# Changing Order of Triple Integration

1. Apr 4, 2016

### njo

1. The problem statement, all variables and given/known data
$$\int_0^5 \int_0^2 \int_0^{4-y^2}\ \, dxdydx$$

Change order to dydxdz

2. Relevant equations

3. 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.

2. Apr 4, 2016

### LCKurtz

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.

3. Apr 4, 2016

### SammyS

Staff Emeritus
You obviously have a typo in the volume element there.

you have dx dy dx .

dx is repeated, no dz.

4. Apr 4, 2016

### HallsofIvy

Staff Emeritus
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 $\int_0^4\int_0^\sqrt{y} F(x, y, z) dzdxdy$.

Last edited by a moderator: Apr 4, 2016