Triple Integral: Is \[\frac{{x^3 }}{3}\] Right?

[the one]

hi everyone
the integral is :
\[<br /> I = \int\limits_0^1 {\int\limits_0^x {\int\limits_0^y {ydzdydx} } } <br /> \]
I'm not sure about the answer , but i think it'll be
\[<br /> \frac{{x^3 }}{3}<br /> \]
am i right ?
thanks

 

[HallsofIvy]

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?

 

[Diffy]

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.

 

[the one]

I knew that i was wrong

 

[Diffy]

I am not so sure about that. I got a different answer. Perhaps you want to show your steps?

 

[the one]

Sorry , It'll be 1/12 (won't it ??)
\[<br /> \begin{array}{l}<br /> \int\limits_0^1 {\int\limits_0^x {\int\limits_0^y {ydzdydx = \int\limits_0^1 {\int\limits_0^x {\left( {\int\limits_0^y {ydz} } \right)} } } } } dydx = \int\limits_0^1 {\int\limits_0^x {y^2 } dydx} \\ <br /> = \int\limits_0^1 {\left( {\int\limits_0^x {y^2 dy} } \right)} dx = \int\limits_0^1 {\frac{{x^3 }}{3}} dx = \left( {\frac{{x^4 }}{{12}}} \right)_0^1 = \frac{1}{{12}} \\ <br /> \end{array}<br /> \]
Thanks

 

[Diffy]

There you go.