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Surface Area in Polar Coordinates

  1. Aug 2, 2006 #1
    Find the surface area of the surface z=cosh(sqrt(x^2+y^2)) above the region in the xy plane given in polar coordinates:
    r is between 0 and theta
    theta is between 2 and 4

    Ok. I used the formula:
    Surface area equals the square root of the partial derivative of x squared plus the partial derivative of y squared plus 1. After doing it out, the (x^2+y^2) turned into 1, leaving me with the square root of 1 + (sinhsqrt(x^2+y^2))^2.
    I took the integral of this from 0 to theta by switching the (x^2+y^2) into just r (polar coordinates). I got that this integral equaled cosh(r)tanh(r) and in turn cosh(theta)tanh(theta). Taking this integral from 2 to 4, I got cosh(4)-cosh(2). I thought I did everything correct, but this is not the answer. What did I do wrong?

    Thanks for any help. I hope you can understand my thought process.
     
  2. jcsd
  3. Aug 2, 2006 #2

    benorin

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    Homework Helper

    [tex]\sqrt{1+\mbox{sinh}^2\left(\sqrt{x^2+y^2}\right) } = \sqrt{\mbox{cosh}^2\left(\sqrt{x^2+y^2}\right) } = \mbox{cosh}\left(\sqrt{x^2+y^2}\right)[/tex]

    hence your SA is given by

    [tex]SA = \iint_R \mbox{cosh}\left(\sqrt{x^2+y^2}\right) \, dA = \int_{2}^{4}\int_{0}^{\theta} \mbox{cosh}(r) r \, dr \, d\theta[/tex]

    where the [tex]r \, dr \, d\theta[/tex] is from the transform to polar. Now integrate by parts...
     
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