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Surface Area of Parameterized Region

  1. Mar 27, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the surface area S of the portion of the hyperbolic paraboloid:

    ## r(u,v) = \langle (u+v),(u-v),uv \rangle ##

    for which:

    ## u^2 + v^2 <= 225 ##


    2. Relevant equations

    (Surface Area for Parameterized Region:)
    ##\int \int ||\frac {\partial r} {\partial u} \times \frac {\partial r} {\partial v}||dA## Where the double integral is over the region R.

    3. The attempt at a solution

    Okay, so first, I found the partial derivatives of u and v with respect to the parameterized function r(u,v):

    ## \frac {\partial r} {\partial u} = \langle 1,1,v \rangle , \frac {\partial r} {\partial v} = \langle 1,-1,u \rangle ##

    After that, I found the cross product:

    ## \langle (u+v), (v-u), -2 \rangle ##

    Then the norm:

    ## \sqrt{2(u^2+v^2+2)}##

    It's at this point that I realized this integrand was pretty rough, and should probably switch to polar:

    ##\sqrt{2r^2+4}##

    Since I got all that set, I need to find some limits, and also convert them to polar coordinates. The limits for u I had as [-3,3], and the limits for v were [##-\sqrt{225-u^2},\sqrt{225-u^2}##]

    So, the new limits for the polar system I received for ##\theta## was [0,2\pi], and for r I got [0, 15], due to the circle being at radius 15.

    Now I set up the integral and evaluate.

    ##\int_0^{2\pi} \int_0^15 \sqrt{2r^2+4} r dr d\theta##

    Using U substitution I get an integral of this form (I plan on plugging the u back in, so I leave the limits the same):

    ##\int_0^{2\pi} \int_0^15 (\frac 1 4) \sqrt{u} du d\theta##

    ##\int_0^{2\pi} (\frac 1 4) [ (\frac 2 3) (2r^2+4)^{\frac 3 2}|_0^{15}]##

    The integral just gets super sloppy here, and I'm not sure if I've even set it up correctly, or if you can even convert to polar from a parameterized function?
     
  2. jcsd
  3. Mar 27, 2016 #2

    Nathanael

    User Avatar
    Homework Helper

    Everything you did looks good to me, but you stopped just one step short of an answer.
     
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