The Divergence of a Polar Vector Function

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SUMMARY

The divergence of the polar vector function \(\vec{v} = (rcos\theta)\hat{r}+(rsin\theta)\hat{\theta}+(rsin\theta cos\phi)\hat{\phi}\) is calculated using the formula \(\nabla\cdot\vec{v}=\frac{1}{r^2}\frac{\partial}{\partial r}(r^2v_r)+\frac{1}{r sin\theta}\frac{\partial}{\partial \theta}(sin\theta v_\theta) + \frac{1}{r sin\theta} \frac{\partial v_\phi}{\partial\phi}\). The correct divergence simplifies to \(5cos\theta\) after resolving the terms, although an initial confusion arose regarding the validity of the Divergence Theorem. The user confirmed that the theorem holds upon re-evaluation, indicating that the discrepancy may lie with Wolfram Alpha's output.

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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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Can you provide a link to exactly what you put into Wolfram Alpha?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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