Question

Divergence theorem to find volume

Answer 1

Use the divergence theorem to show that\[\large \frac{1}{3}\int \int\int_s \hat{n} \cdot rdS = V\]

Where S is a closed surface enclosing a region of volume V,
\(\large \hat{n}\) is a unit vector normal to the surface S, and \(\large r = xi + yj + zk\)

Use this to find the volume of a rectangular parallelepiped with sides a,b,c

Answer 2

I'm a bit confused here, there is no force field to work with, well, as far as I can see at the moment

Answer 3

@phi mind taking a look?

Answer 4

i think the left hand side must be a double integral

Answer 5

\[\large \frac{1}{3} \iint_S \hat{n} \cdot r~dS = V(S) \]

you want to prove this using divergence theorem ?

Answer 6

Ahh yes forgive me, hit copy and paste one too many times lol

And yes, proving that via the divergence theorem

Answer 7

define your vector field same as position vector

Answer 8

\[\vec{F}= \langle x, y, z\rangle \]

Answer 9

Well that will be the \(\large r = xi + yk + zk\)

Wait! \(\large \vec{F} = \vec{r}\) ??

Answer 10

\[\large \frac{1}{3} \iint_S \hat{n} \cdot r~dS = \frac{1}{3} \iint_S \hat{n} \cdot \vec{F}~dS = \dfrac{1}{3} \iiint \mathrm{div}(\vec{F}) dV \]

Answer 11

Omg yes! of course the field would be the position vector because the field is the position at any certain time....argh!

Well this just became very easy >.< lol you always make it seem so simple @ganeshie8

Answer 12

they could have simply used F instead of r
i think they don't want it to be obvious :O

Answer 13

Lol div grad curl and all that just wants to be mischievous XD

So lets see here

\[\large \frac{1}{3}\int \int_s Div(\vec{F})dV \]

div F = \(\large del \cdot F\) = 3

So therefor \[\large \int\int\int_s dV = V\text{of parallelpiped } = abc\]

Answer 14

i suppose that double integral was a typo

Answer 15

other than that, yeah its straight forward once you see it :)

Answer 16

Lol yes...see I really should just use the equation editor instead of copy and paste XD

Thank you @ganeshie8 :)

Answer 17

np :) while you're doing these problems, use `\iint` for double integral and `iiint` for triple integral when limits are not needed

Answer 18

\[\iiiint\]

Answer 19

yeah `iiiint` also works ^

Answer 20

:O latex is wonderful!! lol