How Do You Calculate the Laplacian in Polar Coordinates?

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To calculate the Laplacian in polar coordinates, start with the standard form of the Laplacian operator in Cartesian coordinates, which is the sum of the second partial derivatives with respect to x and y. The chain rule is essential for transforming these derivatives into polar coordinates, where you express the derivatives in terms of r and θ. This process involves applying the chain rule twice for both x and y, followed by combining the results. The discussion emphasizes that this method can be tedious but is necessary for understanding the transformation. Ultimately, mastering this calculation is a valuable exercise in mathematical rigor.
davon806
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Homework Statement


I am trying to calculate the laplacian in polar coordinates but I failed.Please see the attached

Homework Equations




The Attempt at a Solution


My solution to this was uploaded in the attached.I was wondering what's wrong with the purple brackets since they shouldn't exist( If you sum A,B,C and D up). Thanks
 

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I don't think you should be messing with unit vectors and their derivatives. The Laplacian is a scalar operator. It looks like you are doing it in two dimensions. Start with $$\nabla ^2=\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}$$
Observe that according to the chain rule
$$\frac{\partial}{\partial x}=\frac{\partial r}{\partial x}\frac{\partial}{\partial r}+\frac{\partial \theta}{\partial x}\frac{\partial}{\partial \theta}$$
and one more time for the second partial derivative. Repeat with y and then add the two results.

This is the "brute force" method and involves tedious algebra. It is the kind of thing that one has to do once in one's life to satisfy oneself that one can do it. :smile:
 
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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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