Fubini's Theorem apply? Volume

[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 $$\int_{ - 1}^{1} \int_{ - 1}^{1} \int_{0}^{x} \ dz \ dx \ dy$$?

 

[Dick]

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

 

[tronter]

How would I fix it?

 

[Dick]

How would I fix it?

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

 

[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?

 

[HallsofIvy]

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.

 

[Dick]

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?

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?

 

[tronter]

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

 

[Dick]

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

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

 

[tronter]

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

 

[Dick]

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

 

[tronter]

Thank you for your help Dick and HallsofIvy.