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Thread: surface integral

  1. #1
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    surface integral

    Hello, I need help with this one.

    Calculate the area of the surface defined by:
    z=sqrt(2xy), where 0<= x <=1 and 0<= y <= x^3

    /Alex
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  2. #2
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    Let $\displaystyle A$ be your area, then

    $\displaystyle A=\iint_Rds,$ where $\displaystyle R = \left\{ {(x,y):0\le x\le1,\,0\le y\le x^3 } \right\}$ and $\displaystyle ds = \sqrt {1 + \left( {\frac{{\partial z}}
    {{\partial x}}} \right)^2 + \left( {\frac{{\partial z}}
    {{\partial y}}} \right)^2 } \,dy\,dx.$

    Can you proceed from there?
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  3. #3
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    Quote Originally Posted by Krizalid View Post
    Let $\displaystyle A$ be your area, then

    $\displaystyle A=\iint_Rds,$ where $\displaystyle R = \left\{ {(x,y):0\le x\le1,\,0\le y\le x^3 } \right\}$ and $\displaystyle ds = \sqrt {1 + \left( {\frac{{\partial z}}
    {{\partial x}}} \right)^2 + \left( {\frac{{\partial z}}
    {{\partial y}}} \right)^2 } \,dy\,dx.$

    Can you proceed from there?
    I've got this far.
    $\displaystyle
    Zx' = \frac{y}{\sqrt{2xy}} $
    $\displaystyle
    Zy' = \frac{x}{\sqrt{2xy}}

    $
    =>

    $\displaystyle
    A = \iint_R \sqrt{1 + (\frac{y}{\sqrt{2xy}})^2 +(\frac{x}{\sqrt{2xy}})^2} dxdy = \iint_R \sqrt{\frac{x^2}{2xy} + \frac{y^2}{2xy} + 1} dxdy$
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by yOdA View Post
    I've got this far.
    $\displaystyle
    Zx' = \frac{y}{\sqrt{2xy}} $
    $\displaystyle
    Zy' = \frac{x}{\sqrt{2xy}}

    $
    =>

    $\displaystyle
    A = \iint_R \sqrt{1 + (\frac{y}{\sqrt{2xy}})^2 +(\frac{x}{\sqrt{2xy}})^2} dxdy = \iint_R \sqrt{\frac{x^2}{2xy} + \frac{y^2}{2xy} + 1} dxdy$
    $\displaystyle \sqrt{\frac {x^2}{2xy} + \frac {y^2}{2xy} + 1} = \sqrt{\frac {x^2 + y^2 + 2xy}{2xy}} = \sqrt{\frac {(x + y)^2}{2xy}} = \frac {x + y}{\sqrt{2xy}} = \frac {\sqrt{x}}{\sqrt{2y}} + \frac {\sqrt{y}}{\sqrt{2x}}$
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  5. #5
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    Careful when writting the integration order.

    I wrote it in the order $\displaystyle dy\,dx$ so that you can evaluate directly the double integral. Since you wrote it in the order $\displaystyle dx\,dy$ you have to reverse integration order, so we require $\displaystyle 0\le y\le1,\,\sqrt[3]y\le x\le1.$

    Evaluating the double integral with $\displaystyle dx\,dy$ or $\displaystyle dy\,dx$ order, both answers are in agreement.
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