Question

Surface Integral - What do I do when divergence is equal to 0? Is this correct? (Attachment below)

Answer 1

Task;

Answer 2

My work so far;

Answer 3

@ganeshie8

Answer 4

You can still integrate 0, you'll get a constant

Answer 5

It is still 0 - as It is not an unspecified integral ?

Answer 6

Been a while since I've done vector calc, so lets see...\[ \vec F = \frac{ \partial (-xz^7) }{ \partial x }+\frac{ \partial (sinxe^{cosx}-yz^7) }{ \partial y }+\frac{ \partial (1/4z^8) }{ \partial z }\]\[\vec F = -z^7 -z^7 + 2z^7 \] \[\vec F = -2z^7+2z^7 = 0\]

Answer 7

I guess, it means that there is no flux entering or leaving

Answer 8

It's not uncommon I don't think, just means no net flux

Answer 9

Okay, but how do I calculate the surface area of this, if divergence is the wrong approach?

Answer 10

Why is it wrong?

Answer 11

Because the surface area cant be 0?

Answer 12

As I said before you can still integrate

Answer 13

What would integral of 0dp from 0 to 4 be then? I take 0 out, I still get 0*integral from 0 to 4 dP. So even if it would be a constant, it would be a constant * 0 = 0

Answer 14

\[\int\limits 0 dx = C\]

Answer 15

It is from 0 to 4

Answer 16

Not unspecified

Answer 17

For example \[\int\limits \int\limits 0 dx dy = xC_1+C_2\]

Answer 18

That's fine, you just have a constant then

Answer 19

I do not think I can solve it with just a constant

Answer 20

Via divergence theorem your integral would be 0

Answer 21

But I'm not 100% sure, lets ask @Michele_Laino

Answer 22

Haha, thanks for checking @IrishBoy123

Answer 23

If that is the case; How would I ... what would be the range for H with only two integrals? Or rather; How?

Answer 24

\(S\) is not closed, so divergence theorem may not be of much use here

Answer 25

however, if you insist, you may use divergence theorem by constructing a closed surface. If you're a bit clever, you might actually work it smoothly too...

Answer 26

I think I am throwing the towel on this task... at least for now

Answer 27

Thanks for the help

Answer 28

I don't know if this picture makes sense, so I'll explain it. The surface integral out of the whole boundary minus the surface integral of these flat pieces is the surface integral of your region:

The reason being that flat surface integrals and closed surface integrals are easier.