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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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Are you building a time machine?
hahahha i wish.
I would travel 3 months forward in time so i can be done calculus IV

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Other answers:

hahaha
what is the question? find the value?
yes exactly
it will be decimal
have you set up the triple integral?
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
then the integral should be F(dot)r_u X r_v/|r_u X r_v|
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
and the bounds can be the same i.e. 0-->2pi for both u and v
is f=|F|?
or was it suppose to be F
no f is the flux .. f is the result of the integral you solve
F is the vector field that you put inside the integral to solve the flux
essentially since they don't give you values for the constants, they should somehow cancel out in the end
sorry, i'm making it more work for you. but no one's answering anyways right?
so you got f?
no, I wish.. aha
its those cross products that literally take up 2 pages of work
because they are all long trigonmetric quantities very prone to mistakes
o yikes i can't commit to that now. i wish, it would be good practice
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
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
thanks for reading by the way though, your efforts are appreciated
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
Okay thank you :)
How can i contact mattt9 to discuss a maths problem?

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