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Math Help - Surface integral over an annulus

  1. #1
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    Surface integral over an annulus

    Hi, would just like to know if what I have done is correct.
    Find the area integral of the surface z=y^2+2xy-x^2+2 in polar form lying over the annulus \frac{3}{8}\leq x^2+y^2\leq1

    The equation in polar form is r^2\sin^2\theta+2r^2\cos\theta\sin\theta-r^2\cos^2\theta+2

    \displaystyle{\int^{\pi}_{-\pi}}\int^1_{\sqrt{\frac{3}{8}}}(r^2\sin^2\theta+2  r^2\cos\theta\sin\theta-r^2\cos^2\theta+2)\, r\, dr\, d\theta

    Integrating with respect to r

    {\frac {55}{256}}\, \left( \sin \left( \theta \right)  \right) ^{2}+{<br />
\frac {55}{128}}\,\cos \left( \theta \right) \sin \left( \theta<br />
 \right) -{\frac {55}{256}}\, \left( \cos \left( \theta \right) <br />
 \right) ^{2}+5/8<br />

    and with respect to \theta, I get the area as

    \frac{5}{4}\pi

    Does this look ok?
    Last edited by bobred; August 31st 2010 at 09:04 AM.
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  2. #2
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    Quote Originally Posted by bobred View Post
    Does this look ok?
    Yes.
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  3. #3
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    Thanks for checking it over I had some post that said there was a problem, that what I was working out was a volume.

    James
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    Quote Originally Posted by bobred View Post
    Thanks for checking it over I had some post that said there was a problem, that what I was working out was a volume.

    James
    That person probably thinks that any triple integral expresses volume and only volume. It's hard for some people to understand that a single, double or triple integral can express both area and volume. After all, a double integral is a simplified triple integral and a single integral is a simplified double integral.
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    Diamond transform

    Quote Originally Posted by AllanCuz View Post
    That person probably thinks that any triple integral expresses volume and only volume. It's hard for some people to understand that a single, double or triple integral can express both area and volume. After all, a double integral is a simplified triple integral and a single integral is a simplified double integral.
    Since it's been over 25 years I've been in college, I'd like to see the formula that converts a double integral into a single integral (which I understand is called a diamond transform).
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  6. #6
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    Quote Originally Posted by bobred View Post
    Find the area integral of the surface z=y^2+2xy-x^2+2 in polar form lying over the annulus \frac{3}{8}\leq x^2+y^2\leq1
    Quote Originally Posted by bobred View Post
    I had some post that said there was a problem, that what I was working out was a volume.
    Oops, I think there is indeed a problem.

    What you did was to find the volume between the annulus and the surface. But it looks as though the question was actually asking for the surface area of that part of the suface lying above the annulus.

    The formula for surface area says that this area is given by \displaystyle\iint_A\sqrt{\bigl(\tfrac{\partial z}{\partial x}\bigr)^2 + \bigl(\tfrac{\partial z}{\partial y}\bigr)^2+1}\,dxdy, where A denotes the annulus.

    If z = y^2 + 2xy - x^2+2 then \tfrac{\partial z}{\partial x} = 2(y-x), \tfrac{\partial z}{\partial y} = 2(y+x), and \bigl(\tfrac{\partial z}{\partial x}\bigr)^2} + \bigl(\tfrac{\partial z}{\partial y}\bigr)^2} = 8(x^2+y^2) = 8r^2. So the answer to the question is

    . . . . . \begin{aligned}\iint_A\sqrt{8r^2+1}\,rdrd\theta &= \int_0^{2\pi}\int_{\sqrt{3/8}}^1\sqrt{8r^2+1}\,rdrd\theta \\ &= \int_0^{2\pi}\Bigl[\tfrac1{24}(8r^2+1)^{3/2}\Bigr]_{\sqrt{3/8}}^1\,d\theta \\ &= \int_0^{2\pi}\!\!\tfrac{19}{24}\,d\theta = \tfrac{19}{12}\pi.\end{aligned}

    I thought that the question was not very clearly worded, but the fact that those 3/2 powers both turn out to be integers makes me fairly sure that "area integral" is supposed to mean "surface area" in this question.
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    Thanks, I had been looking for an expression of the form (x^2+y^2)=r^2 because the trig expression was horrendous.

    James
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    Quote Originally Posted by wonderboy1953 View Post
    Since it's been over 25 years I've been in college, I'd like to see the formula that converts a double integral into a single integral (which I understand is called a diamond transform).
    You've been out of college for 25 years and you're alias is wonderBOY?!
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  9. #9
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    Quote Originally Posted by AllanCuz View Post
    Quote Originally Posted by wonderboy1953 View Post
    Since it's been over 25 years I've been in college, I'd like to see the formula that converts a double integral into a single integral (which I understand is called a diamond transform).
    You've been out of college for 25 years and you're alias is wonderBOY?!
    That didn't come as a surprise to me. Wonderboy1953 clearly wasn't born yesterday.
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    Quote Originally Posted by Opalg View Post
    That didn't come as a surprise to me. Wonderboy1953 clearly wasn't born yesterday.
    I was never one for numbers!
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    Quote Originally Posted by AllanCuz View Post
    I was never one for numbers!
    Since you were never one for numbers, what brings you to a math website?

    Since this is the second time (besides this thread) that I haven't gotten an answer to my question leads me to believe that no such formula exists to transform a double integral into a single integral. Can you prove me wrong?
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    In some cases doesn't Green's theorem do that?
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    Quote Originally Posted by Danny View Post
    In some cases doesn't Green's theorem do that?
    Doesn't that deal with line integrals?
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  14. #14
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    Green's Theorem relates a surface integral to a line integral. Since surface integrals often decompose into a double iterated integral, and line integrals often decompose into a single integral, you could say that Green's Theorem can reduce some double integrals into a single integral. I'm not sure there is a way, short of actually computing the inner integral, to reduce all double integrals to single integrals.
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    Quote Originally Posted by Ackbeet View Post
    Green's Theorem relates a surface integral to a line integral. Since surface integrals often decompose into a double iterated integral, and line integrals often decompose into a single integral, you could say that Green's Theorem can reduce some double integrals into a single integral. I'm not sure there is a way, short of actually computing the inner integral, to reduce all double integrals to single integrals.
    Finally an answer Ackbeet.
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