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Volume via Triple Integrals.

  1. Nov 6, 2011 #1
    The solid enclosed by the cylinder [itex]x^2 + y^2 = 9[/itex] and the planes y + z = 5 and z=1.

    The biggest part for me (usually) is just being able to find my limits of integration for these problems (any suggestions about that would also be greatly appreciated). I think I found the correct limits for this problem...

    [tex]
    \iiint dV
    [/tex]

    [tex]
    \int_{-3}^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \int_{1}^{5-y} dzdydx
    [/tex]

    [tex]
    \int_{-3}^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} (4-y) dydx
    [/tex]

    At this point I start to get lost. Should I switch it to polar coordinates? I tried to do that from the last step above and it came out wrong. Here's my first step into the polar coordinate switch...

    [tex]
    \int_0^{2\pi} \int_0^1 (4-rsin\theta)rdrd\theta
    [/tex]

    Does this look like I'm headed in the right direction? This chapter is completely confusing me.
     
  2. jcsd
  3. Nov 6, 2011 #2

    SammyS

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    Why change coordinates?

    That's a basic integral.

    What is [itex]\displaystyle \int(4-y)\,dy\,?[/itex]
     
  4. Nov 6, 2011 #3
    I'm doing integral[0,2pi], integral[0,3], integral[1,5-rsintheta] of 1 dz r dr dtheta
    Sorry for the mess. Don't know how to display the integral sign. I got 36 pi.
     
  5. Nov 6, 2011 #4
    Yeah, those limits make sense. The answer is [itex]36\pi[/itex], so you got it. My integrals started looking insane so I figured -rightly- that I was doing something wrong. Thanks.

    Have any general advice for finding the limits? That seems to be my biggest weak-point.
     
  6. Nov 6, 2011 #5
    I know that basic integral, but the limits around it make it really intimidating because you'd have to end up using trig-subs. Right? (Thanks for the help, btw)
     
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