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Homework Help: Trouble with definite integral

  1. Apr 18, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex]\int _{0}^{2\,\pi}\!\int _{0}^{1}\! 0.5\,{r}^{ 3.5}\sin \left( \theta
    \right) {\cos}^{2}+ 0.5\,{r}^{ 2.5}{\sin}^{2}+1/5\,{r}^{5}{\cos}^{2}{
    dr}\,{d\theta}[/tex]

    2. Relevant equations



    3. The attempt at a solution

    So I first integrate the r, which was relatively straight forward, the tricky part was the theta, I had to use the double angle formula? then evaluating, which for the end answer, I got

    [tex]\frac{39}{20}\Pi-\frac{3}{4}[/tex]

    which looks kinda wrong to me....

    tips and hints appreciated
     
  2. jcsd
  3. Apr 18, 2010 #2

    LCKurtz

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    It is bad form to start a new thread instead of continuing the thread that generated the question. Your integrand is wrong. Your integrand should be the dot product of these two vectors:

    [tex]\vec F = \langle r^2\cos\theta\sin\theta,r\sin\theta,r^2\cos^2\theta\rangle[/tex]
    [tex]d\vec S = \langle\frac 1 2 \sqrt r \cos\theta,\frac 1 2 \sqrt r \sin\theta,r\rangle[/tex]
     
  4. Apr 18, 2010 #3
    Hi LCKurtz,

    I did try to continue the old thread which generated this question, but no further response was given, and I was not sure whether I'm allowed to double post into my own topic. So I took the chance and posted another thread, I do apologize for that and will remember not to do so next time.

    Anyway.

    That integrand I posted up to IS the dot product of the 2 vectors you stated above.

    But isn't the vector [tex]d\vec S = \langle\frac 1 2 \sqrt r \cos\theta,\frac 1 2 \sqrt r \sin\theta,r\rangle[/tex][/QUOTE] suppose to be the normal([tex]\overline{n}[/tex])? and Ds is the element of integration?(in this case r dr d(theta))

    So the dot product I got is

    [tex]\left[ \begin {array}{c} 0.5\,{r}^{ 2.5} \left( \cos \left( \theta
    \right) \right) ^{2}\sin \left( \theta \right) \\ \noalign{\medskip}
    0.5\,{r}^{ 1.5} \left( \sin \left( \theta \right) \right) ^{2}
    \\ \noalign{\medskip}{r}^{3} \left( \cos \left( \theta \right)
    \right) ^{2}\end {array} \right] [/tex]

    or more correctly

    [tex] 0.5\,{r}^{ 2.5} \left( \cos \left( \theta \right) \right) ^{2}\sin
    \left( \theta \right) + 0.5\,{r}^{ 1.5} \left( \sin \left( \theta
    \right) \right) ^{2}+{r}^{3} \left( \cos \left( \theta \right)
    \right) ^{2}[/tex]

    (Note: the reason why I have the coefficients and powers in decimals rather than fractions, is that Maple will evaluate it for me if I were to put it into fractions.)

    So after putting in Ds=r dr dtheta, and the limits of integration, I yield the Integrand

    [tex]\int _{0}^{2\,\pi}\!\int _{0}^{1}\! \left( 0.5\,{r}^{ 2.5} \left(
    \cos \left( \theta \right) \right) ^{2}\sin \left( \theta \right) +
    0.5\,{r}^{ 1.5} \left( \sin \left( \theta \right) \right) ^{2}+{r}^{
    3} \left( \cos \left( \theta \right) \right) ^{2} \right) r{dr}\,{d
    \theta}[/tex]

    then I multiplied the r through each term and finally get the Integrand I got up on post 1.

    any further tips and hints appreciated, and I apologize again about posting the twice.

    Thanks.
     
  5. Apr 19, 2010 #4

    LCKurtz

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    OK, maybe I overlooked something. That last integral is correct. But I thought I had already explained to you how to do the theta integrals: u substitution on the first one and the double angle (or half angle, whichever they are called) formulas for the sin2 and cos2 ones.
     
  6. Apr 19, 2010 #5
    hi again LCKurtz,

    I went back to my old thread and started questions b) and c) again which are located here

    https://www.physicsforums.com/showthread.php?p=2672658#post2672658

    at which I got these Integrals.

    For b), I had to calculate the Integrals for the surface S, which is also S1+S2, therefor I got these as my integrals.

    S1

    [tex]\int _{0}^{2\,\pi}\!\int _{0}^{1}\! 0.5\,{r}^{ 3.5} \left( \cos
    \left( \theta \right) \right) ^{2}\sin \left( \theta \right) + 0.5\,
    {r}^{ 2.5} \left( \sin \left( \theta \right) \right) ^{2}+{r}^{4}
    \left( \cos \left( \theta \right) \right) ^{2}{dr}\,{d\theta}[/tex]

    S2
    [tex]
    \int _{0}^{2\,\pi}\!\int _{0}^{1}\!-{r}^{4} \left( \cos \left( \theta
    \right) \right) ^{2}{dr}\,{d\theta}[/tex]

    S1+S2

    [tex]\int _{0}^{2\,\pi}\!\int _{0}^{1}\! 0.5\,{r}^{ 3.5} \left( \cos
    \left( \theta \right) \right) ^{2}\sin \left( \theta \right) + 0.5\,
    {r}^{ 2.5} \left( \sin \left( \theta \right) \right) ^{2}{dr}\,{d
    \theta}[/tex]

    And for the volume Integral c), I got

    [tex]\int _{0}^{2\,\pi}\!\int _{0}^{1}\! 0.5\,{r}^{ 3.5} \left( \cos
    \left( \theta \right) \right) ^{2}\sin \left( \theta \right) + 0.5\,
    {r}^{ 2.5} \left( \sin \left( \theta \right) \right) ^{2}{dr}\,{d
    \theta}[/tex]


    The question states that the volume integral should be the same as the integral of S1+S2 when I have evaluated it, but when I do, it does not give me the same answer?

    any help, hints would be much appreciated.
     
  7. Apr 20, 2010 #6

    LCKurtz

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    The equation I was referring to was:
    [tex]
    \int _{0}^{2\,\pi}\!\int _{0}^{1}\! \left( 0.5\,{r}^{ 2.5} \left(
    \cos \left( \theta \right) \right) ^{2}\sin \left( \theta \right) +
    0.5\,{r}^{ 1.5} \left( \sin \left( \theta \right) \right) ^{2}+{r}^{
    3} \left( \cos \left( \theta \right) \right) ^{2} \right) r{dr}\,{d
    \theta}
    [/tex]

    When I told you it was correct I overlooked that r outside the parentheses. It should not be there. You don't need that extra r when you use the cross product to calculate dS.

    Both of the above integrals have that extra r in them. Remove the extra r and they should both give you the correct answer.

    <snip> Calculate them separately.

    The volume integral shouldn't be nearly that complicated and it should be a triple integral, not a double integral. Set up the triple integral of [itex] \nabla \cdot F [/tex] for the enclosed volume in cylindrical coordinates and post it.
     
  8. Apr 20, 2010 #7
    The extra r you're speaking of came from the element of volume dS which is r dr dtheta? then I multiplied through the r into the integral? or was I not meant to do that?

    lastly, I have no idea why the integral came out on here the way it has, but this was my triple integral:

    [tex]\int _{0}^{2\,\pi}\!\int _{0}^{1}\!\int _{1-\sqrt {r}}^{0}\!{r}^{3}
    \cos \left( \theta \right) \left( \sin \left( \theta \right)
    \right) ^{2}+r\sin \left( \theta \right) {dz}r{dr}\,{d\theta}[/tex]
     
  9. Apr 20, 2010 #8

    LCKurtz

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    In general when you have a surface defined parametrically as

    [tex]\vec R(u,v) = \langle x(u,v), y(u,v), z(u,v)\rangle[/tex]

    the element of surface are is defined by

    [tex]dS = |\vec R_u \times \vec R_v|\, dudv[/tex]

    with no extra u or v. In the case of polar coordinates we get used to writing [itex]rdrd\theta[/itex] as a shortcut. But when you have a surface parameterized as [itex]R(r,\theta)[/itex] and calculate

    [tex]dS = |\vec R_r \times \vec R_\theta|[/tex]

    the r gets built in. Try it with polar coordinates in the plane, which you would parameterize as

    [tex]\vec R(r,\theta) = \langle r\cos\theta, r\sin\theta, 0\rangle[/tex]

    Calculate dS from that and you will see the r is built in. That is in fact where the r we love so well in the formula [itex]rdrd\theta[/itex] comes from.

    If I recall correctly, the vector field you started with was F = <xy, y, x2>. When you take its divergence and parameterize it you don't get anything nearly that complicated. And are your limits in the positive direction?
     
  10. Apr 20, 2010 #9
    Oh! Thank you very much for clearing that up for me, we never got taught when there is no r in dV, we never really get taught that stuff, basically just the way to approach questions, which I'm still not very good at.

    Anyway, in regards to the triple integral, I calculated it as it follows.

    [tex]
    \nabla
    [/tex] = [tex]
    \left[\frac{d}{dx}\frac{d}{dy}\frac{d}{dz}\right]
    [/tex]

    = [tex]
    \left[\frac{d(xy)}{dx}\frac{d(y)}{dy}\frac{d(x^2)}{dz}\right]
    [/tex]

    =[tex]
    \left[\,y,1,0\right]
    [/tex]

    then my F is

    [tex]
    \left[ \begin {array}{c} xy\\ \noalign{\medskip}y
    \\ \noalign{\medskip}{x}^{2}\end {array} \right]
    [/tex]

    taking the dot product of the 2, gives me

    Triple Integral (x*y^2+y) dV

    and substituted the cylindrical coordinates x=rcos(theta), y=rsin(theta) into the triple integral, and thus yielding the triple integral I posted above.

    was I suppose to make the substitution BEFORE the dot product?

    Thanks for your reply.
     
  11. Apr 20, 2010 #10

    LCKurtz

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    You are supposed to calculate [itex]\nabla \cdot F[/itex]. This is a dot product which is a scalar, not a vector.
     
  12. Apr 21, 2010 #11
    Oh right...so I would just have 1+y then wouldn't I?
     
  13. Apr 21, 2010 #12

    LCKurtz

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    Bingo. Now remember the correct formula for dV in cylindrical coordinates. Unlike the surface differential, you aren't calculating it directly from a parameterization which builds in the r.

    At this point hopefully you can get the correct numbers for both sides of the divergence theorem.
     
  14. Apr 21, 2010 #13
    Great! I now have S1+S2=Volume Integral,

    However, for S1+S2, I got a positive answer and for the volume integral I got a negative answer. I assume the limits for my volume integral are in the wrong direct? if so, how will I know which direct they're suppose to be and how do I fix this?

    Thanks.
     
  15. Apr 21, 2010 #14

    LCKurtz

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    Make sure you integrated in the positive direction for all three variables in the volume integral.
     
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