Projection of surface area elements in vector calculus

[jj364]

Homework Statement

(i) Find the normal, n, at a general point on the surface S1 given by; x²+y²+z = 1 and z > 0.

(ii) Use n to relate the size dS of the area element at a point on the surface S1 to its
projection dxdy in the xy-plane.

The Attempt at a Solution

To find n initially I have just done the grad giving 2x,2y,1

Then with the projection of the surface element I have done that dS=$\hat{n}$dS and in this case because it is the xy plane it is just in the $\hat{k}$ direction.

$\hat{n}$=$\frac{1}{\sqrt{1+4x+4y}}$

and so I thought dxdy=$\frac{dS}{\sqrt{1+4x+4y}}$

I'm really not sure about this so and pointers would be a great help, thanks.

Also, sorry about the poor LaTeX!

 

[TSny]

Then with the projection of the surface element I have done that dS=$\hat{n}$dS and in this case because it is the xy plane it is just in the $\hat{k}$ direction.

I think I'm following you here, but I'm not sure what your two "it"s are referring to.

$\hat{n}$=$\frac{1}{\sqrt{1+4x+4y}}$

This is just the k-component of $\hat{n}$, right?
Check your normalization factor here, it's not quite correct.
Otherwise, I think you're on the right track.

 

[jj364]

Yes sorry, wasn't being very clear, I did mean the k component of n hat there yes. Ok I see the problem with the normalisation factor there, should be 4x² and 4y², thank you very much!