Triple integration: f(x,t,z) = (2x - 1)yz^2, S bounded by...  TOPIC_SOLVED

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Triple integration: f(x,t,z) = (2x - 1)yz^2, S bounded by...

Postby redrick on Wed Apr 08, 2009 7:24 pm

I am having difficulty setting up triple integrals. Perhaps an illustration would help.

Suppose f(x,t,z) = (2x - 1)yz^2, S is the solid in the first octant bounded by x + y = 1 and z^2 = x^2 = y^2.

I really don't understand how to set this up. Could anyone explain it to me? Thanks for any help you can be. :clap:
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Re: Triple integration: f(x,t,z) = (2x - 1)yz^2, S bounded by...

Postby Martingale on Wed Apr 08, 2009 9:46 pm

redrick wrote:I am having difficulty setting up triple integrals. Perhaps an illustration would help.

Suppose f(x,t,z) = (2x - 1)yz^2, S is the solid in the first octant bounded by x + y = 1 and z^2 = x^2 = y^2.

I really don't understand how to set this up. Could anyone explain it to me? Thanks for any help you can be. :clap:



you obviously have at least one typo in the above post...I would guess 3 typos.
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Re: Triple integration: f(x,t,z) = (2x - 1)yz^2, S bounded by...

Postby redrick on Wed Apr 08, 2009 10:18 pm

You are right Martingale, there were 2 mistakes. :oops: I fixed them.

I am having difficulty setting up triple integrals. Perhaps an illustration would help.

Suppose f(x,y,z) = (2x - 1)yz^2, S is the solid in the first octant bounded by x + y = 1 and z^2 = x^2 + y^2.

I think when I am freaked out a bit, :shock: I make simple mistakes. :wink: Red
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Re: Triple integration: f(x,t,z) = (2x - 1)yz^2, S bounded by...

Postby Martingale on Wed Apr 08, 2009 10:46 pm

redrick wrote:You are right Martingale, there were 2 mistakes. :oops: I fixed them.

I am having difficulty setting up triple integrals. Perhaps an illustration would help.

Suppose f(x,y,z) = (2x - 1)yz^2, S is the solid in the first octant bounded by x + y = 1 and z^2 = x^2 + y^2.

I think when I am freaked out a bit, :shock: I make simple mistakes. :wink: Red



looks like...
0<z<(x^2+y^2)^(1/2)

0<y<1-x

0<x<1
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Re: Triple integration: f(x,t,z) = (2x - 1)yz^2, S bounded by...

Postby redrick on Thu Apr 09, 2009 2:22 am

Martingale wrote:
looks like...
0<z<(x^2+y^2)^(1/2)

0<y<1-x

0<x<1

I am not sure what you are saying at all. x + y = 1 means that x = 1 - y and y = 1-x

and z^2 = x^2 + y^2 means that z = (x^2 + y^2)^1/2

I don't understand where you are getting the less than symbols and what that has to do with the integration process?
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Re: Triple integration: f(x,t,z) = (2x - 1)yz^2, S bounded by...  TOPIC_SOLVED

Postby Martingale on Thu Apr 09, 2009 2:28 am

redrick wrote:
Martingale wrote:
looks like...
0<z<(x^2+y^2)^(1/2)

0<y<1-x

0<x<1

I am not sure what you are saying at all. x + y = 1 means that x = 1 - y and y = 1-x

and z^2 = x^2 + y^2 means that z = (x^2 + y^2)^1/2

I don't understand where you are getting the less than symbols and what that has to do with the integration process?


you are not integrating on the boundary. so if you are integrating over the region that is in the 1st octant and is bounded by the curves



and




then

is the largest can be and 0 is the smallest...therefore ...and so on...
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