Fubini's Theorem apply? Volume

In summary, finding the volume of the region bounded by x = -1, x = 1, y = -1, y = 1, and z = x involves adding another limit and ensuring that z>0. The first integral provided was incorrect as it only calculated the volume of a rectangular solid. The second integral was also incorrect as it had a lower limit of z=0, which was not a part of the original boundaries.
  • #1
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Find the volume of the region bounded by [tex] x = - 1, \ x = 1 [/tex], [tex] y = - 1, y = 1 [/tex] and [tex] z = x [/tex].

Is this equaled to [tex] \int_{0}^{1} \int_{ - 1}^{1} \int_{ - 1}^{1} dx \ dy \ dz [/tex]?

Can it also equal [tex] \int_{ - 1}^{1} \int_{ - 1}^{1} \int_{0}^{x} \ dz \ dx \ dy [/tex]?
 
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  • #2
Neither is right. Your limits don't enclose any bounded region.
 
  • #3
How would I fix it?
 
  • #4
tronter said:
How would I fix it?

Add another limit. Are you sure you copied all of the limits from the problem you want to solve?
 
  • #5
It also has to be above the [tex] x-y [/tex] plane. Yeah I copied it correctly. So maybe the problem is wrong? You could also take the determinant right?
 
  • #6
The first integral is completely wrong: it is the volume of the rectangular solid [itex]0\le z\le 1[/itex], [itex]-1\le y\le 1[/itex], [itex]-1\le x\le 1[/itex].

For the second integral, where did you get the lower limit z= 0? That is what Dick is suggesting you may have missed.
 
  • #7
tronter said:
It also has to be above the [tex] x-y [/tex] 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?
 
  • #8
If it is above the [itex] x-y [/itex] plane, then [itex] z [/itex] cannot be negative. Its bounded by the [itex] x-y [/itex] plane.
 
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  • #9
tronter said:
If it is above the [itex] x-y [/itex] plane, then [itex] z [/itex] cannot be negative. Its bounded by the [itex] x-y [/itex] plane.

And if z can't be negative, then your lower x limit isn't -1 either.
 
  • #10
Then it would be [tex] \int_{-1}^{-1} \int_{0}^{1} \int_{0}^{x} \ dz \ dx \ dy [/tex]?
 
  • #11
If you add the restriction z>0 as well, yes.
 
  • #12
Thank you for your help Dick and HallsofIvy.
 

What is Fubini's Theorem?

Fubini's Theorem is a mathematical theorem that allows for the calculation of the volume of a three-dimensional object by integrating over two dimensions.

How does Fubini's Theorem apply to volume calculations?

Fubini's Theorem states that if a function is integrable over a region R of the xy-plane, then the volume of the solid below the surface z=f(x,y) and above R is equal to the double integral of f(x,y) over R. This means that by integrating the function over two dimensions, we can calculate the volume of a three-dimensional object.

What is the formula for applying Fubini's Theorem to find volume?

The formula for applying Fubini's Theorem is: V = ∬R f(x,y) dA, where V is the volume, R is the region of integration, f(x,y) is the function representing the height of the solid at each point (x,y), and dA is the differential element representing the area of the region R.

What are some common applications of Fubini's Theorem in science?

Fubini's Theorem is commonly used in physics and engineering to calculate the volume of complex objects, such as irregularly shaped containers or structures. It is also used in fluid dynamics to calculate the volume of a fluid in motion.

Are there any limitations to using Fubini's Theorem for volume calculations?

While Fubini's Theorem is a powerful tool for calculating volume, it does have some limitations. It can only be applied to regions that are bounded by simple curves, such as circles or straight lines. It also assumes that the function f(x,y) is integrable over the region R, which may not always be the case for more complex functions.

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