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Thread: Surface integral - hemisphere

  1. #1
    Mar 2013
    Western Australia

    Surface integral - hemisphere

    Below is a question I have for an assignment. It is saying that number 2 is wrong. Cannot for the life of me work out why. Please help

    Let A be the area of the hemisphere of radius 8 which given by:

    x2+y2+z2=64 where 0z4
    1. Indicate the correct parameterisation of the surface:
    (x,y,z)=S(u,v)=64(cosusinv,sinusinv,cosv); where 0u2π,π3vπ2
    (x,y,z)=S(u,v)=8(cosusinv,sinusinv,cosv); where 0u2π,π3vπ2
    (x,y,z)=S(u,v)=8(cosusinv,sinusinv,cosv); where 0u2π,0vπ3
    (x,y,z)=S(u,v)=8(cosusinv,sinusinv,cosv); where 0v2π,π3uπ2
    2. Find the normal to the surface N(u,v)=SuSv=
    64( -cos u sin^2 v, -sin u sin^2 v, -sin v cosv)
    Answer is a Vector
    3. Evaluate the area A=
    Answer is a Number
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  2. #2
    MHF Contributor
    Sep 2012

    Re: Surface integral - hemisphere

    Hey darekau.

    What exactly is π3 and π2? Also what directions are u and v? (For example is u in the east/west and v in the north/south directions?)
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