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Math Help - Vector Calculus

  1. #1
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    Vector Calculus

    Hi.

    Calculate the integral:

    \iint\limits_S\ f(x,y,z)dS

    f(x, y, z) = xy

    r(u, v) = (u - v, u + v, 2u + v + 1) u[0, 1] ; v[0, u]

    What i didn't get is how to calculate the limits of the double integral, how to transform v[0, u] in v[0, number].

    A.: \sqrt {14} / 6

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    I've tried to calculate in this way (but it didn't worked)

    A = \int_0^u \int_0^1 (u^2 - v^2) \left ||{{\vec{r_u}.\vec{r_v}}} \right || dudv
    Last edited by PedroMinsk; June 21st 2011 at 11:16 AM.
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  2. #2
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    Re: Vector Calculus

    You need to know the region in the (u, v)-plane that you're integrating over. This should be specified in the problem. What you've written, u[0,1];v[0,u], doesn't make any sense.
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  3. #3
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    Re: Vector Calculus

    Quote Originally Posted by PedroMinsk View Post
    Hi.

    Calculate the integral:

    \iint\limits_S\ f(x,y,z)dS

    f(x, y, z) = xy

    r(u, v) = (u - v, u + v, 2u + v + 1) u[0, 1] ; v[0, u]

    What i didn't get is how to calculate the limits of the double integral, how to transform v[0, u] in v[0, number].

    A.: \sqrt {14} / 6

    ______________________________

    I've tried to calculate in this way (but it didn't worked)

    A = \int_0^u \int_0^1 (u^2 - v^2) \left ||{{\vec{r_u}.\vec{r_v}}} \right || dudv
    As ojones said, this makes no sense. The "outside" integral has an upper limit of "u" and so the integral will be a function of u, not a constant. Did you mean
    \int_{u= 0}^1\int_{v= 0}^u (u^2- v^2) ||\vec{r_u}\times\vec{r_v}||dudv?
    (not dot product!)

    That should work.
    Last edited by HallsofIvy; June 23rd 2011 at 08:32 AM.
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  4. #4
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    Jun 2011
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    Re: Vector Calculus

    I'm sorry. I solved it now. It was so simple...

    Thanks.
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