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Math Help - Find the surface area S

  1. #1
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    Find the surface area S

    S is the upper cap cut from the sphere x^2+y^2+z^2=25 by the cylinder x^2+y^2=16

    Ok, so

    \displaystyle \iint_{R} \frac{|\nabla f|}{|\nabla f * p|}dA

    R: x^2+y^2\leq 16 on the xy-plane[/tex]

    P is a vector normal to R, so p=k

    \nabla f=2xi+2yj+2zk

    |\nabla f|=2\sqrt{2}

    |\nabla f * p|=2z

    So, now I have

    \displaystyle \iint_{R} \frac{2\sqrt{2}}{2z}dA

    After factoring out constants and rewriting z as a function of x and y I have

    \displaystyle \sqrt{2}\iint_{R}\frac{1}{\sqrt{25-x^2-y^2}}dA

    Now, converting to polar coordinates, I get

    \displaystyle \sqrt{2}\int_{0}^{2\pi}\int^{4}_{0}\frac{1}{\sqrt{  25-r^2}}rdrd\theta

    Ok, I let u=25-r^2. Then, du=-2rdr

    Which then gives me


    \displaystyle \sqrt{2}\int_{0}^{2\pi}\frac{1}{2}\int^{4}_{0}\fra  c{-1}{\sqrt{u}}dud\theta

    Then I get

    \displaystyle \sqrt{2}\int_{0}^{2\pi}[ -\sqrt{25-r^2}}]^{4}_{0} d\theta

    Now, I have

    \displaystyle \sqrt{2}\int_{0}^{2\pi}2d\theta

    So

    \frac{\sqrt{2}}{2}[\theta]^{2\pi}_{0}

    =\frac{\sqrt{2}}{2}(2\pi)

    The answer in the book is 20\pi. can someone help me find my mistake(s)? Thanks.
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  2. #2
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    Quote Originally Posted by downthesun01 View Post
    S is the upper cap cut from the sphere x^2+y^2+z^2=25 by the cylinder x^2+y^2=16

    Ok, so

    \displaystyle \iint_{R} \frac{|\nabla f|}{|\nabla f * p|}dA

    R: x^2+y^2\leq 16 on the xy-plane[/tex]

    P is a vector normal to R, so p=k

    \nabla f=2xi+2yj+2zk

    |\nabla f|=2\sqrt{2}
    Well, here's a problem: |2xi+ 2yj+ 2zk|= \sqrt{4x^2+ 4y^2+ 4z^2}= 2\sqrt{x^2+ y^2+ z^2}= 2\sqrt{25}= 10, not " 2\sqrt{2}".

    |\nabla f * p|=2z

    So, now I have

    \displaystyle \iint_{R} \frac{2\sqrt{2}}{2z}dA

    After factoring out constants and rewriting z as a function of x and y I have

    \displaystyle \sqrt{2}\iint_{R}\frac{1}{\sqrt{25-x^2-y^2}}dA

    Now, converting to polar coordinates, I get

    \displaystyle \sqrt{2}\int_{0}^{2\pi}\int^{4}_{0}\frac{1}{\sqrt{  25-r^2}}rdrd\theta

    Ok, I let u=25-r^2. Then, du=-2rdr

    Which then gives me


    \displaystyle \sqrt{2}\int_{0}^{2\pi}\frac{1}{2}\int^{4}_{0}\fra  c{-1}{\sqrt{u}}dud\theta

    Then I get

    \displaystyle \sqrt{2}\int_{0}^{2\pi}[ -\sqrt{25-r^2}}]^{4}_{0} d\theta

    Now, I have

    \displaystyle \sqrt{2}\int_{0}^{2\pi}2d\theta

    So

    \frac{\sqrt{2}}{2}[\theta]^{2\pi}_{0}

    =\frac{\sqrt{2}}{2}(2\pi)

    The answer in the book is 20\pi. can someone help me find my mistake(s)? Thanks.
    Last edited by HallsofIvy; December 9th 2010 at 02:41 AM.
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  3. #3
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    Thanks. I'll try reworking it now
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  4. #4
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    Also, check your u substitution

    when you are using your u substitution, note:

    1/sqrt(u) = u^-^1^/^2 so when you are taking it's anti-derivative, you will gain a 2 as well. It should look like this: 2u^1^/^2. After taking the derivative of this value you will be back to your original answer. This will also be what gets the 5/2 back to being a whole 5, which then allows it to become a 10 again, resulting in your answer of 20pi as per the text.
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