Explaining the Concept of Triple Integrals in Calculus

[karush]

15.6.4 Evaluate the iterated integral
$$\int_0^1\int_y^{2y}\int_0^{x+y}
6xy\, dy\, dx\, dz$$OK this is an even problem # so no book answer
but already ? by the xy

 

[topsquark]

Re: 326 tripel integrals with 6xy

15.6.4 Evaluate the iterated integral
$$\int_0^1\int_y^{2y}\int_0^{x+y}
6xy\, dy\, dx\, dz$$OK this is an even problem # so no book answer
but already ? by the xy

Start with the inner integral:
[math]\int _0^{x + y} 6xy ~dy[/math]

We already have a problem. If this were a Physics problem then I would be just fine. But in a Math context the variable y you are integrating over and the y in the limit are really two different variables. Did you derive this from a problem or did you find it listed this way?

Let's make it:
[math]\int _0^{x + y} 6xy' ~ dy'[/math]

x is constant over the y' integral so...
[math]\int _0^{x + y} 6xy' ~ dy' = 6x \int _0^{x + y} y' ~ dy' = 6x \cdot \dfrac{1}{2} y'^2 \left . \right | _0^{x + y} = 3x(x + y)^2[/math]

Now try the x integral. Remember, the y is constant for this one.

-Dan

 

[Opalg]

Re: 326 tripel integrals with 6xy

15.6.4 Evaluate the iterated integral
$$\int_0^1\int_y^{2y}\int_0^{x+y}
6xy\, dy\, dx\, dz$$OK this is an even problem # so no book answer
but already ? by the xy

It looks to me as though the notation here clashes with what most mathematicians would use. I would reverse the order of $dy\, dx\, dz$, so that the integral becomes $$\int_0^1\left(\int_y^{2y}\left(\int_0^{x+y} 6xy\, dz\right)\, dx\right)\, dy.$$ Treat $x$ and $y$ as constants when doing the inner integral. Then treat $y$ as a constant when doing the $dx$ integral. Finally, do the outer integral as a function of $y$ going from $0$ to $1$

 

[HallsofIvy]

Re: 326 tripel integrals with 6xy

Opalg, if you are going to "reverse the order of dxdydz" then you have to reverse the order of the integrals as well. That is, the limits on the integral with respect to z are 0 to 1, not "0 to x+ y" and, in fact that just gives \left[6xyz\right]_{z=0}^1= 6xy so the integral quickly reduces to \int_y^{2y}\int_0^{x+y} 6xy dxdy.

As other said, we should not have the same letters representing limits of integration and "dummy" variables of integration as well. If this really

Is supposed to be an integral it would be better written, using "u" and "v" as the variables of integration, \int_y^{2y}\int_0^{x+y} 6uv dudv= 6\int_y^{2y}v\left(\int_0^{x+y} udu\right) dv= 6\int_y^{2y}v\left(\frac{1}{2}u^2\right)_0^{x+y} dv= 3\int_y^{2y}u(x+y)^2 du= 3(x+y)^2\left(\frac{1}{2}u^2\right)_y^{2y}= \frac{3}{2}(x+ y)^2(4y^2- y^2)= \frac{9}{2}y^2(x+ y)^2.​

 

[Opalg]

Re: 326 tripel integrals with 6xy

Opalg, if you are going to "reverse the order of dxdydz" then you have to reverse the order of the integrals as well. That is, the limits on the integral with respect to z are 0 to 1, not "0 to x+ y" and, in fact that just gives \left[6xyz\right]_{z=0}^1= 6xy so the integral quickly reduces to \int_y^{2y}\int_0^{x+y} 6xy dxdy.

As other said, we should not have the same letters representing limits of integration and "dummy" variables of integration as well. If this really

Is supposed to be an integral it would be better written, using "u" and "v" as the variables of integration, \int_y^{2y}\int_0^{x+y} 6uv dudv= 6\int_y^{2y}v\left(\int_0^{x+y} udu\right) dv= 6\int_y^{2y}v\left(\frac{1}{2}u^2\right)_0^{x+y} dv= 3\int_y^{2y}u(x+y)^2 du= 3(x+y)^2\left(\frac{1}{2}u^2\right)_y^{2y}= \frac{3}{2}(x+ y)^2(4y^2- y^2)= \frac{9}{2}y^2(x+ y)^2.​

I disagree with that, and I do indeed believe that whoever wrote this problem intended that the $z$ integral should go from $0$ to $x+y$, not from $0$ to $1$. I believe that this is the only reasonable way of interpreting it. Otherwise, you would be left with an integral where the dummy variable of integration also occurs as one of the limits of integration.

As I implied in my previous comment, I believe that whoever set this question must be using a convention in which the order of the symbols $$\int_0^1\int_y^{2y}\int_0^{x+y}$$ matches up with the symbols $dy\, dx\, dz$ in the same order, namely $$\int_0^1 dy$$, $$\int_y^{2y}dx$$, $$\int_0^{x+y}dz$$.

 

[karush]

OK, I learned some very valuable things with this
I did notice on other exercises that the dxdydz was not always in this order so I presume the the way the order the integrals are given has to match so we always go from the inside out as we do with composites.

I know I am posting here really a lot but tomorrow I turn 75 and this forum has been very good for me
when I took math before the internet was not in existence like it is today and even calculators could not do integrals etc. there was no learning center to go to for help. if you made an appt with the teacher it was just a 5 min thing that often raised just more questions. actually much of the help I got here. I was in turn able to help some high school students. So its been a very strong sense of being useful which many seniors fail to experience. Also, I have tried about 10 different forums but I always come back to this one. it definitely is the most user friendly with live latex and other features. And the help is a lot more extensive.

probably future posts I will just post a cropped screenshot of the problem so that what the intention is clear.

 

[HallsofIvy]

Yes, \int_a^b\int_c^d\int_e^f \phi(x,y,z) dxdydz means \int_{z=a}^b \left(\int_{y=c}^d \left(\int_{x= e}^f \phi(x,y,z) dx\right)dy\right)dz

 

[topsquark]

O
...tomorrow I turn 75...

Happy birthday!

-Dan