The Divergence of a Polar Vector Function

In summary, the conversation discusses finding the divergence of a given function, with the use of the Divergence Theorem. The attempt at a solution involves using the given equations to simplify the function, which results in an answer of 5cosθ-sinφ. However, the answer should only be 5cosθ, as the Divergence Theorem does not hold with the second term included. Upon further checking, it appears that Wolfram Alpha's answer may be incorrect.
  • #1
transmini
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Homework Statement


Find the divergence of the function ##\vec{v} = (rcos\theta)\hat{r}+(rsin\theta)\hat{\theta}+(rsin\theta cos\phi)\hat{\phi}##

Homework Equations


##\nabla\cdot\vec{v}=\frac{1}{r^2}\frac{\partial}{\partial r}(r^2v_r)+\frac{1}{r sin\theta}\frac{\partial}{\partial \theta}(sin\theta v_\theta) + \frac{1}{r sin\theta} \frac{\partial v_\phi}{\partial\phi}##

The Attempt at a Solution


##\nabla\cdot\vec{v}=\frac{1}{r^2}\frac{\partial}{\partial r}(r^3 cos\theta)+\frac{1}{r sin\theta}\frac{\partial}{\partial \theta}(rsin^2\theta) + \frac{1}{r sin\theta} \frac{\partial}{\partial\phi}(rsin\theta cos\phi)##
##\nabla\cdot\vec{v}=\frac{1}{r^2}(3r^2cos\theta)+\frac{1}{r sin\theta}(2rsin\theta cos\theta) + \frac{1}{r sin\theta} (-rsin\theta sin\phi)##
##\nabla\cdot\vec{v}=3cos\theta+2cos\theta-sin\phi##
##\nabla\cdot\vec{v}=5cos\theta-sin\phi##

Except, the answer shouldn't have the second term in it at all and should just be ##5cos\theta## (I don't know 100% certain what the answer is, just the wolfram alpha gives ##5cos\theta## and that the Divergence Theorem doesn't hold with the second term and would hold if that term were gone like in wolfram's answer)

Where did I make a mistake here? Thanks in advance

Edit: Okay I spoke too soon on the Divergence Theorem not holding, as it appears to hold now that I double checked that problem's work. So is there just a problem with wolfram's answer?
 
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  • #2
Can you provide a link to exactly what you put into Wolfram Alpha?
 

What is meant by "Polar Divergence of a Vector"?

The polar divergence of a vector is a measurement of how much the vector field spreads out or converges at a given point. It is a scalar value that indicates whether the vector field is expanding or contracting at that point.

How is the polar divergence of a vector calculated?

The polar divergence of a vector is calculated by taking the dot product of the vector field with the unit vector in the radial direction and then taking the derivative with respect to the radial distance. This can also be written as the sum of the partial derivatives of the vector components with respect to each coordinate.

What does a positive polar divergence value indicate?

A positive polar divergence value indicates that the vector field is expanding at that point. This means that the vectors are moving away from each other and the overall flow of the field is outward.

What does a negative polar divergence value indicate?

A negative polar divergence value indicates that the vector field is contracting at that point. This means that the vectors are moving towards each other and the overall flow of the field is inward.

What are some real-world applications of polar divergence of a vector?

The polar divergence of a vector is used in various fields such as fluid dynamics, electromagnetism, and meteorology. It can be used to analyze the flow of fluids, the distribution of electric and magnetic fields, and the movement of air masses in weather patterns. It is also useful in image processing and computer graphics for edge detection and feature extraction.

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