1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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?

  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]


    [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

    [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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook