Question

Idk I think i may just ::scream::; I have another annoying problem that I simply cant figure out

Answer 1

Find the vector flux of G=2yi+3xj-zk through the cylinder x^2+y^2=4 and \[2\le z \le -2\]

Answer 2

the cylindar is again simple enough for us; the normal is <x/2,y/2,0>

Answer 3

the rotation is again from 0 to 2pi

Answer 4

\[\int_{0}^{2pi}\int_{-2}^{2}G.N\ adzdt\]

Answer 5

where a is our Normal length .... for some reason

Answer 6

G= <2y, 3x, -z>
N=<x/2, y/2, 0>
----------------
xy+3xy/2+0 = 5xy/2

5xy/2 * 2 = 5xy

x=2cos(t), y=2sin(t)

\[\int\int20sin(t)cos(t)dzdt\]

Answer 7

and with any luck, i did that correctly :)

Answer 8

20 sin cos = 10 sin(2t) if it helps

Answer 9

\[5\int\int 2sin(2t)dzdt\]

Answer 10

we agree?

Answer 11

kkk got to that point

Answer 12

yes i agree LOL Now i am finally getting it YAY

Answer 13

im just remembering it from that video i linked you to last time :)

Answer 14

hmmm like y did u keep the 2 in?
like i know its an identity but like its a constant so we can remove it

Answer 15

yeah, i was thinking it be easier in to integrate sin2t with; but if we do z first its kinda pointless to begin with

Answer 16

4sin2t +4sin2t = 8sin2t

pull out the 4 and leave the 2

20 int 2sin2t dt = 20(-cos(2t)), from 0 to 2pi

Answer 17

-20 cos(4pi) - - 20cos(0) = 0 .... we got a answer key for this one? to dbl chk

Answer 18

noppppppeeee not this time. Its not from the textbook :(

Answer 19

hmm, id have to review that video one last time to see that im remembering it correctly

Answer 20

it seems to me to be correct

Answer 21

Like we followed teh same process as the previous example

Answer 22

ok, i reviewed the tape ... they still call it a tape right?

Answer 23

lol ummmm tape=cassette

Answer 24

G= <2y, 3x, -z>
N=<x/2, y/2, 0>
----------------
xy+3xy/2+0 = 5xy/2 this is good still

the x=2c and y=2s is good as well

5.2c.2s
------- = 10 s c = 5 s 2t
2

Answer 25

hmmm i didnt follow this to be honest

Answer 26

ohh its 5sin(2t) k got that

Answer 27

\[\int_{0}^{2pi}\int_{-2}^{2}5sin(2t)\ 2\ dzdt\]

\[\int_{0}^{2pi}\int_{-2}^{2}10sin(2t)\ \ dzdt\]

\[\int_{0}^{2pi}10(2)sin(2t)-10(-2)sin(2t)\ \ dt\]

\[\int_{0}^{2pi}40sin(2t)\ \ dt\]

\[-20cos(4pi)+20cos(0)=0\]

Answer 28

i keep getting zero :)

Answer 29

Then 0 it is:D

Answer 30

Thanks amistre like u saved the day

Answer 31

lol, only if its right ;)