.\vec F## not ##\nabla \times \vec F##. The triple integral with dV represent the volume. Maybe you can work out your method and see if you get the same answer?
     
    Last edited: May 24, 2012
  11. May 24, 2012 #10
    Yes, that answer is correct. But I don't think your solution is. They ask you to use Gauss' law to find [itex]\int\int\int_{E}∇\bullet\vec{F}dV[/itex]. Not to evaluate the triple integral directly. So you have to use Gauss' law:

    [itex]\int\int_{S}\vec{E}\bullet d\vec{S}=\frac{1}{ε_{0}}\int\int\int_{E}ρdV[/itex]

    Along with the following Maxwell equation:

    [itex]\frac{ρ}{ε_{0}}=∇\bullet\vec{E}[/itex]

    In our case, [itex]\vec{E}=\vec{F}[/itex]. So really, we have:

    [itex]\int\int\int_{E}∇\bullet\vec{F}dV=\int\int_{S}\vec{F}\bullet d\vec{S}[/itex]

    In other words, the triple integral is equal to the outward flux of F across the closed surface S. In part a, you calculated the outward flux through the paraboloid. But now you have to close the paraboloid with the disk [itex]x^{2}+y^{2}≤2^{2}[/itex] on the xy-plane. So really, all you have to do is find the flux (with proper orientation) through the disk, and add it to the flux from part a. It's a rather quick computation.
     
    Last edited: May 24, 2012
  12. May 24, 2012 #11

    sharks

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    I assure you that the problem is described in my class notes under the heading "Gauss Divergence Theorem" and the method is as i have described. I just don't understand how my notes could diverge so much from what all of you are suggesting. My most pressing question to my tutor right now would be "Why the apparently unorthodox methods and techniques?" but it's already too late. This is a really dire situation for me, as for the entire academic year (ends this week) i've been learning to solve those problems the way i've already shown in this thread. At this point, i'm probably going to leave the calculus section of my exam paper blank.

    So, with a big sign, i want to ask -- what book do you recommend that i start to learn everything from scratch from? I assume that i'm going to repeat the yearly maths module. I have seen some similarities from what all of you have suggested, in Thomas' Calculus 12th edition. Is there a better book for self-learning the right techniques and theorems? At the beginning of the academic year, last August, my tutor had suggested "Calculus: One and Several Variables" 10th edition by Salas and Hille. But i preferred to concentrate fully on my tutor's notes, since the techniques, variables, notations, etc, taught in class were somewhat different compared to the Salas book.
     
    Last edited: May 24, 2012
  13. May 24, 2012 #12
    I still think part C is supposed to be done using a surface integral.

    Anyways, I learned from James Stewart's Calculus: Concepts and Contexts (Not sure of the edition, 4 I think). There's one for single variable calculus and another for multivariable. I found it to be pretty good at explaining things. Much better than my professors in any case. You shouldn't be so worried though. Most of your work was right. And learning different techniques is different than learning from scratch. You seem to have a good understanding already.
     
  14. May 24, 2012 #13

    HallsofIvy

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    What you wrote, in post #7:
    [tex]\oint \vec{F}\cdot d\sigma= \int\int \nabla\times\vec{F}\cdot d\vec{S}[/tex]
    is exactly what I wrote, in post #6:
    [tex]\int\int \nabla\times\vec{F}\cdot d\vec{S}= \oint \vec{F}\cdot d\sigma[/tex]
    whether each part is on the left or right does not matter at all.

    And the point is that, because they are equal, you can use either side to find the other.
    Since you are told to "use Stoke's theorem" to find the integral over the surface, that means that you use the other side, not just integrate directly. That would not be "using Stoke's theorem".
     
    Last edited: May 24, 2012
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