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Luigi0210

  • one year ago

Stoke's Theorem:

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  1. Luigi0210
    • one year ago
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    Use Stoke's theorem to solve \(\large \int_{C} F~ dr \) where F(x,y,z) =<xz, 2xy, 3xy> and C is the boundary of part of plane 3x+y+z=3 in the first octant.

  2. Astrophysics
    • one year ago
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    I had trouble understanding Stoke's and I've always wanted to go over it again and learn it a better way...mhm lets see if we can figure this out \[\int\limits_{C} \vec F \cdot d \vec r = \int\limits \int\limits_S curl \vec F \cdot d \vec S\] right

  3. Luigi0210
    • one year ago
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    Not sure, I'm a bit confused because of these: http://prntscr.com/7t4ipd

  4. Luigi0210
    • one year ago
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    Same thing?

  5. dan815
    • one year ago
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    yes all same thing

  6. Astrophysics
    • one year ago
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    Yes, you should understand the notation and what it means then I think it would be more clear hmm. I'm thinking we can pick any surface S with the boundary C.

  7. Astrophysics
    • one year ago
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    This would be a nice question to observe haha, I sort of have an idea, but I think @dan815 @ganeshie8 @Empty would be better in teaching this, plus I can learn from them to xD.

  8. dan815
    • one year ago
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    |dw:1436995421444:dw|

  9. Astrophysics
    • one year ago
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    Projection?!

  10. dan815
    • one year ago
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    we gotta see those intersections with xz yz and xy plane

  11. dan815
    • one year ago
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    we can see the 3 intersections with the 3 axis by setting x,y=0 y,z=0 x,z=0

  12. dan815
    • one year ago
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    |dw:1436996249413:dw|

  13. dan815
    • one year ago
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    thre u go not integrate over that surface area

  14. dan815
    • one year ago
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    |dw:1436996425783:dw|

  15. dan815
    • one year ago
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    F is a function of x,y so we want to refine our new surface area in terms of x and y too

  16. dan815
    • one year ago
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    ds=|F_x X Fy| dxdy

  17. dan815
    • one year ago
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    i dont remember that exactly, u can work it out though

  18. dan815
    • one year ago
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    |dw:1436996594088:dw|

  19. dan815
    • one year ago
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    there fore ds= |fx X fy| dxdy

  20. dan815
    • one year ago
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    your domain will be that triangle

  21. dan815
    • one year ago
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    |dw:1436996772371:dw|

  22. dan815
    • one year ago
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    |dw:1436996788233:dw|

  23. dan815
    • one year ago
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    |dw:1436996930190:dw|

  24. dan815
    • one year ago
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    finally note how Gx and Gy vectors are formed

  25. dan815
    • one year ago
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    |dw:1436997056399:dw|

  26. dan815
    • one year ago
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    |dw:1436997128852:dw|

  27. dan815
    • one year ago
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    and finally u can look up a proof for why the determinant of the 2 vectors gives you the area composed by them

  28. dan815
    • one year ago
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    but for starters u can take the formula at face cvalue how u dot v = |U||V| cos theta and U X V = |U||V| sin theta <--- area

  29. dan815
    • one year ago
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    as |V| sin theta is the height

  30. dan815
    • one year ago
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    so that would the formula for the area of a parallelogram

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