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I was having trouble showing that the following closed surface integral:
$$\iint_S F \cdot dS = 0$$

With the vector field $$F = \langle x,y,-2z\rangle$$ over the cone with a lid on top:

$$\ z^2 = x^2+y^2$$ $$\ 0≤z≤5 $$

I've tried parameterizing the cone as $$\ G(u,v) = (v\cos(u), v\sin(u),v) $$ and the normal vector $$\ n = (v\cos(u), v\sin(u), -v)$$

Dot multiplying $$\ F(G(u,v)) \cdot n $$

and doing the surface integral

$$\int_0^5\!\int_0^{2\pi}\ \ F(G(u,v)) \cdot\ n \ du\,dv$$

This does not end up being 0. can anyone explain why this isn't 0? Are my bounds correct? Does it have to do with the top lid? When I use divergence Theorem the answer is 0. Please help,Thanks in advance.

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  • $\begingroup$ You just parametrized the "sides" of the cone, and, as you pointed out, you are missing the top lid. So you just computed the flux across the cone, and not across the top lid as well. $\endgroup$ – Malkoun May 16 '17 at 20:24
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    $\begingroup$ You are not integrating over the full surface. In order to be a closed surface, you must also include the circular "top" of the cone, at z= 5, which is given by $x^2+ y^2= 25$. $\endgroup$ – user247327 May 16 '17 at 20:24
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    $\begingroup$ @user247327, yes indeed. I agree. You probably meant though $z=5$ and $x^2+y^2 \leq 25$ (in order to get the whole disk, and not just the boundary circle). $\endgroup$ – Malkoun May 16 '17 at 20:27

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