Triple Integral

  1. hi everyone
    the integral is :
    I = \int\limits_0^1 {\int\limits_0^x {\int\limits_0^y {ydzdydx} } }
    I'm not sure about the answer , but i think it'll be
    \frac{{x^3 }}{3}
    am i right ??????
  2. jcsd
  3. HallsofIvy

    HallsofIvy 41,260
    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. 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. I knew that i was wrong
    Last edited: Jun 6, 2008
  6. I am not so sure about that. I got a different answer. Perhaps you want to show your steps?
  7. Sorry , It'll be 1/12 (won't it ??)
    \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}} \\
  8. There you go.
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