Question

Find the flux through the surface. I cant make it possible, need help setting up.
F(x, y, z) = ze^(xy)i − 3ze^(xy)j + (xy)k, through the parrellogram x=u+v, y=u-v, z=1+2u+v, for u in [0,2] and v in [0,1]... please help

Answer 1

so i take it this is thru 6 surfaces?

Answer 2

Looks like one surface to me...

Answer 3

no its through the parametric parallelogram in u and v

Answer 4

are the vectors u and v posted correctly?

Answer 5

those arnt vectors

Answer 6

http://www.webassign.net/scalcet7/16-7-005.gif

Answer 7

murmur ... well that narrows down the context at least :)

Answer 8

that is the parallogram (ignore the int they just reused the surface), this is from a calc 3 course

Answer 9

been to far gone for me to make a quick assessment then :/

Answer 10

what i can recall is something of the sort:
\[\int F\cdot n~dS\]

Answer 11

where n is a gradient of the surface that is being fluxed thru

Answer 12

\[dS=\sqrt{(x')^2+(y')^2+(z')^2}~dA\]

Answer 13

\[ dS= \sqrt{14}\], then i try to solve the int, replacing z with \[3/2x+1/2y+1\]and its way too hard to do

Answer 14

\[\frac{ 3 }{ 2 }x + \frac{ 1 }{ 2 }y + 1 = z\] because \[z = 1 + 2u +v, x = u + v, y = u - v\]

Answer 15

It's difficult even after performing the dot-product and separating it into three separate integrals?

Answer 16

How can i do that? all terms contain x y and z

Answer 17

You should be dotting the vector field F with the vector field (r_u X r_v).

Answer 18

\[\int\limits \int\limits_S \vec{F} \cdot d\vec{S}=\int\limits\int\limits_D \vec{F}\cdot (\vec{r_u} \times \vec{r_v})\ dA\]

Answer 19

\[|r _{u} X r_{v}| = \sqrt{14}\] then if i plug in u and v for x,y,z i get\[\sqrt{14} \int\limits_{0}^{1} \int\limits_{0}^{2} -2(1+2u+v)e^{(u^2+v^2)}+u^2v^2dudv\]

Answer 20

which i dont know how to solve

Answer 21

No. You don't multiply by the magnitude of r_u X r_v. You take the two vector fields and dot them:\[(ze^{xy}\hat{i}-3ze^{xy}\hat{j}+xy\hat{k})\cdot (3\hat{i}+1\hat{j}-2\hat{k})\](That's the vector I got for r_u X r_v, but I didn't double check it).

Answer 22

thats the correct vector

Answer 23

still, can you see that its possible now? that integral looks very difficult

Answer 24

OH I SEE! the exponential function cancels out, thank you! :)

Answer 25

Hm... I'm getting -20/3, but that's not the right answer. Are you getting 4?

Answer 26

i got -20/3

Answer 27

Hm... My book has the same problem in it, and my book says the answer is 4.

Answer 28

what book do u use yeah -20/3 is wrong i just verfied that on my hmwrk

Answer 29

James Stuart. Calculus, 7th Edition.

Answer 30

me tooodd i looked but couldnt find the prob, 4 is right tho

Answer 31

thanks again

Answer 32

Yeah, I just wish I knew what we were doing wrong. Or if it's a typo?

Answer 33

Hey, I redid this problem and I figured out why we were getting wrong answers.
\[xy=(u+v)(u-v)=u^2-v^2\]We were both using:\[xy \ne u^2+v^2\]><