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Triple Integral

  1. Jun 6, 2008 #1
    hi everyone
    the integral is :
    [tex]\[
    I = \int\limits_0^1 {\int\limits_0^x {\int\limits_0^y {ydzdydx} } }
    \][/tex]
    I'm not sure about the answer , but i think it'll be
    [tex]\[
    \frac{{x^3 }}{3}
    \][/tex]
    am i right ??????
    thanks
     
  2. jcsd
  3. Jun 6, 2008 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Go back and learn the basics again. Since there is an integral with respect to dx, the result cannot possibly be a function of x. The result here must be a number. Did you forget to do the final integral?
     
  4. Jun 6, 2008 #3
    As halls said, pay very close attention to what variable you are integrating with respect to. If you have a different variable within the integrand treat it as a constant both while integrating and evaluating.
     
  5. Jun 6, 2008 #4
    I knew that i was wrong
     
    Last edited: Jun 6, 2008
  6. Jun 6, 2008 #5
    I am not so sure about that. I got a different answer. Perhaps you want to show your steps?
     
  7. Jun 6, 2008 #6
    Sorry , It'll be 1/12 (won't it ??)
    [tex]\[
    \begin{array}{l}
    \int\limits_0^1 {\int\limits_0^x {\int\limits_0^y {ydzdydx = \int\limits_0^1 {\int\limits_0^x {\left( {\int\limits_0^y {ydz} } \right)} } } } } dydx = \int\limits_0^1 {\int\limits_0^x {y^2 } dydx} \\
    = \int\limits_0^1 {\left( {\int\limits_0^x {y^2 dy} } \right)} dx = \int\limits_0^1 {\frac{{x^3 }}{3}} dx = \left( {\frac{{x^4 }}{{12}}} \right)_0^1 = \frac{1}{{12}} \\
    \end{array}
    \][/tex]
    Thanks
     
  8. Jun 6, 2008 #7
    There you go.
     
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