I need to setup the integral in three different ways.. solid bounded between the planes x=1, y=0, and 2x+y=6, bounded below by z=0 and bounded above by z=sqrt(x). Its just setting up the different orders of integrals, no solving.

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I need to setup the integral in three different ways.. solid bounded between the planes x=1, y=0, and 2x+y=6, bounded below by z=0 and bounded above by z=sqrt(x). Its just setting up the different orders of integrals, no solving.

Mathematics
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\[\int\limits_{1}^{3}\int\limits_{0}^{6-2x}\int\limits_{0}^{\sqrt{x} } f(x, y, z) dz dx dy\] Is this at all correct?
you want triple or double integral?

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Other answers:

triple
Sorry for the confusion.
sorry I am in the middle of something. I can't help you right now. I'll be back if you patient :).
Thanks!
it is almost right but the order should be dz dy dx and also f(x,y,z)=1
so would it be correct if it was dz dy dx? I need to find 2 more orders that's not the same, but will ultimately bring me to the same answer. I don't need to solve for f(x,y,z), just change the orders around..
the next easiest is in the order of dz dx dy
\[\int\limits_{0}^{6}\int\limits_{1}^{(6-y)/2}\int\limits_{0}^{\sqrt{x}} f(x,y,z) dz dx dy\] Would that be correct?
what is the intersection between x=1 and 2x+y=6?
3
If only there is a one on one chat thing..
2(1)+y=6 so y=4
so the limit for dy is from 0 to 4
OH. okay so it would be \[\int\limits_{0}^{4}\int\limits_{1}^{(6-y)/2}\int\limits_{0}^{\sqrt{x}} dz dx dy\]?
yes :)
now how would i do something like dx dz dy?
Would this work out also?\[\int\limits_{1}^{\sqrt{3}}\int\limits_{0}^{z^2}\int\limits_{0}^{6-2x} dy dx dz\]?
let see
I'm so lost... I know that its not right..

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