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A double integral

  1. Dec 15, 2006 #1

    quasar987

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    It's about finding the area of the paraboloid z=x²+y² beneath z=2. The area integral is

    [tex]\int\int_{\{(u,v): u^2+v^2<2\}}\sqrt{1+4(u^2+v^2)}dudv[/tex]

    A polar change of variable seems to fits nicely:

    [tex]=\int_0^{\sqrt{2}}\int_0^{2\pi}\sqrt{1+4r^2}rd\theta dr[/tex]

    Then the change of variable [itex]\xi=1+4r^2[/itex] brings the area to

    [tex]2\pi\int_1^9\sqrt{\xi}\frac{d\xi}{8}=\frac{2\cdot 2 \pi}{3 \cdot 8}[\xi^{3/2}]_1^9=\frac{13\pi}{3}[/tex]

    Looks good?
     
    Last edited: Dec 16, 2006
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  3. Dec 15, 2006 #2

    D H

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    You got the right answer. However, you lost a factor of two on the last line of math and somehow gained that factor of two back again to get the correct answer. Those equalities are not equalities as written.
     
    Last edited: Dec 15, 2006
  4. Dec 15, 2006 #3

    quasar987

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    My friends found the area by "setting" z=u²+v² and just evaluating

    [tex]=\int_0^{2}\sqrt{1+4z}dz[/tex]

    They found 13/3 (with no pi). I really can'T find any kind of justification for this integral but they say this is how the teaching assistant did a similar problem.

    Any ways, thanks for confirming D H.

    Btw - I think I just spotted a mistake in my OP. Instead of

    [tex]2\pi\int_1^9\sqrt{\xi}\frac{d\xi}{8}=\frac{2\pi}{3 \cdot 8}[\xi^{3/2}]_1^9=\frac{13\pi}{3}[/tex]

    it should be

    [tex]2\pi\int_1^9\sqrt{\xi}\frac{d\xi}{8}=\frac{2\cdot 2 \pi}{3 \cdot 8}[\xi^{3/2}]_1^9=\frac{26\pi}{3}[/tex]

    (I edited the OP)
     
    Last edited: Dec 16, 2006
  5. Dec 16, 2006 #4

    quasar987

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    Oh I see. Ok, we're back as 13/3, but with the right amount of two's in the last line.
     
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