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Volume of Container - Triple Integral

  1. Jul 13, 2009 #1
    1. The problem statement, all variables and given/known data

    A container has a vertical height of 1m, a circular base with radius 1m and a circular top with radius 2m. Use a triple integral and spherical coordinates to evaluate the volume of the container.

    2. The attempt at a solution

    If we set up the problem so that the centre of the container's base is at (z,r)=(0,0) it is clear that the region of integration can be viewed as a union of two Type II regions D1 and D2 where

    [tex]D_1=\{(r,\theta ,z)\|0\leq\theta\leq2 \pi , 0\leq z\leq 1 \mbox{and} 0\leq r\leq 1\}[/tex]
    [tex]D_2=\{(r,\theta ,z)\|0\leq\theta\leq2 \pi , 0\leq z\leq 1 \mbox{and} z+1\leq r\leq 2\}[/tex]

    So that

    [tex]\mbox{Volume of Container}=\int_0^{2\pi}\int_0^1\int_{z+1}^2 r drdzd\theta +\int_0^{2\pi}\int_0^1\int_0^1 r drdzd\theta [/tex]
    [tex]=\int_0^{2\pi}\int_0^1 [\frac{1}{2}r^2]_{z+1}^2dzd\theta + \pi[/tex]
    [tex]=\frac{1}{2}\int_0^{2\pi}\int_0^1 [4-(z^2+2z+1)]dzd\theta + \pi[/tex]
    [tex]=\frac{1}{2}\int_0^{2\pi}\int_0^1 [3-z^2-2z]dzd\theta + \pi[/tex]
    [tex]=\frac{1}{2}\int_0^{2\pi} [3z-\frac{1}{3}z^3-z^2]_0^1 d\theta + \pi[/tex]
    [tex]=\frac{5}{6}\int_0^{2\pi}d\theta + \pi[/tex]
    [tex]=\frac{5}{3}\pi + \pi[/tex]
    [tex]=\frac{8}{3}\pi[/tex]


    This is the first problem of its kind that I'm attempting and I'm not sure if I've done everything correctly. I'm slightly suspicious of my limits of integration...Could someone please check this and tell me if it's correct or not? I've attached an image of the regions D1 and D2.

    Thanks!
    phyz
     

    Attached Files:

  2. jcsd
  3. Jul 13, 2009 #2
    You said spherical. So why have you use cylindrical?
     
  4. Jul 14, 2009 #3
    Thanks :redface:

    However, as I've gone through the effort of typing this all out, had the question asked for cylindrical coordinates, would this have been correct?
     
  5. Jul 14, 2009 #4
    D1 is good.

    The integration for D2 is "good," but as you suggested, the limits are wrong.

    Look at your picture and draw a dot on the z-axis between z=0 and z=1 to represent a generic z value satisfying 0<z<1.

    Now draw a horizontal ray emanating from your dot, going to the right. Your ray will enter D2 at r=f(z) and will exit D2 at r=g(z).

    Now f(z)=1 (the constant function). What is g(z)? Your drawing is mislabeled, but your work shows that you know it is g(z)=z+1.

    Thus, 1<r<z+1 would be the correct limits for r.
     
  6. Jul 14, 2009 #5
    Oh, ok cool. Thanks for the help! :smile:
     
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