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

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

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.

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

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.

you obviously have at least one typo in the above post...I would guess 3 typos.

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

You are right Martingale, there were 2 mistakes. 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, I make simple mistakes. Red

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

redrick wrote:You are right Martingale, there were 2 mistakes. 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, I make simple mistakes. Red

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

0<y<1-x

0<x<1

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

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?

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

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

$x+y=1$

and

$z^2=x^2+y^2$

then $z=\sqrt{x^2+y^2}$

is the largest $z$ can be and 0 is the smallest...therefore $0\leq z\leq \sqrt{x^2+y^2}$ ...and so on...