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MathSofiya

  • 3 years ago

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 \[\int_0^{\pi}(cost\hat i +sint\hat j-t\hat k)\cdot(\hat i+cos(t) \hat j -sint \hat k)dt\] \[\int_0^{\pi}(cost+sintcost+tsint)dt\] \[\int_0^{\pi} cost dt+\int_0^{\pi} sintcostdt+\int_0^{\pi}tsintdt\]

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

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

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

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

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

  6. mukushla
    • 3 years ago
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    almost done

  7. MathSofiya
    • 3 years ago
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    0?

  8. mukushla
    • 3 years ago
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    yes

  9. MathSofiya
    • 3 years ago
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    cos Pi is -1 though

  10. MathSofiya
    • 3 years ago
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    so negative pi should be the answer?

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

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

  13. mukushla
    • 3 years ago
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    i meant 0 for middle integral

  14. mukushla
    • 3 years ago
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    answer is \(\pi\) ?

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

  16. MathSofiya
    • 3 years ago
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    YAY!

  17. mukushla
    • 3 years ago
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    We are DONE.....;-D

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