Changing Order of Triple Integration

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Homework Help Overview

The discussion revolves around changing the order of integration in a triple integral, specifically from the order \(dx\,dy\,dz\) to \(dy\,dx\,dz\). The integral in question has limits that are primarily constants, which raises questions about the correct setup for the new order of integration.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants express confusion regarding the limits of integration, particularly when they are constants rather than functions. There is a suggestion to sketch or describe the volume being integrated to clarify the limits. Some participants also note the presence of a potential typo in the original integral's notation.

Discussion Status

The discussion is ongoing, with participants exploring the implications of the limits of integration and the correct interpretation of the volume. Guidance has been offered regarding the need to carefully describe the volume and the relationships between the variables in the integration limits.

Contextual Notes

There is mention of the necessity for the outermost integral limits to be constants, while the innermost limits can depend on the other variables. This highlights the importance of understanding the relationships between the variables when changing the order of integration.

njo
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Homework Statement


\int_0^5 \int_0^2 \int_0^{4-y^2}\ \, dxdydx

Change order to dydxdz

Homework Equations

The Attempt at a Solution


I'm confused mainly because the limits are mostly numbers, not functions. I graphed the limits in @D and #d and this is what I got: \int_0^{4-y^2} \int_0^5 \int_0^2\ \, dydxdz

Not sure if its correct because I have a function on the outer integral, I can't check my work.

I'd appreciate any help. Thanks.
 
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njo said:

Homework Statement


\int_0^5 \int_0^2 \int_0^{4-y^2}\ \, dxdydx

Change order to dydxdz

Homework Equations

The Attempt at a Solution


I'm confused mainly because the limits are mostly numbers, not functions. I graphed the limits in @D and #d and this is what I got: \int_0^{4-y^2} \int_0^5 \int_0^2\ \, dydxdz

Not sure if its correct because I have a function on the outer integral, I can't check my work.

I'd appreciate any help. Thanks.

It is clearly incorrect because you have variables in the outer integral limits. You need to start by sketching, or at least describing carefully, the volume being enclosed.
 
njo said:

Homework Statement


\int_0^5 \int_0^2 \int_0^{4-y^2}\ \, dxdydx

Change order to dydxdz
You obviously have a typo in the volume element there.

you have dx dy dx .

dx is repeated, no dz.
 
The limits on the "innermost" integration can be functions of the other two variables. The limits on the "middle" integration can be functions only of the one remaining variable and the limits of integration on the "outermost" integral must be constants. For example, integrating "dydzdx" we must have something like \int_a^b\int_{c(x)}^{d(x)}\int_{e(x,y)}^{f(x,y)} F(x,y,z) dydzdx.

With \int_0^2\int_0^{x^2}\int_0^{x+ y} F(x, y, z) dzdydx then x varies from 0 to 2, for each x, y varies from 0 to x^2, and, for each x and y, z varies from 0 to x+ y. If you were to graph the first two, it would be a "curved triangle" with one side the x-axis, one side the line x= 2, and the third side the curve y= x^2 which crosses x= 2 at (2, 4). If we only wanted to reverse dx and dy, to dzdxdy, note that, overall, y goes from 0 to 4 and that, for every y, x goes from 0 on the left to y= x^2, so x= \sqrt{y}. The integral, in that order, would be \int_0^4\int_0^\sqrt{y} F(x, y, z) dzdxdy.
 
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