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Math Help - Surface Area in Polar Coordinates

  1. #1
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    Surface Area in Polar Coordinates

    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.
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  2. #2
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    Quote Originally Posted by MikeWakefield32
    r is between 0 and theta
    You mean,
    0\leq r \leq \theta?

    That is not possible, cuz the region is unbounded!
    (The curve produces an Archimedean Spiral).
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  3. #3
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    1theta it says. I didn't think that mattered.
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  4. #4
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    My fault the region is bounded. Because you gave theta a limitation. Sorry.
    ---
    Then the integration is,
    \int_A \int \sqrt{1+f_x^2+f_y^2} dA
    Where, A is the region you described. Since A is simpler expressed in polar coordinates we will use polar integration.
    ---
    The surface is,
    f(x,y)=\cosh \sqrt{x^2+y^2}
    Then,
    f_x=\frac{x}{\sqrt{x^2+y^2}} \sinh \sqrt{x^2+y^2}
    Then,
    f_y=\frac{y}{\sqrt{x^2+y^2}} \sinh \sqrt{x^2+y^2}
    The next part is two square each one,
    f_x^2=\frac{x^2}{x^2+y^2}\sinh^2 \sqrt{x^2+y^2}
    f_y^2=\frac{y^2}{x^2+y^2}\sinh^2 \sqrt{x^2+y^2}
    Add them,
    f_x^2+f_y^2=\frac{x^2+y^2}{x^2+y^2}\sinh^2 \sqrt{x^2+y^2}=\sinh^2 \sqrt{x^2+y^2}
    Finally add 1,
    1+f_x^2+f_y^2=1+\sinh^2 \sqrt{x^2+y^2}
    But,
    \cosh^2 u=1+\sinh^2 u
    Therefore,
    1+f_x^2+f_y^2=\cosh^2 \sqrt{x^2+y^2}
    Thus, the integration is,
    \int_A \int \sqrt{\cosh^2 \sqrt{x^2+y^2}} dA
    This simplifies as,
    \int_A \int |\cosh \sqrt{x^2+y^2}| dA
    But, \cosh u >0 thus,
    \int_A \int \cosh \sqrt{x^2+y^2} dA
    Now, finally you can use polar substituion,
    \sqrt{x^2+y^2}\rightarrow r
    \int_2^4 \int_0^{\theta} r \cosh r dr, d\theta
    Can you solve it from here?
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  5. #5
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    Quote Originally Posted by ThePerfectHacker
    Can you solve it from here?
    Let me continue.

    To solve this integral the problem is to find,
    \int r \cosh r dr
    You gonna use integration by parts.
    Let, u=r and v'=\cosh r
    Then, u'=1 and v=\sinh r
    Using parts,
    r\sinh r -\int \sinh r dr
    Thus,
    r\sinh r - \cosh r
    Thus,
    \int_2^4 \int_0^{\theta} r\cosh r dr\, d\theta is,
    \int_2^4 r\sinh r-\cosh r \big|_0^\theta d\theta
    Thus,
    \int_2^4 \theta \sinh \theta - \cosh \theta +1 d\theta
    The problem now is to find,
    \int \theta \sinh \theta d\theta
    Let,
    u=\theta and v'=\sinh \theta
    Then,
    u'=1 and v=\cosh \theta
    Thus, by parts,
    \theta \cosh \theta -\int \cosh \theta d\theta
    Thus,
    \theta \cosh \theta - \sinh \theta
    Thus, you have, (entire integral)
     \theta \cosh \theta -\sinh \theta -\sinh \theta +\theta\big|_2^4
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