# Rearranging a triple integral - vector calc

• Who Am I
In summary, the student attempted to rearrange a triple integral by changing the order of integration. They looked at it as a graph and realized that this way gives you a prism. However, they are still confused and need help to visualize the second integral.
Who Am I
Rearranging a triple integral -- vector calc

## Homework Statement

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$

## Homework Equations

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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.

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.

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.

HallsofIvy said:
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.

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.

## 1. What is a triple integral?

A triple integral is a type of mathematical integration that is used to calculate the volume of a three-dimensional region. It involves integrating a function over a three-dimensional region, and the result is a numerical value.

## 2. How do you rearrange a triple integral?

Rearranging a triple integral involves changing the order of integration. This can be done by using the properties of integrals and changing the limits of integration for each variable. It is important to carefully consider the region of integration and the order in which the variables are integrated to get the correct result.

## 3. What is the purpose of rearranging a triple integral?

Rearranging a triple integral can make it easier to solve and can also provide insight into the geometry of the region being integrated. By changing the order of integration, the integrand may become simpler, making it easier to evaluate. It can also help in visualizing the region and understanding its boundaries.

## 4. What are some common mistakes when rearranging a triple integral?

Some common mistakes when rearranging a triple integral include changing the limits of integration incorrectly, forgetting to change the integrand, and not considering the correct order of integration. It is important to carefully follow the steps and double-check the calculations to avoid these mistakes.

## 5. When is rearranging a triple integral necessary?

Rearranging a triple integral is necessary when the original order of integration is not suitable for solving the problem or when it is needed to gain a better understanding of the region being integrated. It is also necessary when using different coordinate systems, such as cylindrical or spherical coordinates, which require a different order of integration.

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