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meggiesmermaid

  • 3 years ago

Evaluate the line integral (x2+y2)dx+2xydy where is the path of the semicircular arc of the circle x2+y2=64 starting at (8,0) and ending at (−8,0) going counterclockwise.

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  1. TuringTest
    • 3 years ago
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    did you want to check an answer, or are you stuck?

  2. meggiesmermaid
    • 3 years ago
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    \[512\int\limits_{0}^{\pi} \cos^2t sint - \sin^3t\] im stuck at how to integrate this.

  3. TuringTest
    • 3 years ago
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    use\[u=\cos t\]and\[\sin^2t=1-\cos^2t\]

  4. meggiesmermaid
    • 3 years ago
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    so du =-sinx

  5. TuringTest
    • 3 years ago
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    how did you get cos^2t*sint for x^2+y^2dx ?

  6. meggiesmermaid
    • 3 years ago
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    i keep getting confused but heres where i started \[\int\limits_{0}^{\pi} ((8cost)^2 + (8sint)^2)(-8sint) + 2(8cost)(8sint)(8cost)dt\]

  7. TuringTest
    • 3 years ago
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    yes, and that can be written\[\int_0^\pi512(\sin^2t+\cos^2t)(-\sin t)dt+\int_0^\pi512(2\cos^2t\sin t) dt\]

  8. meggiesmermaid
    • 3 years ago
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    \[-512\int\limits_{0}^{?\pi} sint dt +1024\int\limits_{0}^{\pi} \cos^2tsint\] which becomes this

  9. TuringTest
    • 3 years ago
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    yes, which you can recombine to form this\[512\int_0^\pi2\cos^2t\sin t-\sin tdt\]

  10. meggiesmermaid
    • 3 years ago
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    so would you use u=cost and du=-sint ?

  11. TuringTest
    • 3 years ago
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    yep

  12. meggiesmermaid
    • 3 years ago
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    or could you do \[sint(\cos^2t-1)\] which becomes \[\sin^3t\] ?

  13. TuringTest
    • 3 years ago
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    and how to you propose to integrate that?

  14. meggiesmermaid
    • 3 years ago
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    oh yeah.haha.

  15. meggiesmermaid
    • 3 years ago
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    \[-512\int\limits_{1}^{-1}- 2u^2du - du \] ?

  16. meggiesmermaid
    • 3 years ago
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    i mean + du

  17. TuringTest
    • 3 years ago
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    yeah, looks good

  18. TuringTest
    • 3 years ago
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    oh wait, how did 512 get negative?

  19. meggiesmermaid
    • 3 years ago
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    im confused because du=-sint and the sint is positive so (-1/-1) would need to be multiplied right?

  20. TuringTest
    • 3 years ago
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    yeah, but you already covered that by changing thethe signs og the terms

  21. TuringTest
    • 3 years ago
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    however the double du notation is highly unorthodox and looks screwy to me, so I'd avoid it

  22. TuringTest
    • 3 years ago
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    \[512\int_0^\pi2\cos^2t\sin t-\sin tdt\]\[u=\cos t\implies du=-\sin tdt\]\[512\int_{1}^{-1}- 2u^2du +\int_1^{-1} -\sin tdt\]

  23. TuringTest
    • 3 years ago
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    sorry I messed up the bounds on the last integral, they should be 0 to pi

  24. meggiesmermaid
    • 3 years ago
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    and the answer is 1706.66666666667?

  25. meggiesmermaid
    • 3 years ago
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    nevermind thats wrong.

  26. TuringTest
    • 3 years ago
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    \[512\int_0^\pi2\cos^2t\sin t-\sin tdt\]\[u=\cos t\implies du=-\sin tdt\]\[512\int_{1}^{-1}- 2u^2du +\int_0^{\pi} -\sin tdt\]let me see what I get...

  27. TuringTest
    • 3 years ago
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    \[512\left(\left.-\frac23u^3\right|_1^{-1}+\left.\cos t\right|_0^\pi\right)\]\[512(\frac43-2)=512(-\frac23)\]but I always could have made a mistake...

  28. meggiesmermaid
    • 3 years ago
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    its right! yay!

  29. TuringTest
    • 3 years ago
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    sweet, I just hope you find your (likely algebra-based) mistake

  30. meggiesmermaid
    • 3 years ago
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    thats what i got to too! thanks so much!

  31. TuringTest
    • 3 years ago
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    welcome!

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