Need help with this Triple Integral

  • #1
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Homework Statement


Triple Integral (x^6e^y)dV bounded by z=1-y^2 z=0 x=-1 x=1



The Attempt at a Solution


So I chose to try to integrate this in the order dydzdx
My bounds for the dy integral were from zero to (1-z)^(1/2)
my bounds for the dz integral were from 0 to 1
and my bounds for the dx integral were from -1 to 1.

I did the whole crazy integral and got 2/7 which my online thing is saying is wrong X_x doh...

Do these bounds look correct?
 

Answers and Replies

  • #2
vanhees71
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I'm not sure which integration region is meant here, because it sounds not very well defined. I guess, it's meant to integrate over a "paraboloic tunnel-like volume" between [itex]x=-1[/itex] and [itex]x=1[/itex] and from [itex]z=0[/itex] to [itex]z=1-y^2 > 0[/itex]. This implies that also [itex]y \in [-1,1][/itex]. Then your integral would read
[tex]\int_{-1}^{1} \mathrm{d} x \int_{-1}^{1} \mathrm{d} y \int_0^{1-y^2} \mathrm{d} z x^6 \exp y.[/tex]
Perhaps this gives the correct result?
 
  • #3
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yes, I believe it does. What would the integral look like if I evaluated dydzdx? would y go from 0 to (1-z)^(1/2) and z go from 0 to 1?
 
  • #4
vanhees71
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Hm, if you want to do the [itex]y[/itex] integral first (which I'd consider a bad idea if I think about it ;-)), then for each [itex]z \in [0,1][/itex] it should run in [itex]y \in [-\sqrt{1-z},\sqrt{1-z}][/itex]. Then the integral would be
[tex]\int_{-1}^{1} \mathrm{d} x \int_0^1 \mathrm{d} z \int_{-\sqrt{1-z}}^{\sqrt{1-z}} \mathrm{d} y \; x^6 \exp y.[/tex]
Mathematica tells me that this gives the same result as it should be, but after the [itex]y[/itex] integral which is a bit easier in this way, the [itex]z[/itex] integral looks pretty cumbersome!
 

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