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Surface integral z^2=2xy

  1. Apr 30, 2010 #1
    Apostol page 429, problem 4

    Is there a better way to set up this problem or have I made a mistake along the way?
    (ie easier to integrate by different parameterization)

    1. The problem statement, all variables and given/known data
    Find the surface area of the surface [tex]z^2=2xy[/tex] lying above the [tex]xy[/tex] plane and bounded by [tex]x=2[/tex] and [tex]y=1[/tex].

    2. Relevant equations
    [tex]
    S=r(T)
    =\bigg(
    X(x,y),Y(x,y),Z(x,y)
    \bigg)
    =\bigg(
    x,y,\sqrt{2xy}
    \bigg)
    [/tex]
    [tex]
    \frac{\partial r}{\partial x}=(1,0,\frac{\sqrt{2y}}{2\sqrt{x}})
    [/tex]
    [tex]
    \frac{\partial r}{\partial y}=(0,1,\frac{\sqrt{2x}}{2\sqrt{y}})
    [/tex]
    [tex]
    \frac{\partial r}{\partial x}\times\frac{\partial r}{\partial y}
    =\bigg(
    -\frac{\sqrt{2y}}{2\sqrt{x}},-\frac{\sqrt{2x}}{2\sqrt{y}},1
    \bigg)
    [/tex]
    [tex]
    \left|\left|\frac{\partial r}{\partial x}\times\frac{\partial r}{\partial t}\right|\right|
    =\sqrt{1+\frac{2y}{4x}+\frac{2x}{4y}}
    [/tex]
    [tex]
    a(S)=\int_0^1 \int_0^2 \sqrt{1+\frac{2y}{4x}+\frac{2x}{4y}}\,dx\,dy
    [/tex]
     
    Last edited: Apr 30, 2010
  2. jcsd
  3. Apr 30, 2010 #2
    Did I perhaps set something up wrong? Could this be parameterized somehow as an Elliptic Paraboloid?
     
    Last edited: Apr 30, 2010
  4. May 1, 2010 #3

    LCKurtz

    User Avatar
    Homework Helper
    Gold Member

    You are on the right track. It's actually easy to continue. Add the terms under the radical and rewrite is as

    [tex]\frac 1 {\sqrt 2}\sqrt{\frac {(x+y)^2}{xy}}[/tex]

    and take the root in the numerator.
     
  5. May 1, 2010 #4
    Yes, thank you. I was actually inquiring if anyone could find a better way to represent the surface area.
     
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