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Math Help - area of ellipsoid

  1. #1
    C.E
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    area of ellipsoid

    Can somebody please explain how to do the following question, I am struggling to get started and also don't know why it is asking for an area as opposed to a volume.

    1.(i). Calculate the area of the ellipsoid (x^2/4) +(y^2/4)+z^2=1 that lies above the xy plane using the parameterisation:
    r(θ,φ)=(2cosφsinθ, 2sinφsinθ, cosθ).
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  2. #2
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    Area = \iint_S \left|\vec r'_\varphi \times \vec r'_\theta\right|d\theta d\varphi

    Surface integral - Wikipedia, the free encyclopedia

    The surface in question is the upper half of an ellipsoid (only the positive part of the z axis).
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  3. #3
    MHF Contributor Calculus26's Avatar
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    See attachment in later post
    Last edited by Calculus26; April 26th 2009 at 02:49 PM.
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  4. #4
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    Quote Originally Posted by Calculus26 View Post
    C.E

    See Attachment

    double check my Calculations
    How on earth did you obtain

     <br />
|| \vec{r}_{\phi} \times \vec{r}_{\theta}|| = 4(\sin^2 \theta + \sin^2 \theta \cos^2 \theta)?<br />
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  5. #5
    MHF Contributor Calculus26's Avatar
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    That's why i said double check my calculations--- i rushed

    what did you come up with for comparison sake?
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  6. #6
    MHF Contributor Calculus26's Avatar
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    Does 4 *sin^4(t) +16sin^2(t)sin^2(t) sound better?
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  7. #7
    MHF Contributor Calculus26's Avatar
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    See new attachment

    again better double check
    Attached Files Attached Files
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  8. #8
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    Quote Originally Posted by Calculus26 View Post
    Does 4 *sin^4(t) +16sin^2(t)sin^2(t) sound better?
    I was thinking \sqrt{ 4 \sin^4 \phi +16\sin^2 \phi \cos^2 \phi }
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  9. #9
    MHF Contributor Calculus26's Avatar
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    What an idiot I am -- forgot the square root
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  10. #10
    C.E
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    Ok, I understand what you have done but have no Idea how to integrate \sqrt{4sin^4(\theta)+16(cos(\theta)sin(\theta))^2}
    I tried reducing it to sin(\theta)\sqrt{1+7cos^2\theta)}
    Then letting u=cos(\theta) it becomes \sqrt{1+7u^2} but I am really stuck at this point.
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  11. #11
    MHF Contributor Calculus26's Avatar
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    If my calculations are correct

    4sin^4(t)+16cos^2(t)sin^2(t) = 4sin^2(t)[sin^2(t)+4cos^2(t)]


    = 4sin^2(t)[1+3cos^2(t)]

    taking sqrt roots

    =2sin(t)[ 1 +3cos^2(t) ]^1/2

    letting u = cos(t) you obtain

    int(1+3u^2)^1/2du

    Use the trig sub u = 1/[3^(1/2)]tan(a) 3^(1/2)du = sec^2(a)da

    You get an integral of the form sec^3(a)

    whose AD is 1/2[ln[sec(a)+tan(a)] +sec(a)tan(a)]
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