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Triple Integral

by the one
Tags: integral, triple
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the one
#1
Jun6-08, 12:10 PM
P: 13
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
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HallsofIvy
#2
Jun6-08, 12:18 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,361
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
#3
Jun6-08, 02:03 PM
P: 443
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
#4
Jun6-08, 02:05 PM
P: 13
Triple Integral

I knew that i was wrong
Diffy
#5
Jun6-08, 02:05 PM
P: 443
I am not so sure about that. I got a different answer. Perhaps you want to show your steps?
the one
#6
Jun6-08, 02:14 PM
P: 13
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
Diffy
#7
Jun6-08, 02:29 PM
P: 443
There you go.


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