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Math Help - Area Integral

  1. #1
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    Question Area Integral

    I have the following question:

    Express as an area integral in polar co-ordinates the area of the surface z = 5 + xy lying over the annulus defined by 2 < x^2 + y^2 < 3. Hence evaluate the area in terms of pi.

    I know i have to get the function:

    g(r,0) = sqrt 1 + (df/dr)^2 + (df/d0)^2


    then evaluate it over the interval 2 to 3.

    The problem is i cant seem to get the required answer 13.628pi. All my answers are nuts!!!


    are the partial derivatives 0^2 and r^2 ?

    and does the 5 in the function get added to the 1 in g(x,y)?

    In other words should I have

    g(r, 0) = sqrt 6 + 0^2 + r^2 ??


    N.B. 0 is meant to be theta.
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  2. #2
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    Quote Originally Posted by iwishiunderstood View Post
    I know i have to get the function:

    g(r,0) = sqrt 1 + (df/dr)^2 + (df/d0)^2

    then evaluate it over the interval 2 to 3.
    Are you sure you do? Your formula is rather more reminiscent of a computation of arc length, or kind of.
    To compute the area of a surface parametrized by (u,v)\mapsto M(u,v), you need to integrate the "surface element" \left\|\frac{\partial M}{\partial u}\times\frac{\partial M}{\partial v}\right\|du\, dv (where \times is the cross product of vectors) over the range of u and v (this is a double integral). In your case, r and \theta replace u and v, and M(r,\theta)=(r\cos\theta,r\sin\theta,5+(r\cos\thet  a)(r\sin\theta)) where r\in[\sqrt{2},\sqrt{3}] and \theta\in[0,2\pi]. I haven't tried much to compute the integral I obtained this way, but it doesn't look very nice anyway...

    I hope you'll be able to carry on using this.

    Laurent.

    (I forgot to mention that a more convenient expression for the surface element is \sqrt{EG-F^2} du\,dv where E=\left\|\frac{\partial M}{\partial u}\right\|, G=\left\|\frac{\partial M}{\partial v}\right\| and F=< \frac{\partial M}{\partial u},\frac{\partial M}{\partial v}>)
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  3. #3
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    definitely sure - I have the notes for the assignment from my tutor and these say(as well as the text book) to find g(x,y) in polar co-ordinates.
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  4. #4
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    OK. In fact, my formula reduces, in polar coordinates, to: \int_0^{2\pi}\int_{\sqrt{2}}^{\sqrt{3} }g(r,\theta)dr\,d\theta, where g(r,\theta)=\sqrt{r^2+r^2 (\partial_r f)^2+(\partial_\theta f) ^2}. This formula with the r^2 is homogeneous to an area, while yours is not ( f is a length, so that \partial_r f is homogeneous to a constant and \partial_\theta f is homogeneous to a length).

    And in the present case f(r,\theta)=5+xy=5+r^2\cos\theta\sin\theta.

    This gives: g(r,\theta)=\sqrt{r^2+r^4}, so the area is 2\pi\int_{\sqrt{2}}^{\sqrt{3}}\sqrt{r^2+r^4}dr=2\p  i\int r\sqrt{1+r^2}dr =\frac{2}{3}\pi [(1+r^2)^{3/2}]_{\sqrt{2}}^{\sqrt{3}}=\frac{2}{3}\pi((1+3)^{3/2}-(1+2)^{3/2})=\frac{2}{3}(8-\sqrt{27})\pi. There must be a mistake, but the idea is there.
    Last edited by Laurent; September 19th 2008 at 02:32 AM.
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  5. #5
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    Quote Originally Posted by Laurent View Post
    There must be a mistake, but the idea is there.
    The mistake is in the statement of the problem: the annulus must be given by 2^2\leq x^2+y^2\leq 3^2, so that the integration is between 2 and 3, and you get that the area is \pi times \frac{2}{3}(10^{3/2}-5^{3/2})=\frac{2}{3}(\sqrt{1000}-\sqrt{125})=\frac{10\sqrt{5}}{3}(2\sqrt{2}-1)\simeq 13.628. This way you have the expected result.
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  6. #6
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    thanks for that - quality bit of help. cheers!!!

    can i just ask how you got the srt r^2 + r^4 in the answer please?
    Last edited by iwishiunderstood; September 19th 2008 at 07:17 AM.
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