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Thread: 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 $\displaystyle -1 \leq \sin r \leq 0$ for $\displaystyle \pi \leq r \leq 2 \pi$ .... ?
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    Quote Originally Posted by mr fantastic View Post
    Your answer is correct.

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

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

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

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

    10x.
    The variable is r:

    Your integrand is $\displaystyle r \sin r$ and you're integrating between $\displaystyle \pi \leq r \leq 2 \pi$. This integrand is always less than or equal to zero for $\displaystyle \pi \leq r \leq 2 \pi$. 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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    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:

    $\displaystyle u = x - y$ .... (1)

    $\displaystyle v = x + y$ .... (2)

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

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

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

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