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anonymous
 4 years ago
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))
anonymous
 4 years ago
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))

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Are you building a time machine?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I would travel 3 months forward in time so i can be done calculus IV

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what is the question? find the value?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0have you set up the triple integral?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no 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
 4 years ago
Best ResponseYou've already chosen the best response.0then the integral should be F(dot)r_u X r_v/r_u X r_v

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0where 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
 4 years ago
Best ResponseYou've already chosen the best response.0and the bounds can be the same i.e. 0>2pi for both u and v

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0or was it suppose to be F

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no f is the flux .. f is the result of the integral you solve

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0F is the vector field that you put inside the integral to solve the flux

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0essentially since they don't give you values for the constants, they should somehow cancel out in the end

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry, i'm making it more work for you. but no one's answering anyways right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0its those cross products that literally take up 2 pages of work

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0because they are all long trigonmetric quantities very prone to mistakes

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0o yikes i can't commit to that now. i wish, it would be good practice

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i 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
 4 years ago
Best ResponseYou've already chosen the best response.0yeah 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 1030 hours of calc a week

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thanks for reading by the way though, your efforts are appreciated

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i 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
 4 years ago
Best ResponseYou've already chosen the best response.0How can i contact mattt9 to discuss a maths problem?
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