Divergence theorem requires a conservative vector field?

In summary, the Divergence Theorem, also known as Gauss's Theorem or Ostrogradsky's Theorem, is a fundamental theorem in vector calculus that relates the surface integral of a vector field over a closed surface to the triple integral of the divergence of the vector field within the enclosed volume. It can only be applied to conservative vector fields, which are vector fields in which the line integral between any two points is independent of the path taken between those points. The Divergence Theorem has many applications in physics, particularly in fluid mechanics and electromagnetism, but it does have limitations. It can only be applied to three-dimensional vector fields and closed surfaces, and the vector field must be well-defined and continuous within the enclosed
  • #1
jesuslovesu
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Can anyone tell me whether or not the divergence theorem requires a conservative vector field? On a practice exam my professor gave a vector field that was nonconservative (I checked the curl) and proceeded to perform the divergence theorem to find the flux.

On one of my homework problems I forgot to check whether the field was conservative and it turns out that field was nonconservative but still got the correct answer. I don't know if that is a fluke or not.


In short: Do you need a conservative vector field in order to use the divergence theorem?
 
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  • #2
jesuslovesu said:
In short: Do you need a conservative vector field in order to use the divergence theorem?

No. Only that the vector function have continuous derivatives.
 
  • #3


Yes, the divergence theorem does require a conservative vector field. This is because the theorem is based on the fundamental theorem of calculus, which states that the integral of a conservative vector field over a closed surface is equal to the integral of the divergence of that field over the volume enclosed by the surface. If the vector field is not conservative, the two integrals will not be equal and the theorem cannot be applied. In your practice exam and homework problems, it is possible that the nonconservative vector fields happened to have a zero divergence, which would still result in the correct answer when using the divergence theorem. However, this is not always the case and it is important to always check for conservative behavior before applying the theorem.
 

Related to Divergence theorem requires a conservative vector field?

1. What is the Divergence Theorem?

The Divergence Theorem, also known as Gauss's Theorem or Ostrogradsky's Theorem, is a fundamental theorem in vector calculus that relates the surface integral of a vector field over a closed surface to the triple integral of the divergence of the vector field within the enclosed volume.

2. What is a conservative vector field?

A conservative vector field is a vector field in which the line integral between any two points is independent of the path taken between those points. In other words, the value of the line integral depends only on the endpoints and not on the path itself.

3. Why does the Divergence Theorem require a conservative vector field?

The Divergence Theorem can only be applied to conservative vector fields because it assumes that the vector field is the gradient of a scalar function, which is a necessary condition for the theorem to hold. In non-conservative vector fields, the divergence may not be well-defined and the theorem cannot be applied.

4. How is the Divergence Theorem used in physics?

The Divergence Theorem has many applications in physics, particularly in fluid mechanics and electromagnetism. In fluid mechanics, it can be used to relate the flow rate of a fluid through a closed surface to its divergence at each point. In electromagnetism, it can be used to relate the electric flux through a closed surface to the charge density within the enclosed volume.

5. Are there any limitations to the Divergence Theorem?

Yes, the Divergence Theorem has some limitations. It can only be applied to three-dimensional vector fields and closed surfaces. In addition, the vector field must be well-defined and continuous within the enclosed volume. If these conditions are not met, the theorem cannot be applied.

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