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Integral 2

  1. Jan 4, 2012 #1
    1. The problem statement, all variables and given/known data
    Evaluate the surface integral [itex]\displaystyle \int \int_\sigma f(x,y,z) dS[/itex] for [itex]f(x,y,z)=(x^2+y^2)zy[/itex] where σ is the portion of the sphere x^2+y^2+z^2=4 and abov plane z=1


    3. The attempt at a solution

    I realise this can be done by parameterising the surface using θ and ∅. However, is it possible to use the other method

    [itex]\displaystyle \int \int_{\sigma} f(x,y,z) dS=\int \int_R f(x,y,g(x,y))\sqrt{z_x^2+z_y^2+1}dA [/itex] (1)

    where [itex]z=g(x,y)=\sqrt{ 4-x^2-y^2}[/itex] (2)?
    Evaluating the RHS of eqn 1 I arrive at

    [itex]\displaystyle \int \int_{\sigma} f(x,y,z) dS=\int \int_R 2x^2y+2y^3dA [/itex]

    Is this correct? If it is, how do I proceed with the limits, that I am not sure of

    Thanks
     
  2. jcsd
  3. Jan 4, 2012 #2

    LCKurtz

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    No, I don't think your last integrand is correct, and you didn't show your work. And once you fix that you need to identify the equation of the circle bounding the xy area. Once you have that you will want to consider whether it is easier in rectangular coordinates xy or polar coordinates.

    [Edit] On re-checking, I think your integrand is correct. So just consider my other comments.
     
    Last edited: Jan 4, 2012
  4. Jan 5, 2012 #3
    if [itex]z=g(x,y)=\sqrt{4-x^2-y^2}[/itex], then projection onto xy plane from the plane z=1 gives

    [itex]1=\sqrt{4-x^2-y^2} \implies 3=\sqrt{x^2+y^2}[/itex]

    I am not sure how to know which way is easier...

    In cartesian coordinates would it be [itex]\displaystyle \int \int_R 2x^2y+2y^3dA[/itex]...how would I define the limits?

    In polar coordinates [itex]\displaystyle \int \int_R 2yr^2dA[/itex] dont know how to deal with the y here...
     
  5. Jan 5, 2012 #4

    LCKurtz

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    So your region in the xy plane is the circle ##x^2+y^2=9## and your dA = dydx. It is an ordinary xy double integral. You will need to solve the equation for y for the inside limits.
    The polar coordinate substitutions are ##x = r\cos\theta,\ y = r\sin\theta,\ dA=rdrd\theta## with appropriate ##r,\theta## limits.
     
  6. Jan 5, 2012 #5
    Oh I see...just before I coninue on..is the equation of circle x^2+y^2=9 correct?

    I write the following equation because we have z=1 as our plane and we have to equate this to our [itex]z=\sqrt{4-x^2-y^2}[/itex]

    [itex]1=\sqrt{4-x^2-y^2} \implies 3=\sqrt{x^2+y^2}[/itex].......?
     
  7. Jan 5, 2012 #6

    LCKurtz

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    Didn't I already answer that??
     
  8. Jan 5, 2012 #7
    I was just repeating what you wrote to emphasise that I undestand it based on my last post. I think I do so I will proceed thanks :-)
     
  9. Jan 5, 2012 #8
    Actually, I hav a typo in the highlighed. It should be [itex]3=x^2+y^2[/itex]

    same again x^2+y^2=3, will proceed...
     
  10. Jan 5, 2012 #9

    LCKurtz

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    Yes, good catch.
     
  11. Jan 5, 2012 #10
    In polar coordinates, I set up the integrand to be

    [itex]\displaystyle \int_{0}^{2\pi} \int_{0}^{\sqrt{3}} 2r^4 sin{\theta}drd\theta[/itex]...?

    in rect coordinates [itex]\displaystyle \int \int 2x^2(\sqrt{3-x^2})+2(\sqrt{3-x^2}^3)dxdy[/itex]

    I dont know how to work the limits for this....
     
  12. Jan 5, 2012 #11
    Sorry, I was just focusing how to do it the conventional way regarding polar and rect coordinates.
    I dont understand your post, but I suspect you have recognised a short cut..?
     
  13. Jan 5, 2012 #12

    LCKurtz

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    Yes, that looks correct.
    That isn't the correct integrand for dydx. You had it correct before:$$
    \iint 2x^2y+2y^3\, dydx$$You have to learn how to put in the correct xy limits. y goes from the lower curve to the upper one and x varies from the low to high limits. If you try it, you will see why you want to do this one in polar coordinates. When you work it out, you will get 0 as Sheriff89 has pointed out, but I don't think you should be looking for such shortcuts until you understand better what you are doing.
     
  14. Jan 6, 2012 #13
    Ok I wont attempt to evaluate the rect coordinate one but I would like to know how the limits are obtained...the below is my guess...the x limits im sure are NOT correct because it still has a variable y whereas the answer should be numeric....

    [itex] \displaystyle \int_{0}^{\sqrt{3-y^2}} \int_{1}^{\sqrt{3-x^2}} 2x^2y+2y^3\, dydx[/itex]
     
  15. Jan 6, 2012 #14

    LCKurtz

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    You are right, that isn't correct not least because the outer limits must be constant, as you have noticed. Your circle is ##x^2 + y^2 = 3##. What do you get for the upper and lower curves if you solve for y? That will give you the dy limits. After that, use the extreme values x can take for the x limits.
     
  16. Jan 7, 2012 #15
    for the y limits

    [itex] - \sqrt {3-x^2}[/itex] and [itex] \sqrt {3-x^2}[/itex]


    x limits

    - sqrt3 and sqrt 3.

    THanks
     
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