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I need to find $\iint_\limits{S}F\cdot\hat{n} d\sigma$ where $\hat{n}$ is the unit outward normal to the surface $S$. Here $S$ is just the conical surface without the base.

If I parameterize the surface as follows : $$\gamma=r \cos\theta\hat{i}+ r\sin\theta\hat{j}+2r\hat{k}$$ then, $$\iint_\limits{S}F\cdot\hat{n}d\sigma=\int_0^{2\pi}\int_0^1F\cdot(\gamma_r\times\gamma_\theta) dr d\theta$$ This gives the answer $-2\pi$. However the correct answer is $2\pi$. So I think I have not taken the correct outward direction for $\hat{n}$.

My question is why is the unit outward normal not given by $\dfrac{\gamma_r\times\gamma_\theta}{|\gamma_r\times\gamma_\theta|}$ but by the negative of it for this problem?

P.S. This is a homework question.

Thank you!

R_D
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1 Answers1

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I did it kind of quick, but also got $-2\pi$. Details: divergence theorem gives the integral $$ \iiint_K(x-2z)\,dxdydz=[\text{symmetry}]=\iiint_K(-2z)\,dxdydz=\\ \int_0^2(-2z)\underbrace{\iint_{2\sqrt{x^2+y^2}\le z} 1 \,dxdy}_{\text{disc area}}\,dz=\int_0^2(-2z)\pi\frac{z^2}{4}\,dz=-\frac{\pi}{2}\int_0^2z^3\,dz=-2\pi. $$

UPDATE: As the OP points out, the top of the set does not belong to the surface. Then we need to take this part off. Calculating the flux through the top (the normal is $\hat n=(0,0,1)$): $$ \iint_{\text{top}}F\cdot\hat n\,d\sigma=\iint_{\text{top}}(-z^2)\,d\sigma=\iint_{\text{top}}-4\,d\sigma=-4\cdot\text{area of the top}=-4\pi. $$ Finally, the total flux is $-2\pi-(-4\pi)=2\pi$.

A.Γ.
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  • Can we really use Gauss divergence theorem(https://en.wikipedia.org/wiki/Divergence_theorem)? According to it we have a compact volume enclosed by a piecewise smooth surface. Here the base of the cone is not included so it doesn't bound a solid. Right? – R_D Nov 05 '15 at 10:08
  • Well, I understood as it was (that's kind of a standard exercise). If it was not, it would be better to clarify the point in the question a bit more. So you mean your surface is just conic and no top? – A.Γ. Nov 05 '15 at 10:37
  • @Rise I edited. – A.Γ. Nov 05 '15 at 10:49
  • Thank you. But I still don't understand why when I do it directly (without use of divergence theorem) I don't get the right answer. What mistake have I made in choosing the outward normal direction? – R_D Nov 05 '15 at 11:20