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MathSofiya Group Title

Evaluate the line integral \[\int_c F\cdot dr\] where c is given by \[\hat r (t)=t\hat i +sint\hat j +cost \hat k\] \[0 \le t le \pi\] and: \[\hat r(t)=\hat i +cost \hat j -sint \hat k\] \[F(x, y, z)=z\hat i+y \hat j-x\hat k\] x=t y=sint z=cost \[\int_0^{\pi}(cost\hat i +sint\hat j-t\hat k)\cdot(\hat i+cos(t) \hat j -sint \hat k)dt\] \[\int_0^{\pi}(cost+sintcost+tsint)dt\] \[\int_0^{\pi} cost dt+\int_0^{\pi} sintcostdt+\int_0^{\pi}tsintdt\]

  • 2 years ago
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  1. MathSofiya Group Title
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    \[sint]_0^{\pi}+\int_0^0 udu+[-tcost+sint]_0^{\pi}\]

    • 2 years ago
  2. mukushla Group Title
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    mukushla is not typing a reply…whats wrong with site...lol let me check it

    • 2 years ago
  3. MathSofiya Group Title
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    LOL...i think mukushla should type a reply

    • 2 years ago
  4. mukushla Group Title
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    its allright but what is that middle integral !!?

    • 2 years ago
  5. MathSofiya Group Title
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    u substitution u =sint du=cost dt sin(pi)=0 sin(0)=0

    • 2 years ago
  6. mukushla Group Title
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    almost done

    • 2 years ago
  7. MathSofiya Group Title
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    0?

    • 2 years ago
  8. mukushla Group Title
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    yes

    • 2 years ago
  9. MathSofiya Group Title
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    cos Pi is -1 though

    • 2 years ago
  10. MathSofiya Group Title
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    so negative pi should be the answer?

    • 2 years ago
  11. MathSofiya Group Title
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    and cos (0) is 1 oh it cancels out!

    • 2 years ago
  12. MathSofiya Group Title
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    no that still doesn't seem right... :(

    • 2 years ago
  13. mukushla Group Title
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    i meant 0 for middle integral

    • 2 years ago
  14. mukushla Group Title
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    answer is \(\pi\) ?

    • 2 years ago
  15. MathSofiya Group Title
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    the first integral would be zero too and then we have left \[[-tcost+sint]_0^{\pi}=\pi\]

    • 2 years ago
  16. MathSofiya Group Title
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    YAY!

    • 2 years ago
  17. mukushla Group Title
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    We are DONE.....;-D

    • 2 years ago
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