Triple Integral

  • Thread starter the one
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  • #1
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Main Question or Discussion Point

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
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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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?
 
  • #3
441
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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.
 
  • #4
13
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I knew that i was wrong
 
Last edited:
  • #5
441
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I am not so sure about that. I got a different answer. Perhaps you want to show your steps?
 
  • #6
13
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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
 
  • #7
441
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There you go.
 

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