Question

Suppose
G
is a vector field with the property that
divG = 5
for
2 ≤ || r || ≤ 8
and that the flux of
G
through the sphere of radius 4 centered at the origin is 19π. Find the flux of
G
through the sphere of radius 6 centered at the origin.

Answer 1

Flux of G
\[\Phi_G=\oint_S\mathbf{\vec G}\cdot\mathrm d\mathbf{\vec S}\]

Answer 2

Gauss's Divergence Theorem
\[\oint_S\mathbf{\vec F}\cdot\mathrm d\mathbf{\vec S}=\iiint_V(\nabla\cdot\mathbf{\vec F})\mathrm d V\]

Answer 3

Surface area of a sphere\[S=4\pi r^2\]
Volume of a sphere\[V = \tfrac43\pi r^3\]

Answer 4

so its going to be 5 * 4/3pi (6)^3?

Answer 5

They're going to be proportional.

Answer 6

@Ldaniel come back to this question

Answer 7

careful you need to subtract the volumes @Ldaniel

Answer 8

@UnkleRhaukus I felt asleep

Answer 9

@ganeshie8 can you help me finish up

Answer 10

so i need to find the flux for radius 6 and radius 4

Answer 11

and get the difference?

Answer 12

Kindof, use divergence theorem for the region between inner sphere and outer sphere

Answer 13

just multiply the divergence by the volume

Answer 14

flux through outer sphere :
\[\large 19\pi + 5 (V_6 - V_4)\]

Answer 15

1440pi - 425/3pi?

Answer 16

nevermind

Answer 17

so i got 3952pi/3

Answer 18

it was mark wrong

Answer 19

http://www.wolframalpha.com/input/?i=19pi%2B5%284%2F3*pi*6%5E3-4%2F3*pi*4%5E3%29

Answer 20

19pi+5(288pi-(85/3)pi)

Answer 21

thanks

Answer 22

i guess is better to type everything in wolframapha to make no mistake haha

Answer 23

lol agree, no point in wasting time proving ur arithmetic efficiency to yourself... let the wolfram do the donkey work where possible ;)

Answer 24

cam you help me with one more thing

Answer 25

il try, post it