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.

Question

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.

watchmath · Answer

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anonymous · Answer

$$\int\limits_{1}^{3}\int\limits_{0}^{6-2x}\int\limits_{0}^{\sqrt{x} } f(x, y, z) dz dx dy$$
Is this at all correct?

watchmath · Answer

you want triple or double integral?

anonymous · Answer

triple

anonymous · Answer

Sorry for the confusion.

watchmath · Answer

sorry I am in the middle of something. I can't help you right now. I'll be back if you patient :).

anonymous · Answer

Thanks!

watchmath · Answer

it is almost right but the order should be dz dy dx and also f(x,y,z)=1

anonymous · Answer

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..

watchmath · Answer

the next easiest is in the order of dz dx dy

anonymous · Answer

$$\int\limits_{0}^{6}\int\limits_{1}^{(6-y)/2}\int\limits_{0}^{\sqrt{x}} f(x,y,z) dz dx dy$$ Would that be correct?

watchmath · Answer

what is the intersection between x=1 and 2x+y=6?