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Math Help - Multiple integral

  1. #1
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    Multiple integral

    How come I get a negative answer?

    What's wrong?

    10x in advance.
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  2. #2
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    Quote Originally Posted by asi123 View Post
    How come I get a negative answer?

    What's wrong?

    10x in advance.
    Your answer is correct.

    You do realise that -1 \leq \sin r \leq 0 for \pi \leq r \leq 2 \pi .... ?
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    Quote Originally Posted by mr fantastic View Post
    Your answer is correct.

    You do realise that -1 \leq \sin r \leq 0 for \pi \leq r \leq 2 \pi .... ?
    By sin r you mean \phi?

    like 0 \leq \phi \leq 2\pi?

    10x.
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  4. #4
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    Quote Originally Posted by asi123 View Post
    By sin r you mean \phi?

    like 0 \leq \phi \leq 2\pi?

    10x.
    The variable is r:

    Your integrand is r \sin r and you're integrating between \pi \leq r \leq 2 \pi. This integrand is always less than or equal to zero for \pi \leq r \leq 2 \pi. So it's unsurprising that the result of the integration is less than zero.
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  5. #5
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    Got ya, 10x.

    I have another one, how am I suppose to solve this one, I cant use polar coordinates, right?

    I have no idea.

    10x in advance.
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  6. #6
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    Quote Originally Posted by asi123 View Post
    Got ya, 10x.

    I have another one, how am I suppose to solve this one, I cant use polar coordinates, right?

    I have no idea.

    10x in advance.
    New questions require new threads.


    You need to switch to a new set of coordinates. The integrand suggests the following transformation:

    u = x - y .... (1)

    v = x + y .... (2)

    (1) and (2) imply x = \frac{u + v}{2} and y = \frac{v - u}{2}.

    Jacobian of the transformation: J(u, v) = \frac{1}{2}.

    The region of integration in the xy-plane transforms into the region in the uv-plane bounded by the lines v = u, ~ v = -u, ~ v = 1 and v = 2.

    So the integral becomes \frac{1}{2} \int_{v=1}^{v=2} \int_{u=-v}^{u=v} e^{u/v} \, du \, dv.
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