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Evaluate integral for surface of a paraboloid

  1. Mar 18, 2014 #1
    1. The problem statement, all variables and given/known data

    Evaluate s∫∫ lxyzl dS, where S is part of the surface of paraboloid
    z = x2 + y2, lying below the plane z = 1


    2. Relevant equations



    3. The attempt at a solution

    since z=1 and x2+y2=z,

    therefore integral becomes 0∫^1 0∫^(1-x2) xyz dy dx

    which solves to 1/8. Apprently this is wrong but I don't know how
     
  2. jcsd
  3. Mar 18, 2014 #2

    LCKurtz

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    Use the X2 icon above the advanced editing box for superscripts: z = x2 + y2.

    You could start by looking up the formula for dS and surface integrals in your text.
     
  4. Mar 18, 2014 #3

    HallsofIvy

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    The surface, [itex]z = x^2 + y^2[/itex], can be written in terms of parameters,u, v, as
    x= u, y= v, [itex]z= u^2+ v^2[/itex] so that any point on the surface can be written in terms of the position vector [itex]\vec{r(u, v)}= u\vec{ix}+ v\vec{j}+ (u^2+ u^2)\vec{k}[/itex].

    The derivatives with respect to the parameters are
    [itex]\vec{r_u}= \vec{i}+ 2u\vec{j}[/itex] and [itex]\vec{r_v}= \vec{j}+ 2v\vec{k}[/itex]
    Those are two vectors tangent to the surface, and their cross product,
    [tex]\vec{r_v}\times\vec{r_u}= 2u\vec{i}+ 2v\vec{j}+ \vec{k}[/tex]
    times dudv, is the "vector differential of surface area". Its length, [itex]dS= \sqrt{4u^2+ 4v^2+ 1}dudv[/itex] is the "differential of surface area".

    So [itex]\int |xyz|dS= \int\int |uv(u^2+ v^2)(\sqrt{4u^2+ 4v^2+ 1}dudv[/itex]
    Seeing all those sum of squares I would be inclined to put it in "polar coordinates with [itex]u= rcos(\theta)[/itex], [itex]v= r sin(\theta)[/itex]. z goes from 0 up to [itex]z= u^2+ v^2= r^2= 2[/itex] so that becomes
    [tex]\int_{r= 0}^\sqrt{2}\int_{\theta= 0}^{2\pi} |r^4cos(\theta)sin(\theta)|\sqrt{4r^2+ 1}r drd\theta[/tex]

    To handle the absolute value, you will need to determine where [itex]cos(\theta)sin(\theta)[/itex] is positive and where negative.
     
  5. Mar 21, 2014 #4
    It's given that z = 1 though
     
  6. Mar 21, 2014 #5

    LCKurtz

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    What kind of response is that? To whom are you responding?

    If you are responding to my post, I asked you if you knew the formula for dS because your integral is completely wrong. And you haven't answered.

    And I'm guessing you don't understand HallsofIvy's reply because you may not have had general parameterizations yet. But until you give us a real reply, and quote to whom you are replying, how are we to know?
     
  7. Mar 21, 2014 #6
    Sorry LCKurtz, I forgot to mention, it is a reply to Hallsofly, as they've assumed z=2 whereas the question states it is 1. In response to your question, I completely understand where I went wrong now, looked up the formula, followed Hallsofly's method and got the answer of (1/420) (125√5 - 1). Thanks a lot guys!
     
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