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How to evaluate a Triple Integral

  1. Aug 14, 2009 #1
    How would I evaluate the integral [tex]\int\int\int_{G}[/tex] [tex]\sqrt{4x^2+9y^2}[/tex] dV,
    where G is the elliptic cylinder 4x2+9y2 [tex]\leq[/tex] 25,
    0 [tex]\leq[/tex] z [tex]\leq[/tex] 6
     
  2. jcsd
  3. Aug 14, 2009 #2
    You can easily integrate for Z, since there's no Z in the integrand. Next, you need to do two changes of variables. The first is simply x'=2x and y'=3y. You will get a very familiar integral (don't forget the jacobian). Then you go in polar coordinates (Mr. Jacob is here too) to integrate a very very simple equation.
     
  4. Aug 17, 2009 #3
    I was given a hint:
    Try modified cylindrical coordinates with[tex]\vartheta[/tex] retaining its usual meanings, but with the meaning of r changed so that r2 = 4x2 + 9y2. You will need to make definitions such as x = a1r cos[tex]\vartheta[/tex]
    and y = a2 sin[tex]\vartheta[/tex], for some suitable numbers a1 and a2. You will need to calculate the jacobian for this mapping.
     
  5. Aug 17, 2009 #4
    Yes you can do it. It's even simpler than what I said but it's exactly the same thing, except that it's in one single step.

    So you have your answer or your stuck there?
     
  6. Aug 18, 2009 #5
    Just a little bit stuck. I tried it but it didn't seem to work out.
     
  7. Aug 19, 2009 #6
    I don't know how to start it. If someone would show me how to do this it would be really appreciated.
     
  8. Aug 19, 2009 #7
    You need to define (x,y,z) with (r,t,z'):

    [tex]
    x = \frac{r}{2}cos\theta
    [/tex]

    [tex]
    y = \frac{r}{3}sin\theta
    [/tex]

    [tex]
    z = z'
    [/tex]

    [tex]
    \left| J \right| = \begin{vmatrix} \frac{1}{2}cos\theta & -\frac{r}{2}sin\theta & 0 \\ \frac{1}{3}sin\theta & \frac{r}{3}cos\theta & 0 \\ 0 & 0 & 1 \end{vmatrix} = \frac{r}{6}
    [/tex]

    Noting that using this transformation, r is really equal to [itex]\sqrt{4x^2 + 9y^2}[/itex], you finally have this integral:

    [tex]
    \int_{r=0}^5 { \int_{\theta=0}^{2 \pi} { \int_{z'=0}^6 { r \frac{r}{6}dz' d\theta dr } } }
    [/tex]
     
  9. Aug 19, 2009 #8
    Of course. It's so obvious. Thanks heaps.
     
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