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Integral 4

  1. Jan 7, 2012 #1
    1. The problem statement, all variables and given/known data

    Evaluate the surface integral



    2. Relevant equations

    f(x,y,z)=y where sigma is part of the hyperboloid y=x^2+z^2 that lies inside cylinder x^2+z^2=4



    3. The attempt at a solution

    For [itex] \displaystyle S= \int \int_R \sqrt(z_x^2+z_y^2+1) dA[/itex]

    I calculate [itex]\displaystyle \sqrt(z_x^2+z_y^2+1)=\sqrt( \frac{x^2}{\sqrt{y-x^2}}+ \frac{1}{\sqrt{y-x^2}}+1)[/itex]
    I have tried simplifying this further but it still looks ugly...any suggestions on how to continue and evalute the surface integral?

    Thanks
     
  2. jcsd
  3. Jan 7, 2012 #2
    I assume you've projected on to the xz-plane, after all that would be the easier option.

    It may be easier to stick with the explicit equation already given for y. So you attain the normal vector

    [itex] n = \frac{(2x,-1,2z)}{\sqrt{4x^2 + 4z^2 + 1}} [/itex]

    Projecting on to the xz-plane we get;

    [itex] |n.j| = \frac{1}{\sqrt{4x^2 + 4z^2 + 1}} [/itex]

    So

    [itex] dS = \frac{dA_y}{|n.j|} = \sqrt{4x^2 + 4z^2 + 1} dA_y [/itex]

    After that it would be ideal to convert to polar co-ordinates. You also seem to have left the original y out of the surface integral.
     
  4. Jan 7, 2012 #3
    I was intending to project onto xy plane becasue we can get z as a function of x and y.... How would one know whether to project onto the standard xy or in your case the xz plane...

    Can you explain this

    Projecting on to the xz-plane we get;

    [itex] |n.j| = \frac{1}{\sqrt{4x^2 + 4z^2 + 1}} [/itex]
     
  5. Jan 7, 2012 #4
    OK, it makes sense to project onto the xz plane

    I set up the integrand to be

    [itex] \displaystyle \int_{0}^{2 \pi} \int_{0}^{2} r sin \theta \sqrt(4r^2+1) r dr d \theta [/itex]...?
     
  6. Jan 7, 2012 #5
    Almost

    [itex] y = x^2 + z^2 = r^2 [/itex]

    So you get [itex] \displaystyle \int_{0}^{2 \pi} \int_{0}^{2} r^2 \times \sqrt(4r^2+1)dS [/itex]

    Where dS is the surface of this circle, which is given by [itex] r dr d \theta [/itex]

    [itex] \displaystyle \int_{0}^{2 \pi} \int_{0}^{2} r^2 \times \sqrt(4r^2+1) \times rdrd \theta [/itex]

    You can use the substitution [itex] u = 4r^2 + 1 [/itex]

    Wondered why this question seemed familiar, is it out of the James Stewart Calculus book?
     
  7. Jan 8, 2012 #6
    Wolfram alpha seems to use u=r^2. Both give different answers.

    My attempt for u=4^r2+1 is du=8rdr implies rdr =du/8 and r^2=(u-1)/4

    The integrand becomes

    [itex] \displaystyle \frac{1}{32} \int_0^2 u^{3/2} -u^{1/2} du[/itex]...?
     
  8. Jan 8, 2012 #7
    Did you change your limits?
     
  9. Jan 9, 2012 #8
    Forgot that. Thanks
     
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