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Surface integral problem: ##\iint_S (x^2+y^2)dS##

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  1. Dec 3, 2014 #1
    1. The problem statement, all variables and given/known data
    ##\iint_S (x^2+y^2)dS##, ##S##is the surface with vector equation ##r(u, v)## = ##(2uv, u^2-v^2, u^2+v^2)##, ##u^2+v^2 \leq 1##

    2. Relevant equations
    Surface Integral. ##\iint_S f(x, y, z)dS = \iint f(r(u, v))\left | r_u \times r_v \right |dA##,
    3. The attempt at a solution
    First, I tried to shift the form of ##\iint_S (x^2+y^2)dS##. ##x^2+y^2##.
    ##x^2+y^2## = ##u^4v^4+2u^2v^2+v^4##, and ##\left | r_u \times r_v \right |## = ##\sqrt{32v^4+64u^2v^2+32u^4}##
    Thus, the initial integral becomes ##\iint_S 2^2sqrt{2}(u^2+v^2)^3 dudv##
    I used polar coordinates, as u = rsin##\theta## and v = ##rsin\theta##. ##0\leq r\leq 1##, ## 0 \leq \theta \leq 2\pi##. As result, the answer came to be ##2^3\sqrt{2}/5*\pi## but the answer sheet says its zero. Am I missing something?
     
  2. jcsd
  3. Dec 3, 2014 #2

    RUber

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    I checked your algebra, and all looks good up to the switch to polar coords.
    Did you try without the substitution?
     
  4. Dec 3, 2014 #3
    I tried, but it was way too complexed. Anyway, if my algebra and switghing to polar coords is not bad, is it presumable that the 'answer sheet' is wrong?
     
  5. Dec 3, 2014 #4

    LCKurtz

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    You know immediately just looking at the problem that the answer must be positive. The integrand is nonnegative. So, yes, the answer sheet is wrong.
     
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