- #1

Master1022

- 611

- 116

- Homework Statement:
- Prove the divergence theorem in curvilinear coordinates

- Relevant Equations:
- Vector identities

Hi,

I was trying to gain an understanding of a proof of the divergence theorem in curvilinear coordinates. I have found these online notes here and am looking at the proof on pages 4-5. The method intuitively makes sense to me as opposed to other proofs which fiddle around with vector identities to get the required expression, however I have two main questions:

$$ (v_u) \cdot (-h_v h_w dw dv) + (v_u + \frac{\partial v_u}{\partial u} (h_u du)) \cdot (h_v h_w dv dw) $$

Current thoughts on why I might be wrong:

- In the grad formula in curvilinear coordinates, each component includes a factor of the reciprocal of the metric coefficient. Should I be including that here? I.e. should I have ## \frac{1}{h_u} \frac{\partial}{\partial u} ## which would cancel out the ## h_u ##?

I probably have some basic understanding errors and any help would be greatly appreciated.

Attached below are images from the notes:

I was trying to gain an understanding of a proof of the divergence theorem in curvilinear coordinates. I have found these online notes here and am looking at the proof on pages 4-5. The method intuitively makes sense to me as opposed to other proofs which fiddle around with vector identities to get the required expression, however I have two main questions:

**1)**Why are they using lengths ## du ## instead of ## h_u du ##? I thought that in curvilinear coordinates, we would have something like this for the u-direction (I put my centre at the bottom left corner of the box):$$ (v_u) \cdot (-h_v h_w dw dv) + (v_u + \frac{\partial v_u}{\partial u} (h_u du)) \cdot (h_v h_w dv dw) $$

Current thoughts on why I might be wrong:

- In the grad formula in curvilinear coordinates, each component includes a factor of the reciprocal of the metric coefficient. Should I be including that here? I.e. should I have ## \frac{1}{h_u} \frac{\partial}{\partial u} ## which would cancel out the ## h_u ##?

**2)**In equation (12), how did the metric coefficients ## h_v ## and ## h_w ## end up inside the partial derivative? From my expression above, the metric coefficients are outside the partial derivative. Given that each metric coefficients is, in a generalized sense, a function of the three variables ## u##, ## v##, and ## w##, then how can we treat ## h_v ## and ## h_w ## as constants and move them around?I probably have some basic understanding errors and any help would be greatly appreciated.

Attached below are images from the notes:

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