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Evaluate using divergence theorem

  1. Jan 19, 2012 #1

    sharks

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    The problem statement, all variables and given/known data
    http://s1.ipicture.ru/uploads/20120120/eAO1JUYk.jpg

    The attempt at a solution
    [tex]\int\int \vec{F}.\hat{n}\,ds=\int\int\int div\vec{F}\,dV[/tex]
    where dV is the element of volume.
    [tex]div\vec{F}=3[/tex]
    Now, i need to find dV which (i assume) is the hardest part of this problem.
    I've drawn the graph in my copybook and it's a paraboloid cut off by the plane z=3, with the outward unit normal vector pointing upward.

    OK, i have 2 ideas for calculating the volume of that spherical shape bounded by the plane z=3:
    1. I could use triple integral and transform to spherical coordinates. [itex]dxdydz={\rho}^2 \sin\phi .d\rho d\phi d\theta[/itex]
    2. Use the flux formula for surface integral to find the surface area of the curved surface and then multiply it by the height along the z-axis which is 4-3=1.

    I tried the first idea but i can't figure it out. So, i'm going with my second idea.
    [tex]\phi(x,y,z)=z-4+x^2+y^2[/tex]
    [tex]∇\vec{\phi}=2x\vec{i}+2y\vec{j}+\vec{k}[/tex]
    [tex]\hat{n}=\frac{2x\vec{i}+2y\vec{j}+\vec{k}}{\sqrt{4x^2+4y^2+1}}[/tex]
    [tex]Flux=\int\int \vec{F}.\hat{n}\,.d \sigma=\int\int \frac{2x^2+2y^2+z}{\sqrt{4 x^2+4y^2+1}}\,.d \sigma[/tex]
    [tex]d\sigma = \sqrt{1+4x^2+4y^2}\,.dxdy[/tex]
    [tex]S.A.=\int\int (2x^2+2y^2+z)\,.dxdy[/tex]
    Finding the points of intersection of the plane z=3 and the paraboloid [itex]z=4-x^2-y^2[/itex], gives [itex]x^2+y^2=1[/itex] and hence r=1.

    Transforming to polar coordinates:
    [tex]S.A.=\int\int (2x^2+2y^2+z)\,.dxdy=\int\int (4+r^2)r\,.drd\theta=\int^{2\pi}_0 \int^1_0 (4r+r^3)\,.drd\theta=\frac{9\pi}{2}[/tex]
    The volume becomes: [itex]\frac{9\pi}{2} \times 1 = \frac{9\pi}{2}[/itex]
    [tex]\int\int\int div\vec{F}\,dV=\int\int\int 3\,dV=3\times \frac{9\pi}{2}=\frac{27\pi}{2}[/tex]
    This is wrong, as (according to my notes) the correct answer is [itex]\frac{3\pi}{2}[/itex]. I have no idea what mistake/s i made.

    EDIT: OK, after thinking more about this, i think i know where i made a mistake. I assumed the height of the paraboloid section to be 1 unit (at its peak), but the height is not constant, so i don't know how to find a value for the height. At this point, i'm thinking about going back to my first idea, which is finding the volume via triple integral using spherical coordinates. Any advice?

    If i use triple integral with spherical coordinates:
    [itex]z=4-x^2-y^2[/itex] becomes [itex]\rho\cos\phi=4-{\rho}^2 {\sin}^2 \phi[/itex]
    This becomes: [itex]{\rho}^2 {\cos}^2 \phi +\rho \cos \phi=4+{\rho}^2[/itex]
    It's a dilemma as i'm stuck with 2 variables.
     
    Last edited: Jan 19, 2012
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  3. Jan 20, 2012 #2

    sharks

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    Any help?
     
  4. Jan 20, 2012 #3

    vela

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    I suggest evaluating the volume integral using cylindrical coordinates since the volume is symmetrical about the z-axis.
     
  5. Jan 20, 2012 #4

    vela

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    Up to here, your work is fine. You've evaluated the surface integral over the upper surface of the solid, but you still need to add the contribution from the bottom surface. In other words, you have
    $$\iint_S \vec{F}\cdot\hat{n}\,ds = \iint_{S_\text{top}} \vec{F}\cdot\hat{n}\,ds + \iint_{S_\text{bottom}} \vec{F}\cdot\hat{n}\,ds$$and you need to still do the second integral on the righthand side. The bottom surface is the unit circle lying in the z=3 plane centered about the z-axis.

    I'm not sure where you're getting the idea that the volume is 9π/2 times anything.

    The point of the problem was to avoid doing the surface integrals by evaluating the volume integral, so this is indeed the tack you want to take. (Since you already did the hard part, you might want to finish off the surface integral calculation just to see it indeed all works out.) However, as I noted in my other post, cylindrical coordinates are better suited to this problem.
     
    Last edited: Jan 20, 2012
  6. Jan 20, 2012 #5

    SammyS

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    Before "bumping" your thread, you should wait 24 hours after you first post it. (It's a rule.)
     
  7. Jan 20, 2012 #6

    SammyS

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    The instructions for this problem state that you are to use the divergence theorem. Therefore, you should evaluate [itex]\displaystyle {\int\!\!\!\!\int\!\!\!\!\int}_{D} \ div(\vec{F})\,dV\,. [/itex]

    I agree with vela's suggestion to use cylindrical coordinates. After all, you're using the divergence theorem to make things easier. Why complicate things by using spherical coordinates? Cartesian coordinates would be easier to use than spherical coordinates.
     
  8. Jan 20, 2012 #7

    sharks

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    Thanks for the advice, vela. I'm going to find the flux of bottom disc.
    SammyS, i apologize for bumping this topic a bit early. I waited 17 hours, but i should have read the rules.:redface:

    Finding: [itex]\iint_{S_\text{bottom}} \vec{F}\cdot\hat{n}\,ds[/itex]
    [tex]\phi(x,y,z)=-z+3[/tex][tex]∇\vec{\phi}=-\vec{k}[/tex][tex]\hat{n}=-\vec{k}[/tex][tex]Flux=\iint \vec{F}.\hat{n}\,.d \sigma=\iint -z\,.d \sigma[/tex][tex]d\sigma = \sqrt{0+0+1}\,.dxdy[/tex][tex]S.A.=\iint (-z)\,.dxdy=-\iint 3\,.dxdy[/tex]
    Converting to polar coordinates:[tex]S.A.=-\int^{2\pi}_0 \int^1_0 3\,.rdrd\theta =-3\pi[/tex]
    [tex]\iint_S \vec{F}\cdot\hat{n}\,ds = \iint_{S_\text{top}} \vec{F}\cdot\hat{n}\,ds + \iint_{S_\text{bottom}} \vec{F}\cdot\hat{n}\,ds=\frac{9\pi}{2}+-3\pi=\frac{3\pi}{2}[/tex]
    This is correct. But now, i'm going to try to work it out as the problem intended using the divergence theorem.
    [tex]\int\int \vec{F}.\hat{n}\,ds=\int\int\int div\vec{F}\,dV[/tex][tex]div\vec{F}=3[/tex]
    Now, i have to find dV. To find the volume of this bell-shaped paraboloid using cylindrical coordinates is a bit odd for me, as i never used cylindrical coordinates for anything other than a perfect cylinder. I agree that the section under the plane z=3 will be as a cylinder, but i'm going for it with what i know. So, converting to cylindrical coordinates:
    [tex]\int^{2\pi}_0 \int^1_0 \int^{4-r^2}_3 r.dzdrd\theta[/tex]
    For θ and r fixed, z varies from z=3 to z=4-r^2
    For θ fixed, r varies from 0 to 1
    θ varies from 0 to 2∏
    The answer is... [itex]\frac{\pi}{2}[/itex]
    [tex]\int\int\int div\vec{F}\,dV=3\times \frac{\pi}{2}=\frac{3\pi}{2}[/tex] which is correct! Thanks, vela and SammyS for your advice.
     
    Last edited: Jan 20, 2012
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