Question

Let F = (−2x + sin(yz)) i  + (4y + sin(xz)) j  + (3z + sin(xy)) k 

Let S be the cube of side length 2 centered at (4, 0, 0) and with faces parallel to the coordinate planes. Let S1 be the face of S which is perpendicular to the x-axis and whose x values are greater than 4.

(a) Find the flux of F through S1, oriented in the positive x-direction.


(b) Find the flux of
F
outward through the remaining parts of the cube (all sides except S1).

Answer 1

is there a special case or do i have to use the long general formula

Answer 2

@ganeshie8

Answer 3

you need to work part a by setting up a flux integral

you can use divergence thm for part b

Answer 4

\[\int\limits \int\limits (-2x+\sin(yz))i +(4y+\sin(xz))j + (-(1/3)\sin(xy))k * (+2i-4j+k)dxdy\]

Answer 5

that would be the integral for part a? @ganeshie8

Answer 6

what would my limits of integration be? 0 to 4 for dx and 0to2 for dy?

Answer 7

can you try -32 for part a once ?

Answer 8

no I don't have tries, it was due Tuesday but I want to understand this material

Answer 9

how did you get -32? did i set up the integral right? @ganeshie8

Answer 10

whats the normal vector of S1 ?

Answer 11

I don't know?

Answer 12

the x value is greater than 4 but I don't know what the normal is @ganeshie8

Answer 13

so do you think my integral is wrong for part a?

Answer 14

@ganeshie8

Answer 15

how did u get 2i-4j+k ?

Answer 16

you need to find the normal vector of surface to find the flux through the surface

Answer 17

i use the flux integral general formula

Answer 18

2i+4j+k is (-f_xi - f_yj +k)

Answer 19

@ganeshie8 is the normal just <4,0,0>

Answer 20

that gives you normal vector of surface
but why are you messing with your Force field ?

Answer 21

Right! the normal is just <4, 0, 0>

Answer 22

the unit nromal is just <1, 0, 0>

Answer 23

i was trying to use the general formula of flux integral

Answer 24

okay parameterize the surface

Answer 25

\[\int\limits_{?}^{?}\int\limits_{?}^{?} F (x,y, f(x,y)) * (-f_x-f_y+k)dxdy\]

Answer 26

this is the formula i was trying to use, where f(x,y) is z= -(1/3) sin(x,y)

Answer 27

@ganeshie8 this formula does not work here?

Answer 28

\[x=(-2+sinyz))+4t ....... y=(4y+\sin(xz))+0t.......... z+(3z+\sin(xy))+0t\]

Answer 29

@ganeshie8

Answer 30

that works always
but it would be ridankulous to use it for S1

Answer 31

your surface is very simply, you're told that it is perpendicular to x axis
so the normal vector is <1,0,0>

Answer 32

you can use below parameterization :

x = 4
y = y
z = z

Answer 33

so that's just doing way to much work?

Answer 34

\[\int\limits_{?}^{?}\int\limits_{?}^{?} F (4,y, z) * (1,0,0)dydz\]

Answer 35

setup the bounds and evaluate ^

Answer 36

so if the cube is lenght 2 and center at (4,0,0) dy is 0 to 2 and dz is 0 to 2 ? @ganeshie8

Answer 37

why is it " * (1,0,0)" ?

Answer 38

@ganeshie8 for part b is it the divF * (volume of the cube) - part a?

Answer 39

\[\hat{n}dS = \langle 1,0,0\rangle dydz \]

Answer 40

you're almost correct about bounds

Answer 41

so we don't need the normal <4,0,0>?

Answer 42

so is it -1 to 1?

Answer 43

http://gyazo.com/22c04d8d064622a3b5ec4843246bb11a

Answer 44

Yes! both y and z bounds would be from -1 to +1

Answer 45

\[\int\limits_{-1}^{1}\int\limits_{-1}^{1} F (4,y, z) * (1,0,0)dydz \]

Answer 46

\[\int\limits_{-1}^{1}\int\limits_{-1}^{1} (-2(4) + \sin(yz))dydz \]

Answer 47

evaluate ^

Answer 48

http://www.wolframalpha.com/input/?i=%5Cint_%7B-1%7D%5E1%5Cint_%7B-1%7D%5E%7B1%7D+%28-8+%2B+sin%28yz%29%29+dy+dz

Answer 49

thats for part a

Answer 50

you're correct about partb
just find the entire flux out of cube using divergence thm and subtract the part a

Answer 51

what would dA be?

Answer 52

curlF = _2i +4j+3k volume of cube= 8

Answer 53

@ganeshie8 so 5* 8 +32

Answer 54

72?

Answer 55

Looks good !

Answer 56

thank you so much @ganeshie8