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Math Help - surface area

  1. #1
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    surface area

    Find surface area:
    z^2=2xy
    x+y\leq1,x\geq0,y\geq0

    Got such integral
    \int\limits_0^1dx\int\limits_{1-x}^0 \sqrt{\frac{x^2+y^2+xy}{xy}}dy

    Any suggestions to solve it?
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by waytogo View Post
    Find surface area:
    z^2=2xy
    x+y\leq1,x\geq0,y\geq0

    Got such integral
    \int\limits_0^1dx\int\limits_{1-x}^0 \sqrt{\frac{x^2+y^2+xy}{xy}}dy

    Any suggestions to solve it?
    V=\iint\limits_T z\ dy\ dx

    =\int\limits_0^1\int\limits_0^{1-x}\sqrt{2xy}\ dy\ dx
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  3. #3
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    Quote Originally Posted by alexmahone View Post
    V=\iint\limits_T z\ dy\ dx

    =\int\limits_0^1\int\limits_0^{1-x}\sqrt{2xy}\ dy\ dx
    I think formula you have written is for volume, but I need a surface area. Formula for that in Cartesian coordinates is S=\iint\limits_D\sqrt{1+(\frac{\partial z}{\partial x})^2+(\frac{\partial z}{\partial y})^2}} dx dy. That is from where my definite integral comes.
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  4. #4
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by waytogo View Post
    I think formula you have written is for volume, but I need a surface area. Formula for that in Cartesian coordinates is S=\iint\limits_D\sqrt{1+(\frac{\partial z}{\partial x})^2+(\frac{\partial z}{\partial y})^2}} dx dy. That is from where my definite integral comes.
    I'm sorry; I misread the question.
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  5. #5
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    Can you tell us how you obtained your integrand \sqrt{\frac{x^2+y^2+xy}{xy}} ?
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  6. #6
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    Thank you, Danny! I rechecked my calculus and found a mistakte. So the actual integrand is \frac{1}{\sqrt{2}}\iint\limits_D\frac{x+y}{\sqrt{x  y}}dxdy=\frac{1}{\sqrt{2}}\int\limits_0^1dx\int\li  mits_0^{1-x}\frac{x+y}{\sqrt{xy}}dy=...=\frac{2}{\sqrt{2}}\i  nt\limits_0^1\frac{\sqrt{1-x}(2x+1)}{3\sqrt{x}}dx. And this is where I have stopped for now, because I cannot deal with this integrand either! Any ideas?
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  7. #7
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by waytogo View Post
    Thank you, Danny! I rechecked my calculus and found a mistakte. So the actual integrand is \frac{1}{\sqrt{2}}\iint\limits_D\frac{x+y}{\sqrt{x  y}}dxdy=\frac{1}{\sqrt{2}}\int\limits_0^1dx\int\li  mits_0^{1-x}\frac{x+y}{\sqrt{xy}}dy=...=\frac{2}{\sqrt{2}}\i  nt\limits_0^1\frac{\sqrt{1-x}(2x+1)}{3\sqrt{x}}dx. And this is where I have stopped for now, because I cannot deal with this integrand either! Any ideas?
    \int\limits_0^1\frac{\sqrt{1-x}(2x+1)}{3\sqrt{x}}dx

    Let  x=u^2 \to 1-x=1-u^2 \to dx=2udu

    \frac{2}{3} \int (2u^2+1) \sqrt{1-u^2}  du = \frac{4}{3} \int u^2 \sqrt{1-u^2}du + \frac{2}{3} \int \sqrt{1-u^2}du

    You should be able to evaluate both of these easily (sin substitution should work).
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  8. #8
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    yeah, it always seems easy, when someone gives a clue. thanks, AllanCuz!
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