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Ed25

  • 2 years ago

Little help with line integral

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  1. Ed25
    • 2 years ago
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  2. Ed25
    • 2 years ago
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    Using Green's theorem I got 2 as result, but Wolfram says it is 5/3. Can somebody check on this?

  3. Ed25
    • 2 years ago
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    @experimentX

  4. myko
    • 2 years ago
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    express y as function of x and substitute into the integral: y=1-x

  5. Ed25
    • 2 years ago
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    I did it already. thats how I get 2

  6. amistre64
    • 2 years ago
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    can you walk me thru your application of greens thrm

  7. Ed25
    • 2 years ago
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    partial derivatives are: dp/dy=-3 and dq/dx=y, (i know that d should be written different)

  8. Ed25
    • 2 years ago
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    do we get dq-dp = y+3

  9. Ed25
    • 2 years ago
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    and then integral is: |dw:1370887152616:dw|

  10. Ed25
    • 2 years ago
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    boundaries for X goes from 0 to 1 and for Y they are 0 and 1-x, am I right?

  11. amistre64
    • 2 years ago
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    y = 0 to y = -x+1 x = 0 to x = 1

  12. amistre64
    • 2 years ago
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    so yes

  13. amistre64
    • 2 years ago
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    well, dxdy or dydx?

  14. amistre64
    • 2 years ago
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    dydx is intx[0,1] inty[0,1-x]

  15. amistre64
    • 2 years ago
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    (1-x)^2 / 2 from 0 to 1

  16. amistre64
    • 2 years ago
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    i forgot to yup the 3 :/

  17. amistre64
    • 2 years ago
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    \[\int_{0}^{1}\frac{(1-x)^2}{2}+3(1-x)~dx\]

  18. Ed25
    • 2 years ago
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    ok, i've found a mistake in my calculation, i'm doing it now directly by parametrization of curves, will see the result

  19. Ed25
    • 2 years ago
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    |dw:1370889421386:dw|

  20. Ed25
    • 2 years ago
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    for C1: x=t, y=0, 0<=t<=1 and Integral is = 1 C2: x=1-t, y=t, 0<=t<=1, int=-1/3 c3: x=0, y=1-t, 0<=t<=1, int=0

  21. Ed25
    • 2 years ago
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    hm?

  22. experimentX
    • 2 years ago
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    about surface integrals, \[ \iint dx dy \] shoudl you give area|dw:1370889753653:dw| choose any way you like to integrate ... if this gives you your required area then your parametrization if correct. just put the function inside and integrate it.

  23. amistre64
    • 2 years ago
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    \[\int_{0}^{1}\int_{0}^{1-y}y+3~dx~dy\] \[\int_{0}^{1}y(1-y)+3(1-y)~dy\] \[\int_{0}^{1}-y^2-2y+3~dy\] \[-\frac{1}{3}-1+3\]

  24. experimentX
    • 2 years ago
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    |dw:1370889905032:dw|

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