Surface Integral

**adam_leeds** · May 8th 2011, 11:33 AM

I thought about doing the divergence theorem, but then quickly noticed nothign differentiated

$\iint_s F.n d\sigma F = zi + j -xyk$ and S is the surface cut from the cylinder ${y}^{2 } + {z}^{ 2} =1, z \geqslant 0$ . by the planes $x =0, x=2$

**Mondreus** · May 8th 2011, 12:46 PM

F is solenoidal and applying the divergence theorem to that surface isn't a problem, so the outward flux is zero.

**adam_leeds** · May 9th 2011, 02:25 AM

Originally Posted by Mondreus

F is solenoidal and applying the divergence theorem to that surface isn't a problem, so the outward flux is zero.

So then the answer will be 0, because your integral will now be multiplied by 0?

Thread: Surface Integral

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