Question

Find the flux of
F = x^3 i  + y^3 j  + z^3k
through the closed surface bounding the solid region
x^2 + y^2 ≤ 4,
0 ≤ z ≤ 7,
oriented outward.

Answer 1

@wio can you help me with this please, let me tell you what I did first

Answer 2

I got divF=3x^2+3y^2+3z^2

Answer 3

\[\int\limits_{0}^{2\pi}\int\limits_{0}^{2}\int\limits_{0}^{7}(r^2+z^2)rdzdr d \theta \]

Answer 4

times 3 and I got 4620pi

Answer 5

but I got it wrong

Answer 6

@dan815 can you help me out :)

Answer 7

well have you integrate r^3+z^2r with w.r.t z?

Answer 8

you know starting with the inner most integral you have there

Answer 9

@freckles where do you get r^3+z^2r from?

Answer 10

\[(r^2+z^2)r =r^3+z^2r \text{ by distributive property }\]

Answer 11

how would I make green's theorem work here

Answer 12

well yeah when I do the integral I get 4620 pi

Answer 13

yeah I forgot it's only for line integrals, whereas this one is a volume since z ranges from 0 to 7.

Answer 14

i'm wondering if i did my limits of integration right

Answer 15

@ganeshie8 can you help me out with this one please

Answer 16

try 1540pi

Answer 17

bounds and everything looks good, you made a mistake while evaluating the integral thats all. btw, at this point you should let wolfram do the dumb work.. important thing is setting up the integral correctly, not evaluating it manually.

http://www.wolframalpha.com/input/?i=%5Cint%5Climits_%7B0%7D%5E%7B2%5Cpi%7D%5Cint%5Climits_%7B0%7D%5E%7B2%7D%5Cint%5Climits_%7B0%7D%5E%7B7%7D%283r%28r%5E2%2Bz%5E2%29%29+dzdrd%5Ctheta

Answer 18

Kind of just relearned something myself, haha. Finding that normal vector.

Answer 19

surface integral ? that would be really painful as we need to find 3 normals and work 3 integrals... but it would be a nice practice for sure :)

Answer 20

but don't I need to multiply 1540 pi by 3 because of the divF = 3(x^2+y^2+z^3)

Answer 21

z^2 not z^3

Answer 22

it was already multiplied

Answer 23

now i see how u got 4620, youhave multiplied 3 two times

Answer 24

upps haha I multiplied it by 3 two times

Answer 25

yeah haha

Answer 26

silly mistake

Answer 27

thanks for checking my work @ganeshie8

Answer 28

np:) are you done with the two problems that you run off from yesterday ?

Answer 29

so I have another question why if the flux id out of the close surface and not through the closed surface

Answer 30

no I'm not done @ganeshie8 I'm stuck with them

Answer 31

think of it as fluid flow

Answer 32

would it be just negative because is out

Answer 33

it helps to think of the given vector field as "velocity field" in 3d when working with divergence

Answer 34

flux is same as the amount of fluid passing through the surface per unit time

Answer 35

so the flux would be the same?