Question

WILL AWARD MEDAL AND FAN!!
Stokes theorem question attached.

Answer 1

@Austin6i6

Answer 2

so i know how to do the curl which seems to be the easy part, but to parameterize the surface is kicking my retrice

Answer 3

(retrice)

Answer 4

@IrishBoy123

Answer 5

@Austin6i6

Answer 6

this is the explination im given but im so lost in the parameterization

Answer 7

do you know what equation 17.7.10 is?

Answer 8

haha no idea, our book ends at ch 13

Answer 9

then there is a simmilar problem with this explination

Answer 10

well the whole point of stokes is to change dr to a ds or vice verca. so its evident that they change it to ds and now are using one of the formulas for ds

Answer 11

parameterise when it is relevant

in this case, start with the curl. what do you get?

Answer 12

e^xk

Answer 13

then dot it with the normal vector

Answer 14

it looks like they are using \[\int\limits_{}^{}\int\limits_{}^{} curl F * K\]

Answer 15

me too

now, you are trying to project that onto the plane using the dot product. what is the normal for the place. the unit normal.

Answer 16

which would be multiplying e^xk be the k component of the normal

Answer 17

yeah so whats the question?

Answer 18

would the unit normal be <0,0,4>

Answer 19

cause its when k is 1, so z would be one

Answer 20

@pmkat14
i'll come back later. there is no point in your taking questions from >1 people. hopefully @Austin6i6 can fix this for you.

laters.

i'll go have a look at the other one you posted. that might be useful.

Answer 21

thanks @IrishBoy123

Answer 22

what did the curl come out to?

Answer 23

e^xk

Answer 24

and to clarify we want the normal of the boundry

Answer 25

yes. i believe so. normally i have equations where i can parameterize it then take the partials then cross em to find the normal but i dont know how youd do that here

Answer 26

your plane is
7x + y + 7z = 7

divide by 7 both sides
that should be you normal based on the equation of a plane

Answer 27

are we looking at the same problem lol

Answer 28

LMAO its the first link posted, but the second one is a different way to approach it, in reference to a different equation

Answer 29

but the equation with 7's is the one

Answer 30

ohh ok so the normal is (1,1/7,1)
how did you get (0,0,4)?

Answer 31

i thought it was only for the z direction so x and y would be 0

Answer 32

@Austin6i6 sorry for the lag i went to smoke a ciggy

Answer 33

with my normal vector, then how do i find the bounds to integrate over?

Answer 34

@IrishBoy123

Answer 35

\[curl \vec{F} \ \bullet \hat{n} = <0, 0, e^x> \ \bullet \ \frac{<7 ,1, 7>}{\sqrt{99}} = \frac{7 e^x}{\sqrt{99}} \]
\[dS = \frac{1}{|\hat{n} \ \bullet \ \hat{k}|} \ dx \ dy = \frac{\sqrt{99}}{7} dx \ dy\]
not suprisingly the numbers cancel as the curl is only in the z axis so this is just the double integral of e^x over the area
\[\int\limits \int\limits e^x dx \ dy\]

Answer 36

\[\int\limits_{y = 0}^{7} \ \ \int\limits_{x = 0}^{ \ \frac{7-y}{7}} e^x \ dx \ dy \ OR \ \int\limits_{x = 0}^{1} \ \ \int\limits_{y = 0}^{ \ 7-7x} e^x \ dy \ dx\]

either way i get 7 (e - 2) = 7e - 14

there may be some typos or glitches [check algebra] but "how" to do it.

Answer 37

drawings look terrible, orig attached....