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mattt9

  • 2 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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  1. helloimjared
    • 2 years ago
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    Are you building a time machine?

  2. mattt9
    • 2 years ago
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    hahahha i wish.

  3. mattt9
    • 2 years ago
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    I would travel 3 months forward in time so i can be done calculus IV

  4. helloimjared
    • 2 years ago
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    hahaha

  5. elica85
    • 2 years ago
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    what is the question? find the value?

  6. mattt9
    • 2 years ago
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    yes exactly

  7. mattt9
    • 2 years ago
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    it will be decimal

  8. elica85
    • 2 years ago
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    have you set up the triple integral?

  9. mattt9
    • 2 years ago
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    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

  10. mattt9
    • 2 years ago
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    then the integral should be F(dot)r_u X r_v/|r_u X r_v|

  11. mattt9
    • 2 years ago
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    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

  12. mattt9
    • 2 years ago
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    and the bounds can be the same i.e. 0-->2pi for both u and v

  13. elica85
    • 2 years ago
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    is f=|F|?

  14. elica85
    • 2 years ago
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    or was it suppose to be F

  15. mattt9
    • 2 years ago
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    no f is the flux .. f is the result of the integral you solve

  16. mattt9
    • 2 years ago
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    F is the vector field that you put inside the integral to solve the flux

  17. mattt9
    • 2 years ago
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    essentially since they don't give you values for the constants, they should somehow cancel out in the end

  18. elica85
    • 2 years ago
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    sorry, i'm making it more work for you. but no one's answering anyways right?

  19. elica85
    • 2 years ago
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    so you got f?

  20. mattt9
    • 2 years ago
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    no, I wish.. aha

  21. mattt9
    • 2 years ago
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    its those cross products that literally take up 2 pages of work

  22. mattt9
    • 2 years ago
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    because they are all long trigonmetric quantities very prone to mistakes

  23. elica85
    • 2 years ago
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    o yikes i can't commit to that now. i wish, it would be good practice

  24. mattt9
    • 2 years ago
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    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

  25. mattt9
    • 2 years ago
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    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

  26. mattt9
    • 2 years ago
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    thanks for reading by the way though, your efforts are appreciated

  27. elica85
    • 2 years ago
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    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

  28. mattt9
    • 2 years ago
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    Okay thank you :)

  29. jeff1949
    • 2 years ago
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    How can i contact mattt9 to discuss a maths problem?

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