anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
watchmath
  • watchmath
|dw:1327285708927:dw|
anonymous
  • anonymous
\[\int\limits_{1}^{3}\int\limits_{0}^{6-2x}\int\limits_{0}^{\sqrt{x} } f(x, y, z) dz dx dy\] Is this at all correct?
watchmath
  • watchmath
you want triple or double integral?

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

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