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Evaluating a Surface Integral

  1. Jun 30, 2013 #1
    1. The problem statement, all variables and given/known data
    [itex]\int\int _{S} \sqrt{1 + x^2 + y^2} dS[/itex]
    Given that S is the surface of which [itex]\textbf{r}(u,v) = u\cdot cos(v)\textbf{i}+u\cdot sin(v)\textbf{j}+v\textbf{k}[/itex] is a parametrization. [itex](0 \leq u \leq 1, 0 \leq v \leq \pi)[/itex]

    2. Relevant equations
    [itex]dS = \left| \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v} \right| du dv[/itex]


    3. The attempt at a solution
    I think the answer is [itex]\frac{4}{3}\pi[/itex], because [itex]dS = \sqrt{1+u^2}du dv[/itex] and [itex]\sqrt{1+x^2+y^2} = \sqrt{1+u^2}[/itex] using the given parameterization.
     
  2. jcsd
  3. Jun 30, 2013 #2

    LCKurtz

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    Is there a question there? Do you expect us to work it out for ourselves just to check your answer? If you would show your work we could tell you at a glance whether or not your answer is correct.
     
  4. Jun 30, 2013 #3
    Alright, this was my full calculation:
    [itex]dS = \left| \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v} \right| du dv
    \\= \left| (cos(v), sin(v), 0) \times (-u\cdot sin(v),u\cdot cos(v), 1) \right| du dv
    \\= \left| (sin(v),-cos(v),u\cdot (cos^2(x) + sin^2(x))) \right| du dv
    \\= |(sin(v),-cos(v),u)| du dv
    \\= \sqrt{cos^2(v)+sin^2(v)+u^2} du dv
    \\= \sqrt{1+u^2}du dv[/itex]

    [itex]\int\int_{S} \sqrt{1+x^2+y^2}dS = \int_0^\pi \int_0^1 \sqrt{1+r_1(u,v)^2+r_2(u,v)^2} \cdot \sqrt{1+u^2} du dv
    \\=\int_0^\pi \int_0^1 \sqrt{1+(u\cdot cos(v))^2+(u\cdot sin(v))^2}\cdot \sqrt{1+u^2} du dv
    \\=\int_0^\pi \int_0^1 \sqrt{1+u^2} \sqrt{1+u^2} du dv
    \\=\int_0^\pi \int_0^1 (1+u^2) dudv
    \\=\int_0^\pi \frac{4}{3}dv
    \\=\frac{4}{3} \pi
    [/itex]
     
  5. Jun 30, 2013 #4

    LCKurtz

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    Very good. Yes it looks correct.
     
  6. Jun 30, 2013 #5
    Thanks! :)
     
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