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Math Help - surface area using double integral

  1. #1
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    surface area using double integral

    my problem is like this: find the surface area of the portion of the cone  z=\sqrt{x^2+y^2} that lies inside the cylinder  x^2+y^2=2x .

    My solution:
     x^2+y^2=2x
    =>  (x-1)^2+y^2=1
    So my circle now has center at (1,0), radius = 1
    =>  y=\sqrt{1-(x-1)^2}

    A= \int_{0}^{2} \int_{-\sqrt{1-(x-1)^2}}^{\sqrt{1-(x-1)^2}} <br />
\sqrt {(\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2 +1}<br />
dydx

    A= \int_{0}^{2} \int_{-\sqrt{1-(x-1)^2}}^{\sqrt{1-(x-1)^2}} <br />
\sqrt 2}<br />
dydx
    A= \int_{0}^{2} 2\sqrt2(\sqrt{1-(x-1)^2}dx

    I stuck at this point, I dont know if I set up a right limit of integration, please help?
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  2. #2
    A Plied Mathematician
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    I agree with your steps so far. I would probably do a series of substitutions: u substitution to get rid of the x-1, and then a trig substitution for the remainder of the integral. Or, after the u substitution, you could simply recognize the remaining integral as the area under a well-known type of curve. That would enable you simply to write down the answer without further computation.
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  3. #3
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    I got it. Thanks
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  4. #4
    A Plied Mathematician
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    You're welcome. Have a good one!
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