Chaning order of integration in double/triple integrals

[mathmanmath]

Hey guys, so I think I can always change the order of integration with double integrals in rectangular coordinates (x,y) and also polar coordinates (by literally just changing the order and leaving the bounds the same). However, I was trying to change the order of integration of triple integrals, and it seemed to me that not every order was possible to obtain since in this particular case there was one variable that did not depend on another one. I was just wondering if it should be possible to compute the integration in 6 different ways or if sometimes it was impossible. In "Calculus 6e" by Edwards and penny there is an example problem of calculating a volume with triple integrals and they claim it can be done in 6 different ways but only chose to show 3 ways. I don´t see how it would be possible to integrate in the order dx dy dz, I tried but failed miserably. I would like to note that this is NOT some kind of homework problem, it is an example problem solved in the book in 3 different ways, but I was trying to see if it was indeed possible to do it in 6 ways in order to study for an exam. I am attaching pictures of the problem, graphs that make it easier to solve and 3 of the alleged 6 solutions. Perhaps someone could help me out here. Thanks so much.

Problem Statement:
http://imgur.com/yGZAx.jpg

Graph:
http://imgur.com/mJk9b.jpg

Solutions:
http://imgur.com/eMskl.jpg

I did not post the images directly because they are quiet large. Thanks again in advance.

 

[Petr Mugver]

Do you mean something like this?

$$V\,\,=\,\,\int_0^1dx\int_{-\sqrt{x}}^{+\sqrt{x}}dy\int_0^{1-x}dz\,\,=\,\,\int_0^1dx\int_{-\sqrt{x}}^{+\sqrt{x}}(1-x)dy\,\,=\,\,\int_0^12(1-x)\,\sqrt{x}\,dx\,\,=\,\,\frac{8}{15}$$

 

[mathmanmath]

Do you mean something like this?

$$V\,\,=\,\,\int_0^1dx\int_{-\sqrt{x}}^{+\sqrt{x}}dy\int_0^{1-x}dz\,\,=\,\,\int_0^1dx\int_{-\sqrt{x}}^{+\sqrt{x}}(1-x)dy\,\,=\,\,\int_0^12(1-x)\,\sqrt{x}\,dx\,\,=\,\,\frac{8}{15}$$

Thanks, but actually i meant the reverse order. Using the format that is used in the book, you would have the three integral signs and then inside dxdydz. I'm actually not familiar with the notation you used... Thanks

 

[HallsofIvy]

Changing the order of integration in triple integrals is not intrinsically different from changing the order in double integrals.

In fact, in the "solution" you link to, they do the problem in three different orders.

If you had, using Peter Mugver's example,
$$\int_{x=0}^1\int_{y= -\sqrt{x}}^{\sqrt{x}}\int_{z= 0}^{1- x} dz dy dx$$

then to "reverse the order of integration" (to dx dy dz), I would note that while x runs from 0 to 1, z runs from 0 to 1- z. The line z= 1- x is the same as x= 1- z and, while x runs from 0 to 1, z runs from 1- 0= 1 to 1-1= 0. To integrate with respect to x first, I would have to have
[tex]\int_{x= 0}^{1- z} dx[/itex].<br /> <br /> Now, I cannot have y going from $-\sqrt{x}$ to $\sqrt{x}$ because I have already integrated with respect to x. But $\sqrt{x}= \sqrt{1- z}$ at the boundary so the next integral would be <br /> [tex]\int_{y= -\sqrt{1- z}}^{\sqrt{1-z}} dy[/itex] <br /> <br /> Finally, of course, z goes between 0 and 1 so we would have <br /> $$\int_{z=0}^1\int_{y=-\sqrt{1- z}}^{\sqrt{1- z}}\int_{x=0}^{1- z}dx dy dz$$[/tex][/tex]

 

[mathmanmath]

Changing the order of integration in triple integrals is not intrinsically different from changing the order in double integrals.

In fact, in the "solution" you link to, they do the problem in three different orders.

If you had, using Peter Mugver's example,
$$\int_{x=0}^1\int_{y= -\sqrt{x}}^{\sqrt{x}}\int_{z= 0}^{1- x} dz dy dx$$

then to "reverse the order of integration" (to dx dy dz), I would note that while x runs from 0 to 1, z runs from 0 to 1- z. The line z= 1- x is the same as x= 1- z and, while x runs from 0 to 1, z runs from 1- 0= 1 to 1-1= 0. To integrate with respect to x first, I would have to have
[tex]\int_{x= 0}^{1- z} dx[/itex].<br /> <br /> Now, I cannot have y going from $-\sqrt{x}$ to $\sqrt{x}$ because I have already integrated with respect to x. But $\sqrt{x}= \sqrt{1- z}$ at the boundary so the next integral would be <br /> [tex]\int_{y= -\sqrt{1- z}}^{\sqrt{1-z}} dy[/itex] <br /> <br /> Finally, of course, z goes between 0 and 1 so we would have <br /> $$\int_{z=0}^1\int_{y=-\sqrt{1- z}}^{\sqrt{1- z}}\int_{x=0}^{1- z}dx dy dz$$[/tex][/tex]

[tex]$$<br /> Wow, seriously, thanks so much for that detailed explanation. I really didn´t know how to do it / thought it wasn't possible. I knew y went from -sqrt(x) to sqrt(x) and that this wouldn´t be of much use in this case, but I didn´t realize I could replace sqrt(x) with sqrt(1-z). Thanks! :)<br /> <br /> Ps: I think you meant "z runs from 0 to 1-x" when you said "z runs from 0 to 1- z"$$[/tex]

 

[mathmanmath]

Changing the order of integration in triple integrals is not intrinsically different from changing the order in double integrals.

In fact, in the "solution" you link to, they do the problem in three different orders.

If you had, using Peter Mugver's example,
$$\int_{x=0}^1\int_{y= -\sqrt{x}}^{\sqrt{x}}\int_{z= 0}^{1- x} dz dy dx$$

then to "reverse the order of integration" (to dx dy dz), I would note that while x runs from 0 to 1, z runs from 0 to 1- z. The line z= 1- x is the same as x= 1- z and, while x runs from 0 to 1, z runs from 1- 0= 1 to 1-1= 0. To integrate with respect to x first, I would have to have
[tex]\int_{x= 0}^{1- z} dx[/itex].<br /> <br /> Now, I cannot have y going from $-\sqrt{x}$ to $\sqrt{x}$ because I have already integrated with respect to x. But $\sqrt{x}= \sqrt{1- z}$ at the boundary so the next integral would be <br /> [tex]\int_{y= -\sqrt{1- z}}^{\sqrt{1-z}} dy[/itex] <br /> <br /> Finally, of course, z goes between 0 and 1 so we would have <br /> $$\int_{z=0}^1\int_{y=-\sqrt{1- z}}^{\sqrt{1- z}}\int_{x=0}^{1- z}dx dy dz$$[/tex][/tex]

[tex]$$<br /> Mmm, actually, after trying to do this out, wouldn't x go from y^2 to 1-z, instead of from 0 to 1-z? I don't think it works if you put from 0 to 1-z...$$[/tex]

 

[Petr Mugver]

Mmm, actually, after trying to do this out, wouldn't x go from y^2 to 1-z, instead of from 0 to 1-z? I don't think it works if you put from 0 to 1-z...

Yes, you are right.