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Find the exact volume of a bounded surface. Multiple Integrals.

  1. Nov 4, 2012 #1
    1. Ok, so the question is.. Find the exact volume of the solid bounded above by the surface [itex]z=e^{-x^2-y^2}[/itex], below by the xy-plane, and on the side by [itex]x^2+y^2=1[/itex].



    2. Alright. So, I know that I can use a double integral to find the volume, and switching to polar coordinates would be simpler.[itex] z=e^{-r^2}[/itex] and [itex]r^2 =1[/itex], therefore [itex]z=e^{-1}[/itex].



    3. I'm not sure how to calculate the double integral. I know volume would be the double integral of the top surface minus the bottom surface, but do i need to split the surface into two separate double integrals and add them together? This should be easy but I can't seem to figure this out.
     
    Last edited: Nov 4, 2012
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  3. Nov 4, 2012 #2

    LCKurtz

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    Not quite. ##r=1## only on the bounding surface. But inside the volume you need to use ##e^{-r^2}##.
    Not sure what is bothering you here. What do you get when you set up the integral in polar coordinates using "top - bottom" surface in the integrand as you suggest?
     
  4. Nov 4, 2012 #3

    HallsofIvy

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    You integrate [itex]e^{-r^2}[itex] with r from 0 to 1, [itex]\theta[/itex] from 0 to [itex]2\pi[/itex]. The "differential of area" in polar coordinates is [itex]r dr d\theta[/itex].
     
  5. Nov 4, 2012 #4
    OK here is what I have so far..

    [itex] \int_0^{2\pi} \int_{-rsinθ}^{rsinθ} (e^{-r^2}-{\frac{1}{e}})rdrdθ + \int_0^{2\pi} \int_{-rsinθ}^{rsinθ} ({\frac{1}{e}}rdrdθ)[/itex]

    I'm not sure if this is right.. I may be overthinking this way too much.. haha

    edit: This is just equal to [itex]\int_0^{2\pi} \int_{-rsinθ}^{rsinθ} (e^{-r^2})rdrdθ [/itex] right?
     
    Last edited: Nov 4, 2012
  6. Nov 4, 2012 #5

    LCKurtz

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    Well, they aren't equal because you dropped the ##\frac 1 e##. But where did that come from anyway?? And your inner limits, the ##r## limits, are incorrect. What is the equation of the circle ##x^2+y^2=1## in polar coordinates? Where did the ##\sin\theta## terms come from?
     
  7. Nov 4, 2012 #6
    ok i had [itex] \int_0^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} (e^{-(x^2+y^2)}) [/itex]
    Then I did this and used [itex]r^2=1[/itex], but this is only on the boundry so this is incorrect.
    [itex]
    \begin{align}
    &= \sqrt{1-x^2}\\
    &= \sqrt{r^2-r^2\cos^2{\theta}}\\
    &= r\sqrt{1-\cos^2{\theta}}\\
    &= r\sin{\theta}
    \end{align}


    [/itex]
     
  8. Nov 4, 2012 #7

    LCKurtz

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    In the integrand ##r## varies throughout the circular region. But think about how ##r,\theta## vary to cover a circle. It is the limiting values that go in the limits. You have already figured out that ##\theta## must go from ##0\rightarrow 2\pi##. Now if you start at the origin and move in any particular ##\theta## direction, what are the lower and higher limits ##r## will take?
     
  9. Nov 4, 2012 #8
    r would go from 0 to 1.
    so the equation would be [itex]\int_0^{2\pi} \int_0^1 (e^{-r^2})rdrd\theta [/itex]?
     
  10. Nov 4, 2012 #9

    LCKurtz

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    Yes. Not only is that correct, it is easier eh?
     
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