Undergrad Calculating Divergence of a Vector Field in Three Dimensions

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To calculate the divergence of a three-dimensional vector field, all three components must be considered, even if one component is zero. In the example given, the vector field includes a term that depends on z, which necessitates taking the partial derivative with respect to z. Although the k-component is zero, the divergence still requires the inclusion of the z-component's derivative in the calculation. Therefore, the correct approach is to compute the partial derivatives with respect to x, y, and z. This ensures an accurate representation of the divergence in three-dimensional space.
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If I have a vector field say ## v = e^{z}(y\hat{i}+x\hat{j}) ##, and I want to calculate the divergence. Do I only take partial derivatives with respect to x and y (like so, ## \frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y} ##) or should I take partial derivatives with respect to x, y and z (like so, ## \frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y} + \frac{\partial A_z}{\partial z} ##). I'm confused as to which one because there is no ## \hat{k} ## unit vector, but z is changing and the graph should therefore be three dimensional.
 
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Where did you get this problem? It could be a typo. Did you check the online errata page associated with the book?

It seems the e^z term is the k-component in which case you take its partial derivative with respect to z.
 
The example was made up however, I remember seeing a question like this and it had me confused.
 
Oh apologies I made an error in my vector field equation, it has been corrected now.
 
Okay so the k-component would be ##0\hat{k}##

Now take your x,y,z partials for the divergence.
 
So since the k component is 0, would that not mean that the divergence is calculated using only the partials of x and y since ## A_z = 0 ##.
 
Yes but you need to remember to do all of them.
 

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