## anonymous 4 years ago 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.

1. watchmath

|dw:1327285708927:dw|

2. 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?

3. watchmath

you want triple or double integral?

4. anonymous

triple

5. anonymous

Sorry for the confusion.

6. watchmath

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

7. anonymous

Thanks!

8. watchmath

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

9. 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..

10. watchmath

the next easiest is in the order of dz dx dy

11. 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?

12. watchmath

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

13. anonymous

3

14. anonymous

If only there is a one on one chat thing..

15. watchmath

2(1)+y=6 so y=4

16. watchmath

so the limit for dy is from 0 to 4

17. anonymous

OH. okay so it would be $\int\limits_{0}^{4}\int\limits_{1}^{(6-y)/2}\int\limits_{0}^{\sqrt{x}} dz dx dy$?

18. watchmath

yes :)

19. anonymous

now how would i do something like dx dz dy?

20. anonymous

Would this work out also?$\int\limits_{1}^{\sqrt{3}}\int\limits_{0}^{z^2}\int\limits_{0}^{6-2x} dy dx dz$?

21. watchmath

let see

22. anonymous

I'm so lost... I know that its not right..