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## Homework Statement

Consider the surface [tex] S_1 [/tex] described by the equations

[tex] x = (1-w)^3cos(u), y = (1-w)^3sin(u), z = w, 0 <= u < 2\pi, 0 <= w < 1 [/tex]

The first few parts of the question were quite simple. Firstly we had to calculate dS and then compute the surface integral for the vector field [tex] \mathbf{a}= (y-z)\mathbf{i} + (z-x)\mathbf{j} + 2z \mathbf{k} [/tex]

The following part was a surface [tex] S_2 [/tex] which is a circle of radius 1 centered at the origin at the z = 0 plane

## The Attempt at a Solution

Now, I'm unsure if I've just missed something blatantly obvious, but I was attempting to to find [tex] d\mathbf{S} [/tex] which I get to be [tex] d\mathbf{S} = -\mathbf {k}dS [/tex]. But when I dot this with [tex] {\boldmath a} [/tex], since z has reduced to zero in the vector field I get 0

Can anyone give me a push in the right direction? Thanks

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