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Homework Help: Double integral

  1. Apr 20, 2007 #1
    Int Int (x*y^3 + 1) dS

    where S is the surface r=1, tetha from 0 to Pi and z from 0 to 2.

    How can I solve this integral? I haven't got a clue.
     
  2. jcsd
  3. Apr 21, 2007 #2
    What a peculiar question that asks you to evaluate an integrand in cartesian coordinates but defines the surface in cylindrical. I would start by converting x*y^3 to cylindrical coordinates. You should be able to see the solution more clearly then.
     
  4. Apr 21, 2007 #3

    HallsofIvy

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    The given surface is half a cylinder. Convert to cylindrical coordinates. Do you know how to find the "differential of surface area"?
     
  5. Apr 21, 2007 #4
    No. But I know that x=rcos(t) and y=rsin(t) in cylindrical coords.
     
  6. Apr 21, 2007 #5

    HallsofIvy

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    Odd, if someone expects you to be able to do a problem like this then surely they expect you to be able to integrate over a surface area! Perhaps you need to review your text.

    Since we are given that r=1, we have [itex]x= cos(\theta)[/itex], [itex]y= sin(\theta)[/itex], and z= z. The "position vector" of any point on the surface is [itex] cos(\theta)\vec{i}+ sin(\theta)\vec{j}+ z\vec{j}[/itex].

    The derivative with respcect to [itex]\theta[/itex] is [itex]-sin(\theta)\vec{i}+ cos(\theta)\vec{j}[/itex] and the derivative with respect to z is [itex]\vec{k}[/itex]. The "fundamental vector product" is the cross product of those two vectors:[itex]cos(\theta)\vec{i}+ sin(\theta)\vec{j}[/itex] and the length of that gives the "differential of surface area". [itex]\sqrt{cos^2(\theta)+ sin^2(\theta)}= 1[itex] so [itex]d\sigma = d\theta dy[/itex].
     
    Last edited by a moderator: Apr 21, 2007
  7. Apr 21, 2007 #6
    My book simply says that

    Int Int (xy^3+1) dS = Int Int dS + 0 = 2*Pi

    I don't understand why...


    I converted into cylindrical coords:

    Int Int (r^5cos(t)sin((t))^3 + r) dr dt

    But what are the limits now?
     
  8. Apr 21, 2007 #7

    HallsofIvy

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    Had a little too much wine with dinner! There should be no "+ 1" in the square root. [itex]d\sigma= d\theta dz[/itex] is the correct differential.

    In these coordinates, xy3+ 1 is [itex]cos(\theta)sin^3(\theta)+ 1[/itex] so you want to find
    [tex]\int_{\theta= 0}^\pi\int_{z=0}^2 cos(\theta)sin^3(\theta)dzd\theta[/itex]
     
  9. Apr 21, 2007 #8
    I understood it now finding some examples in my book. Thank you so much!
     
    Last edited: Apr 21, 2007
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