# Fubini's Theorem apply? Volume

1. Mar 5, 2008

### tronter

Find the volume of the region bounded by $$x = - 1, \ x = 1$$, $$y = - 1, y = 1$$ and $$z = x$$.

Is this equaled to $$\int_{0}^{1} \int_{ - 1}^{1} \int_{ - 1}^{1} dx \ dy \ dz$$?

Can it also equal $$\int_{ - 1}^{1} \int_{ - 1}^{1} \int_{0}^{x} \ dz \ dx \ dy$$?

2. Mar 5, 2008

### Dick

Neither is right. Your limits don't enclose any bounded region.

3. Mar 5, 2008

### tronter

How would I fix it?

4. Mar 5, 2008

### Dick

Add another limit. Are you sure you copied all of the limits from the problem you want to solve?

5. Mar 5, 2008

### tronter

It also has to be above the $$x-y$$ plane. Yeah I copied it correctly. So maybe the problem is wrong? You could also take the determinant right?

6. Mar 5, 2008

### HallsofIvy

Staff Emeritus
The first integral is completely wrong: it is the volume of the rectangular solid $0\le z\le 1$, $-1\le y\le 1$, $-1\le x\le 1$.

For the second integral, where did you get the lower limit z= 0? That is what Dick is suggesting you may have missed.

7. Mar 5, 2008

### Dick

It could be above the xy plane, if you say so. But nothing in the original boundaries forces that to be true. Take the determinant of what?

8. Mar 5, 2008

### tronter

If it is above the $x-y$ plane, then $z$ cannot be negative. Its bounded by the $x-y$ plane.

Last edited: Mar 5, 2008
9. Mar 5, 2008

### Dick

And if z can't be negative, then your lower x limit isn't -1 either.

10. Mar 5, 2008

### tronter

Then it would be $$\int_{-1}^{-1} \int_{0}^{1} \int_{0}^{x} \ dz \ dx \ dy$$?

11. Mar 5, 2008

### Dick

If you add the restriction z>0 as well, yes.

12. Mar 5, 2008

### tronter

Thank you for your help Dick and HallsofIvy.