Question

Compute the flux of the vector field
F
through the surface S.
F = 2r
and S is the part of the surface
z = x^2 + y^2
above the disk
x^2 + y^2 ≤ 4,
oriented downward.

Answer 1

I keep getting pi but its wrong

Answer 2

show ur work

Answer 3

ok

Answer 4

\[\int\limits(x i+yj+(x^2+y^2)k * (2 x i +2yj - k)dxdy\]

Answer 5

\[\int\limits (2x^2 +2y^2 -x^2 -y^2)dxdy\]

Answer 6

\[\int\limits(x^2+y^2) dxdy\]

Answer 7

\[\int\limits_{0}^{2\pi} \int\limits_{0}^{2} r^2rdrd \theta\]

Answer 8

I got 8pi this time but its wrong

Answer 9

@ganeshie8 what do you think i'm doing wrong

Answer 10

try -8pi

Answer 11

\[\int\langle x i+yj+(x^2+y^2\rangle \cdot \langle -2 x i -2yj+ k \rangle ~dxdy \]

Answer 12

cuz the surface is oriented downward

Answer 13

\[\iint\langle x i+yj+(x^2+y^2\rangle \cdot \langle -2 x i -2yj+ k \rangle ~dxdy \]

btw its good to use double integral whenever you mean a double integral

Answer 14

well that why is positive

Answer 15

-8pi didnt work ?

Answer 16

http://www.webassign.net/ebooks/hhcalc6/hughes-hallett9780470888643/c19/math/math394.gif

Answer 17

this is the formula

Answer 18

it wont open for me

Answer 19

the formula youre using is for oriented upward

Answer 20

let me lookup my notes

Answer 21

have u tried -8pi ?

Answer 22

\[\int\limits \int\limits F(x,y, f(x,y) ) * (f_x i +f_yj -k)dxdy \]

Answer 23

this is the formula for oriented downward

Answer 24

nvm, you're right upward : <-fx, -fy, 1>
downward <fx, fy, -1>

Answer 25

I'm wondering if my integral points are right

Answer 26

hey F = 2r right?

Answer 27

you're using F=r instead it seems

Answer 28

yeah F=2r

Answer 29

that's why I'm going from 0 to 2

Answer 30

what has your region got to do anything with vectorfield ?

Answer 31

nothing, i'm confuse

Answer 32

try 16pi

Answer 33

what does the

Answer 34

I only have one more try

Answer 35

your bounds depend only on the shadow in xy plane

Answer 36

0->2
0->2pi

give you the full disk of radius 2 in xy plane

Answer 37

they are not related to given vector field

Answer 38

F = 2r = <2x, 2y, 2z>

Answer 39

so it's\[\int\limits_{0}^{2\pi} \int\limits_{0}^{1}\]

Answer 40

\[\int\limits_{0}^{2\pi}\int\limits_{0}^2\langle 2x i+2yj+2(x^2+y^2\rangle \cdot \langle 2 x i +2yj- k \rangle ~dxdy \]

Answer 41

so why is it \[\int\limits_{0}^{}\]

Answer 42

\[\int\limits_{0}^{?}\]

Answer 43

0 to 2

Answer 44

\[\int\limits_{0}^{2}\]

Answer 45

becuae 0f <4?

Answer 46

how do u represent the disk x^2+y^2 <= 4 in polar ?

Answer 47

so the radius is 2

Answer 48

so what about if \[x^2 +x^2 \le1\]

Answer 49

that would make it \[\int\limits_{0}^{1}\]

Answer 50

thats right for representing the radius "r", it ranges from 0 to 1 if you are on disk of radius 1

Answer 51

so what's making it 0 to 2 in this example? why not 0 to 4?

Answer 52

whats the radius of circle x^2+y^2 = 4 ?

Answer 53

oh yeah 2 haha the diameter is 4

Answer 54

right

Answer 55

so \[\int\limits \int\limits 2(x^2+y^2)dxdy\]

Answer 56

\[\int\limits_{0}^{2\pi}\int\limits_{0}^{2} 2r^2rdr d \theta\]

Answer 57

looks good to me