Surface integral z^2=2xy

  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. Did I perhaps set something up wrong? Could this be parameterized somehow as an Elliptic Paraboloid?
     
    Last edited: Apr 30, 2010
  4. LCKurtz

    LCKurtz 8,393
    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. Yes, thank you. I was actually inquiring if anyone could find a better way to represent the surface area.
     
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