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meggiesmermaid Group Title

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.

  • one year ago
  • one year ago

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  1. TuringTest Group Title
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    did you want to check an answer, or are you stuck?

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

    • one year ago
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    use\[u=\cos t\]and\[\sin^2t=1-\cos^2t\]

    • one year ago
  4. meggiesmermaid Group Title
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    so du =-sinx

    • one year ago
  5. TuringTest Group Title
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    how did you get cos^2t*sint for x^2+y^2dx ?

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

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

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

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

    • one year ago
  10. meggiesmermaid Group Title
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    so would you use u=cost and du=-sint ?

    • one year ago
  11. TuringTest Group Title
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    yep

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

    • one year ago
  13. TuringTest Group Title
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    and how to you propose to integrate that?

    • one year ago
  14. meggiesmermaid Group Title
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    oh yeah.haha.

    • one year ago
  15. meggiesmermaid Group Title
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    \[-512\int\limits_{1}^{-1}- 2u^2du - du \] ?

    • one year ago
  16. meggiesmermaid Group Title
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    i mean + du

    • one year ago
  17. TuringTest Group Title
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    yeah, looks good

    • one year ago
  18. TuringTest Group Title
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    oh wait, how did 512 get negative?

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

    • one year ago
  20. TuringTest Group Title
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    yeah, but you already covered that by changing thethe signs og the terms

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

    • one year ago
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    \[512\int_0^\pi2\cos^2t\sin t-\sin tdt\]\[u=\cos t\implies du=-\sin tdt\]\[512\int_{1}^{-1}- 2u^2du +\int_1^{-1} -\sin tdt\]

    • one year ago
  23. TuringTest Group Title
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    sorry I messed up the bounds on the last integral, they should be 0 to pi

    • one year ago
  24. meggiesmermaid Group Title
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    and the answer is 1706.66666666667?

    • one year ago
  25. meggiesmermaid Group Title
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    nevermind thats wrong.

    • one year ago
  26. TuringTest Group Title
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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...

    • one year 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...

    • one year ago
  28. meggiesmermaid Group Title
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    its right! yay!

    • one year ago
  29. TuringTest Group Title
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    sweet, I just hope you find your (likely algebra-based) mistake

    • one year ago
  30. meggiesmermaid Group Title
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    thats what i got to too! thanks so much!

    • one year ago
  31. TuringTest Group Title
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    welcome!

    • one year ago
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