# Rearranging a triple integral - vector calc

1. Mar 27, 2012

### Who Am I

Rearranging a triple integral -- vector calc

1. The problem statement, all variables and given/known data
Change the order of integration in the following integral to dy dz dx.
$\int^{1}_{0}$$\int^{2x}_{0}$$\int ^{x+y}_{0}F(x,y,z) dz dy dx$

2. Relevant equations

N/A

3. Attempt at solution

Well, the overall thing that I did was look at it as a graph where I got an integral with endpoints 0 to 2x, 0 to x and 0 to 1 in dy dz dx respectively (o to 2x for y), which gave half the volume if integrating both over F(x,y,z,)=1. I then looked at it on the graph again and realized that this way gives you a prism.

It's confusing because you have two functions in terms of dy and dz that 2x "hoovers" over.

This is from my vector calc class.

2. Mar 28, 2012

### clamtrox

Re: Rearranging a triple integral -- vector calc

You should really draw a figure! It's easy to figure out the range for z, but it's a little trickier to find f(z) in 0 < y < f(z). A picture will help with this.

3. Mar 28, 2012

### HallsofIvy

Staff Emeritus
Re: Rearranging a triple integral -- vector calc

Good! Yes, in the xy-plane, we have x ranging from 0 to 1 and, for each x, y ranging from 0 to 2x so we have the right triangle with vertices at (0, 0), (1, 0), and (1, 2), the hypotenuse being the line y= 2x.

Then, for each point (x, y) in the xy-plane, z ranges from 0 up to the plane z= x+ y. We can note that in the "corner" (1, 2), z= 3, the larges value z takes on. If we project that to the xz-plane, where y= 0, we have corners at (x, z)= (0, 0), (1, 0), and (1, 1), the hypotenuse of that triangle being z= x. That tells us that x ranges from 0 to 1 (the "outer" integral is still with respect to x so still has the same limits) and that, for each x, z ranges from 0 to x.

Now, in three dimensions, the "upper boundary" is still z= x+ y which we can write as y= z- x.

4. Mar 28, 2012

### Who Am I

Re: Rearranging a triple integral -- vector calc

Well, I'm still rather confused, because when you're differentiating with respect to y with y first, when you draw the picture, you are going to 2x. When looking down in dy, you're looking down at zx plane, you can see that in this context you get the plane y=2x "hovering" over the plane y=0 and y=z-x within the bounds of the original integral. So, I came up with the idea that I may have to write two separate integrals and add them because of this fact that you have two different functions to differentiate. But I have been having issues visualizing and setting up the second integral, because like I said, that original function I had made a prism that accounts for exactly half of the volume of the entire shape, but it's not correct to describe the entire shape as twice the volume, especially if you're integrating a function as a function of those variables.

But I think that if you add the two separate integrals, it is then properly accounting for the two different volumes. But that is what I also had to do with an integral bound by the plane z=x+10, the cone z=(y2+x2)1/2 and over a disk on the xy plane with radius one in spherical coordinates.

But I don't know how to set the other triangle up in this case, and I think that it would work to separate the integral because it worked in the last case.