Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Triple integral for bounded regions

  1. May 11, 2015 #1
    Checking my steps and answer. Thanks in advance!

    Compute [tex]\int_0^3 \int_0^2 \int_1^3 xyz\ dz\ dy\ dx[/tex].

    [tex]\int_0^3 \int_0^2 \frac{xyz^2}{2} \Big|_1^3 = \frac{9xy}{2}-\frac{xy}{2} = \frac{8xy}{2} = 4xy[/tex]
    [tex]\int_0^3 2xy^2 \Big|_0^2[/tex]
    [tex]\int_0^3 8x\ 4x^2 \Big|_0^3 = 36[/tex]
     
    Last edited by a moderator: May 11, 2015
  2. jcsd
  3. May 11, 2015 #2
    Yes, it's right.
     
  4. May 11, 2015 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Because the limits of integration on all integrals are constants, and the integrand is a simple product, this can be done as
    [tex]\left(\int_0^3 x dx\right)\left(\int_0^3 dy\right)\left(\int_1^3 z dz\right)[/tex]
    [tex]\left(\frac{1}{2}x^2\right)_0^3\left(\frac{1}{2}y^2\right)_0^2\left(\frac{1}{2}z^2\right)_1^3[/tex]
    [tex]\left(\frac{9}{2}- 0\right)\left(2- 0\right)\left(\frac{9}{2}- \frac{1}{2}\right)[/tex]
    [tex]\left(\frac{9}{2}\right)\left(2\right)\left(4\right)= 9(4)= 36[/tex]

     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook