# Triple Integral: Is $\frac{{x^3 }}{3}$ Right?

• the one

#### the one

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 ?
thanks

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?

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.

I knew that i was wrong

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

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

There you go.