anonymous
  • anonymous
let f be the outward flux of the vector field F = (c_1x,c_2y,c_3z) over the torus: x=cosu(1.86+0.9cosv); y=sinu(1.86+0.9cosv); z=0.9sinv; for 0<=u<=2pi; 0<=v<=2pi, where c_1, c_2, c_3 are positive constraints. then the value of cot(f/(c_1+c_2+c_3))
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Are you building a time machine?
anonymous
  • anonymous
hahahha i wish.
anonymous
  • anonymous
I would travel 3 months forward in time so i can be done calculus IV

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anonymous
  • anonymous
hahaha
anonymous
  • anonymous
what is the question? find the value?
anonymous
  • anonymous
yes exactly
anonymous
  • anonymous
it will be decimal
anonymous
  • anonymous
have you set up the triple integral?
anonymous
  • anonymous
no its a ridiculous amount of work. and it should only be double? im not changing the coordinates. when you reparametrize it you get r(u,v) =x(u,v)i +y(u,v)j +z(u,v)k where x(u,v), etc. are those functions for x y and z above
anonymous
  • anonymous
then the integral should be F(dot)r_u X r_v/|r_u X r_v|
anonymous
  • anonymous
where thats the vector cross product operator and r_u and r_v are the partial derivatives of r with respect to u and v
anonymous
  • anonymous
and the bounds can be the same i.e. 0-->2pi for both u and v
anonymous
  • anonymous
is f=|F|?
anonymous
  • anonymous
or was it suppose to be F
anonymous
  • anonymous
no f is the flux .. f is the result of the integral you solve
anonymous
  • anonymous
F is the vector field that you put inside the integral to solve the flux
anonymous
  • anonymous
essentially since they don't give you values for the constants, they should somehow cancel out in the end
anonymous
  • anonymous
sorry, i'm making it more work for you. but no one's answering anyways right?
anonymous
  • anonymous
so you got f?
anonymous
  • anonymous
no, I wish.. aha
anonymous
  • anonymous
its those cross products that literally take up 2 pages of work
anonymous
  • anonymous
because they are all long trigonmetric quantities very prone to mistakes
anonymous
  • anonymous
o yikes i can't commit to that now. i wish, it would be good practice
anonymous
  • anonymous
i figured out r_u X r_v.. but its taking the absolute value that im just too scared to try. because you have to square each i j k term and put them under a root sign
anonymous
  • anonymous
yeah I understand aha, this was really a shot in the dark trying to get help on here. I can't imagine anyone wanting to attempt this question. It's the last one I have left on my assignment. Our calculus professor is straight up unfair, literally 10-30 hours of calc a week
anonymous
  • anonymous
thanks for reading by the way though, your efforts are appreciated
anonymous
  • anonymous
i was more interested in someone else answering, i only went to calc III. good luck, there are a lot of helpful people on here. i would post again so they see it
anonymous
  • anonymous
Okay thank you :)
anonymous
  • anonymous
How can i contact mattt9 to discuss a maths problem?

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