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Thread: Evaluating a double integral over a region

  1. #1
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    Evaluating a double integral over a region

    Hey, I am trying to solve this problem, but I'm not sure if I'm on the right track.
    Evaluate
    $\displaystyle
    \iint_D \frac{1}{(x^2 + y^2)^{\frac{n}{2}}}dA
    $
    where $\displaystyle n$ is an integer and $\displaystyle D$ is the region bounded by the circles with center the origin and radii $\displaystyle r_n$ and $\displaystyle R_n$, $\displaystyle 0<r_n<R_n$.
    What I have so far is the idea to use polar coordinates. In this case we would have
    $\displaystyle
    D = \{(r, \theta)\, |\, r_n\leq r \leq R_n\mbox{ , }0\leq\theta\leq2\pi\}
    $
    (i added the n subscripts to I can differentiate between $\displaystyle r_n$ and $\displaystyle r$)
    I transformed the function using $\displaystyle x = r\cos\theta$ and $\displaystyle y = r\sin\theta$. The resulting expression is
    $\displaystyle
    \int^{2\pi}_0\int^{R_n}_{r_n}\frac{1}{(r^2\cos^2 \theta + r^2\sin^2\theta)^{\frac{n}{2}}}\,r\,dr\,dx
    $
    Do you think this is a good approach?
    Also, now I don't really know how to solve this integral (how to start). Any help? Thank you.
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  2. #2
    Member Walagaster's Avatar
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    Re: Evaluating a double integral over a region

    Quote Originally Posted by nuebie View Post
    Hey, I am trying to solve this problem, but I'm not sure if I'm on the right track.
    Evaluate
    $\displaystyle
    \iint_D \frac{1}{(x^2 + y^2)^{\frac{n}{2}}}dA
    $
    where $\displaystyle n$ is an integer and $\displaystyle D$ is the region bounded by the circles with center the origin and radii $\displaystyle r_n$ and $\displaystyle R_n$, $\displaystyle 0<r_n<R_n$.
    What I have so far is the idea to use polar coordinates. In this case we would have
    $\displaystyle
    D = \{(r, \theta)\, |\, r_n\leq r \leq R_n\mbox{ , }0\leq\theta\leq2\pi\}
    $
    (i added the n subscripts to I can differentiate between $\displaystyle r_n$ and $\displaystyle r$)
    I transformed the function using $\displaystyle x = r\cos\theta$ and $\displaystyle y = r\sin\theta$. The resulting expression is
    $\displaystyle
    \int^{2\pi}_0\int^{R_n}_{r_n}\frac{1}{(r^2\cos^2 \theta + r^2\sin^2\theta)^{\frac{n}{2}}}\,r\,dr\,dx
    $
    Do you think this is a good approach?
    Also, now I don't really know how to solve this integral (how to start). Any help? Thank you.
    Start by simplifying that denominator. What is $\cos^2\theta + \sin^2\theta$?
    Thanks from nuebie
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  3. #3
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    Re: Evaluating a double integral over a region

    Thanks! Here's what I got:
    $\displaystyle
    \frac{1}{(r^2 \cos^2 \theta + r^2 \sin^2 \theta)^{\frac{n}{2}}} = \frac{1}{(r^2 (\cos^2 \theta + \sin^2 \theta))^{\frac{n}{2}}} = \frac{1}{(r^2)^{\frac{n}{2}}} = \frac{1}{r^n}
    $
    Now to solve the integral:
    $\displaystyle
    \int^{2\pi}_0\int^{R_n}_{r_n}\frac{1}{r^n}\mbox{ }r\,dr\,d\theta = \int^{2\pi}_0\int^{R_n}_{r_n}\frac{1}{r^{n-1}}\mbox{ }dr\,d\theta = \int^{2\pi}_0\int^{R_n}_{r_n}r^{1-n}\mbox{ }dr\,d\theta = \int^{2\pi}_{0}\frac{1}{2-n}\Big[r^{2-n}\Big]^{R_n}_{r_n}\,d\theta = \int^{2\pi}_{0}\frac{1}{2-n}(R_n^{2-n} - r_n^{2-n})\,d\theta = \frac{1}{2-n}(R_n^{2-n} - r_n^{2-n}) \Big[\theta\Big]^{2\pi}_0 = \frac{2\pi}{2-n}(R_n^{2-n} - r_n^{2-n})
    $
    Does this look correct? Thanks!
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  4. #4
    Member Walagaster's Avatar
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    Re: Evaluating a double integral over a region

    Quote Originally Posted by nuebie View Post
    Thanks! Here's what I got:
    $\displaystyle
    \frac{1}{(r^2 \cos^2 \theta + r^2 \sin^2 \theta)^{\frac{n}{2}}} = \frac{1}{(r^2 (\cos^2 \theta + \sin^2 \theta))^{\frac{n}{2}}} = \frac{1}{(r^2)^{\frac{n}{2}}} = \frac{1}{r^n}
    $
    Now to solve the integral:
    $\displaystyle
    \int^{2\pi}_0\int^{R_n}_{r_n}\frac{1}{r^n}\mbox{ }r\,dr\,d\theta = \int^{2\pi}_0\int^{R_n}_{r_n}\frac{1}{r^{n-1}}\mbox{ }dr\,d\theta = \int^{2\pi}_0\int^{R_n}_{r_n}r^{1-n}\mbox{ }dr\,d\theta = \int^{2\pi}_{0}\frac{1}{2-n}\Big[r^{2-n}\Big]^{R_n}_{r_n}\,d\theta = \int^{2\pi}_{0}\frac{1}{2-n}(R_n^{2-n} - r_n^{2-n})\,d\theta = \frac{1}{2-n}(R_n^{2-n} - r_n^{2-n}) \Big[\theta\Big]^{2\pi}_0 = \frac{2\pi}{2-n}(R_n^{2-n} - r_n^{2-n})
    $
    Does this look correct? Thanks!
    Yes, as far as it goes. But you need to consider the case where $n=2$ separately because it's different.
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  5. #5
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    Re: Evaluating a double integral over a region

    That's a good remark, thank you
    For the case when $\displaystyle n=2$ I got
    $\displaystyle
    2\pi \ln\Bigg|\frac{R_n}{r_n}\Bigg|
    $
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  6. #6
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    Re: Evaluating a double integral over a region

    Quote Originally Posted by nuebie View Post
    That's a good remark, thank you
    For the case when $\displaystyle n=2$ I got
    $\displaystyle
    2\pi \ln\Bigg|\frac{R_n}{r_n}\Bigg|
    $
    Since $n=2$, this is for the specific case where $R_n = R_2$ and $r_n = r_2$.
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