# Limit change on triple integral

1. Apr 18, 2005

### trancefishy

ok, so i've got this triple integral: $$\int_{0}^{2} \int_{0}^{y^3} \int_{0}^{y^2} f(x,y,z) dz\, dx\, dy\$$

what i want to do is get the other five integrals that are equivalen. i've got correctly 3 of them, but, for the life of me, cannot get dy dz dx and dy dx dz to work out.

i've bumbled around with $$\int_{0}^{8} \int_{0}^{x^{2/3}} \int_{x^{1/3}}^{2} f(x,y,z) dy\, dz\, dx\$$
and
$$\int_{0}^{8} \int_{0}^{4} \int_{x/z}^{2} f(x,y,z) dy\, dz\, dx\$$
among many others, and can't seem to get it pegged down.

if i integrate to check (just having a constant in the beginning) the correct answer seems to be 32/3. anyways, ANY clues to this, would help me a ton. i'm positive i'm missing something blatant or just have the region thought out incorrectly.

2. Apr 18, 2005

### Theelectricchild

Ill give you a very general suggestion first, then you can try it out. If it's not working I'll take a look at it furthur...

Draw out your spacial region as best as you can--- and the idea is that you draw what that region looks like on the specific coordinate planes such as xy xz etc... then you can make the appropriate change of the limits for the specific differential element.

3. Apr 18, 2005

### whozum

Another tip: The outermost integral will always have constant limits, and the inner integrals usually translate 2 functions of each variable into ones of the next integral. If you didnt udnerstand that, just pay attention to the first hint I gave you, the second part is confusing.

4. Apr 19, 2005

### trancefishy

i'm pretty familiar with those things, i've done this on others, and i just cannot figure out what is sticking me on this particular problem. i've tried what i *thought* was all possible ways (for dy anyhow). let's see, the region is $$z=y^2, x=y^3, 0<y<2$$ (those are "...or equal to" sings). it's mostly getting dy correct. i've tried combining the functions, i'm still quite stuck. x/z as a lower limit is not working as i hoped it would, either.

5. Apr 19, 2005

### whozum

More appropriately, as z goes from 0 to y^3, and x goes from 0 to y^3, y goes to 0 and 2.

Start rearranging that.

6. Apr 19, 2005

### saltydog

Hello Trance,

I've looked at this one and I've concluded you can't evaluate this by evaluating dy first and still remain with a single triple integral. dz and dx are ok first but not dy. The reason I say this is because the region being integrated over with respect to a "dy" is not smooth but rather has an edge. However I'm no wiz and may stand to be corrected. I'll stick by my claim until proven otherwise.

Salty

Oh yea, if someone comes up with a solution, I wish to verify that the proposed solution returns the same value as:

$$\int_{0}^{2} \int_{0}^{y^3} \int_{0}^{y^2} (x^2+y^2+z^2) dz\, dx\, dy\ =\frac{8096}{45}$$

Don't wish to be a pain, just precise.

Last edited: Apr 19, 2005
7. Apr 19, 2005

### Theelectricchild

were we told what f(x,y,z) is?

8. Apr 19, 2005

### saltydog

It doesn't matter what the integrand is. I just used that one for a test. The triple integral over the indicated volume should be the same no matter how you integrate it.

9. Apr 19, 2005

### trancefishy

i've just left the integrand as the constant "1", and all my correct answers came out as 32/3. i have been beginning to suspect this is not possible to keep it as a single triple integral. i have tried, serioulsy, every possibly combination of things, i must have done a dozen integrals in all different ways, and none are correct.

so far, i have entertained the idea or parametricizing the whole thing, and also splitting it in two. thanks though, at least i'm not the only one who does not see something that can start with dy.

10. Apr 20, 2005

### trancefishy

well, i solved this. it was problematic, and so, i graphed this thing in maple, so i could spin around a lot more easily, and see what it was i was doing. i am really glad maple exists, or it may have taken me a little longer and a lot more drawing to get it right.... to anyone who may care, here is one of the solutions (of the two that start with dy). if you see one you can figure out the other, obviously.

$$\int_{0}^{4} \int_{0}^{8} \int_{z^{1/2}}^{2} f(x,y,z) dy\, dx\, dz\ - \int_{0}^{4} \int_{z^{3/2}}^{8} \int_{z^{1/2}}^{x^{1/3}} f(x,y,z) dy\, dx\, dz\$$

on one hand, i sort of wish i had someone to give me a better hint, as it would have saved me over 4 hours of time (i think i spent about 6 on this one alone, if not more). though, i did get a sweet feeling of accomplishment. and also, a "well, that was actually pretty easy"

the key was to graph $$x=z^{3/2}$$ in maple, and then look at the answers of several wrong integrals, to really get a feel for how the triple integral behaves, and how you define regions with it.

11. Apr 20, 2005

### whozum

12. Apr 20, 2005

### dextercioby

Last edited: Nov 22, 2006
13. Apr 20, 2005

### trancefishy

computers are notorious for that sort of thing. i'm terrible careful using maple, and generally use it just for graphing things. i'd never even dream of using it for something like integration.

thanks, both of you, for checking my work though, i'm glad to see that it was confirmed correct by human and almost correct by machine, and i learned to trust maple even less (even the graphs are incorrect sometimes, it tries to make everything a continuous function...)

14. Apr 20, 2005

### whozum

Whats wrong with maple
I like maple :(

15. Apr 21, 2005

### trancefishy

don't get me wrong, maple is totally sweet, but, as the adage goes, only trust it as far as you can throw it.

speaking of which, anyone have a really cool trick they can make maple do, like your favourite graph or anything else really interesting?

16. Apr 21, 2005

### dextercioby

There are a lotta things he can do.But sometimes,he's just plain stupid.

Daniel.

17. Apr 21, 2005

### whozum

18. Apr 21, 2005

### dextercioby

And what graph is that?

Daniel.

19. Apr 21, 2005

### whozum

I dont remember, lol.