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Math Help - 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 A be your area, then

    A=\iint_Rds, where R = \left\{ {(x,y):0\le x\le1,\,0\le y\le x^3 } \right\} and ds = \sqrt {1 + \left( {\frac{{\partial z}}<br />
{{\partial x}}} \right)^2  + \left( {\frac{{\partial z}}<br />
{{\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 A be your area, then

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

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

    <br />
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.
    <br />
Zx' = \frac{y}{\sqrt{2xy}}
    <br />
Zy' = \frac{x}{\sqrt{2xy}} <br /> <br />
    =>

    <br />
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
    \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
    Math Engineering Student
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    Careful when writting the integration order.

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

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