1. Not finding help here? Sign up for a free 30min 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!

Surface Integral

  1. May 9, 2013 #1
    1. The problem statement, all variables and given/known data
    Find integral I = ∫∫xz^2 dydz + (x^2y − z^3) dzdx + (2xy + y^2z) dxdy (Integrate over A)
    if A is half a sphere(radius is a). Sphere is given with equation z=(a-x^2-y^2)^1/2 and z=0.

    2. Relevant equations
    The excercise is in 2 parts , find it with just integrating and b) applying gauss's law.

    3. The attempt at a solution
    I just cant understand how i get it ... Every way i can think of , gives me wrong answer, I have to find scalar. If i substitute x from the sphere equation , then in integration bounds it still remains ?
    I know this aint much to go on but help me. Just tell me what i can substitute so i can find this ingtegral or is it even possible ? I dont need whole excercise, i can integrate myself.
    sorry for bad english.
     
  2. jcsd
  3. May 10, 2013 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I would do this as three separate integrals: Doing the "dydz" integral y will go from -a to a and, for each y, z will go from 0 to [itex]\sqrt{a^2- y^2}[/itex]. And, of course, for each y and z, [itex]x= \sqrt{a^2- y^2- z^2}[/itex]. The first integral is
    [tex]\int\int xz^2 dydz= \int_{y=-a}^a\int_{z= 0}^\sqrt{a^2- y^2} z^2\sqrt{a^2- y^2- z^2}dzdy[/tex]
    and similarly for the other two integrals.
     
  4. May 10, 2013 #3
  5. May 11, 2013 #4
    Did it with spherical coordinates now . Tthis is so impossible :(
    http://www.upload.ee/image/3302135/20130511_041747.jpg [Broken]
    http://www.upload.ee/image/3302136/20130511_041718.jpg [Broken]
     
    Last edited by a moderator: May 6, 2017
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Surface Integral
  1. Surface integrals (Replies: 10)

Loading...