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Change the order of triple integration

  • #1

Homework Statement



Find all 5 other orders of intergration


Homework Equations



[tex]\int_{0}^{1}\int_{0}^{x^2}\int_{0}^{y}dzdydx[/tex]

The Attempt at a Solution



I am really confused as to how to represent it graphicaly so I don't have any visuals to help. Can you guys help me? Thank you SO much!
 
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Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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Use [ tex ] and [ /tex ] without the spaces:
[tex]\int_{0}^{1}\int_{0}^{x^2}\int_{0}^{y}dzdydx[/tex]

Draw an xy- coordinate system. x goes from 0 to 1 so pencil in vertical lines at x= 0 and x= 1. For every x, y goes from 0 to [itex]x^2[/itex] so draw the graph y= x^2. In the xy-plane the figure is bounded by [itex]y= x^2[/itex], y= 0, and x= 1. For every x and y, z goes from 0 to y. That's a plane at 45 degrees cutting through x-axis.


To change the order of integration to, say, dxdydz, note that y reaches a max of 1 (at x= 1) and so z can have a max of 1. The "outer integral", with respect to z, has limits z= 0 and z= 1. For each z, the equation z= y becomes y= z. y ranges from 0 to z. Finally, since [itex]y= x^2[/itex] gives [itex]x= \sqrt{y}[/itex] (x is positive, remember), x ranges from 0 to [itex]\sqrt{y}[/itex].

I realize now, this is an error. with y going from 0 to [itex]x^2[/itex], the region of integration is from below and to the right of the graph of [itex]y= x^2[/itex], the integration in x should be from [itex]\sqrt{y}[/itex] to 1.
 
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  • #3
Thank you for getting me started.
I drew 3 separate graphics to help me.

I still am confused on how to find all 5. How do you go from one to the other? I am not certain i am keeping the right variables
 
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  • #4
So basicaly, you are saying one of the answers would be

[tex] \int_{0}^{1}\int_{0}^{z}\int_{0}^{sqrt(y)}dxdydz [/tex]

Then how do i proceed to find, let's say, dz dx dy?

I find it hard to visualize, once again. I find the graphic really helps, and I can't seem to get it right in 3D

Thank you for you help

A very confused Canadian student ;)
 
  • #5
Wow...i have been trying to figure it out since i first posted this, 2 hours ago. I really don't understand how to proceed. Any help would be much appreciated. Our teacher doesn't TEACH, he just scramble notes on the board.

It can't be that complicated? With double intergrals i am doing fine....
 
  • #6
So basicaly, you are saying one of the answers would be

[tex] \int_{0}^{1}\int_{0}^{z}\int_{0}^{sqrt(y)}dxdydz [/tex]

Then how do i proceed to find, let's say, dz dx dy?

I find it hard to visualize, once again. I find the graphic really helps, and I can't seem to get it right in 3D

Thank you for you help

A very confused Canadian student ;)
I no expert by any means, but I don't think this dxdydz works. Just to check, try having wolfram alpha or wolfram alpha's triple integrator evaluate it with f(x,y,z) equaling, say, 1 (or do it by hand, whichever), and compare that evaluation to your evaluation of f(x,y,z) with dzdydx bounds.

If they don't match, try letting

sqrt(y)≤x≤1
z≤y≤1
0≤z≤1

for your dxdydz bounds.
 
  • #7
After a rather long and arduous amount of work I get these other four bounds:

dxdzdy:
0≤y≤1
0≤z≤y
sqrt(y)≤x≤1

dydxdz:
0≤z≤1
sqrt(z)≤x≤1
0≤y≤z

dzdxdy:
0≤y≤1
sqrt(y)≤x≤1
0≤z≤y

dydzdx:
0≤y≤z
0≤z≤x^2
0≤x≤1
 
  • #8
557
1
It really get complicated, but in 3d it's still manageable. It got me hard and arduous work too.
We have 6 representation altogether. The last two are actually the trickier because you have to split the volume in 2 and manage 2 integrals then to sum them.

What I get is:
[tex]\int_{0}^{1} \int_{x^2}^{1} \int_{0}^{y} dz\ dy\ dx[/tex]

[tex]\int_{0}^{1} \int_{0}^{\sqrt y} \int_{0}^{x^2} dz\ dx\ dy[/tex]

[tex]\int_{0}^{1} \int_{y}^{1} \int_{0}^{\sqrt z} dx\ dz\ dy[/tex]

[tex]\int_{0}^{1} \int_{0}^{z} \int_{0}^{\sqrt y} dx\ dy\ dz[/tex]

[tex]\int_{0}^{1} \int_{\sqrt z}^{1} \int_{z}^{1} dy\ dx\ dz + \int_{0}^{1} \int_{0}^{\sqrt z} \int_{x^2}^{1} dy\ dx\ dz[/tex]

[tex]\int_{0}^{1} \int_{x^2}^{1} \int_{\sqrt x}^{1} dy\ dz\ dx + \int_{0}^{1} \int_{0}^{x^2} \int_{x}^{1} dy\ dz\ dx[/tex]

One must absolutely try to visualize the volume and to make intersections with the basic planes [tex]x=0, x=1, y=0, y=1, z=0, z=1[/tex] to get an idea of how it works.

I hope it's ok, but it was the first of this kind, so I'm not really sure.
 
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  • #9
I think you're right about needing to divide up the region in the last two. However, I also think that you're bounds on dzdxdy are not correct. If you integrate with 1 as your integrand, you find 1/5 as your answer. However, if you use the dzdydx bounds with the same integrand your answer is 1/10.
 
  • #10
557
1
I think you're right about needing to divide up the region in the last two. However, I also think that you're bounds on dzdxdy are not correct. If you integrate with 1 as your integrand, you find 1/5 as your answer. However, if you use the dzdydx bounds with the same integrand your answer is 1/10.
Yep. I edited the 1st and the 2nd.
In fact now there is more "simmetry" between the 1-2 3-4 5-6 couples.
Thanks.
 
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  • #11
Guys thank you so much for your help.
I think that is why i blocked (amongst other reasons!!)
I couldn't figure it out without adding two

I am so thankful: you guys are better online than the actual teacher ;)
 
  • #12
It really get complicated, but in 3d it's still manageable. It got me hard and arduous work too.
We have 6 representation altogether. The last two are actually the trickier because you have to split the volume in 2 and manage 2 integrals then to sum them.

What I get is:
[tex]\int_{0}^{1} \int_{x^2}^{1} \int_{0}^{y} dz\ dy\ dx[/tex]

[tex]\int_{0}^{1} \int_{0}^{\sqrt y} \int_{0}^{x^2} dz\ dx\ dy[/tex]

[tex]\int_{0}^{1} \int_{y}^{1} \int_{0}^{\sqrt z} dx\ dz\ dy[/tex]

[tex]\int_{0}^{1} \int_{0}^{z} \int_{0}^{\sqrt y} dx\ dy\ dz[/tex]

[tex]\int_{0}^{1} \int_{\sqrt z}^{1} \int_{z}^{1} dy\ dx\ dz + \int_{0}^{1} \int_{0}^{\sqrt z} \int_{x^2}^{1} dy\ dx\ dz[/tex]

[tex]\int_{0}^{1} \int_{x^2}^{1} \int_{\sqrt x}^{1} dy\ dz\ dx + \int_{0}^{1} \int_{0}^{x^2} \int_{x}^{1} dy\ dz\ dx[/tex]

One must absolutely try to visualize the volume and to make intersections with the basic planes [tex]x=0, x=1, y=0, y=1, z=0, z=1[/tex] to get an idea of how it works.

I hope it's ok, but it was the first of this kind, so I'm not really sure.
The only problem is the given integral is
[tex]\int_{0}^{1} \int_{0}^{x^2} \int_{0}^{y} dz\ dy\ dx[/tex]
and you wrote
[tex]\int_{0}^{1} \int_{x^2}^{1} \int_{0}^{y} dz\ dy\ dx[/tex]
 
  • #13
557
1
The only problem is the given integral is
[tex]\int_{0}^{1} \int_{0}^{x^2} \int_{0}^{y} dz\ dy\ dx[/tex]
and you wrote
[tex]\int_{0}^{1} \int_{x^2}^{1} \int_{0}^{y} dz\ dy\ dx[/tex]
Arghm, another big mistake. I focussed on the method rather than on the actual problem data. Then the integrals should change accordingly.

[tex]\int_{0}^{1} \int_{0}^{x^2} \int_{0}^{y} dz\ dy\ dx[/tex]

[tex]\int_{0}^{1} \int_{\sqrt{y}}^{1} \int_{0}^{x^2} dz\ dx\ dy[/tex]

[tex]\int_{0}^{1} \int_{\sqrt{z}}^{1} \int_{0}^{x^2} dy\ dx\ dz[/tex]

[tex]\int_{0}^{1} \int_{0}^{x^2} \int_{z}^{1} dy\ dz\ dx[/tex]

[tex]\int_{0}^{1} \int_{z}^{1} \int_{\sqrt{y}}^{1} dx\ dy\ dz[/tex]

[tex]\int_{0}^{1} \int_{0}^{y} \int_{\sqrt{z}}^{1} dx\ dz\ dy[/tex]
 

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