Results 1 to 5 of 5

Math Help - Surface Area in Polar Coordinates

  1. #1
    Newbie
    Joined
    Aug 2006
    Posts
    3

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    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).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2006
    Posts
    3
    1theta it says. I didn't think that mattered.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    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?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: February 14th 2011, 05:17 PM
  2. Area in Polar Coordinates 2
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 5th 2010, 10:31 PM
  3. Replies: 1
    Last Post: August 12th 2009, 09:26 AM
  4. Polar coordinates area
    Posted in the Calculus Forum
    Replies: 4
    Last Post: May 4th 2009, 08:09 AM
  5. Area in Polar Coordinates
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 3rd 2008, 01:23 AM

Search Tags


/mathhelpforum @mathhelpforum