anonymous
  • anonymous
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 \[\int_0^{\pi}(cost\hat i +sint\hat j-t\hat k)\cdot(\hat i+cos(t) \hat j -sint \hat k)dt\] \[\int_0^{\pi}(cost+sintcost+tsint)dt\] \[\int_0^{\pi} cost dt+\int_0^{\pi} sintcostdt+\int_0^{\pi}tsintdt\]
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
\[sint]_0^{\pi}+\int_0^0 udu+[-tcost+sint]_0^{\pi}\]
anonymous
  • anonymous
mukushla is not typing a reply…whats wrong with site...lol let me check it
anonymous
  • anonymous
LOL...i think mukushla should type a reply

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anonymous
  • anonymous
its allright but what is that middle integral !!?
anonymous
  • anonymous
u substitution u =sint du=cost dt sin(pi)=0 sin(0)=0
anonymous
  • anonymous
almost done
anonymous
  • anonymous
0?
anonymous
  • anonymous
yes
anonymous
  • anonymous
cos Pi is -1 though
anonymous
  • anonymous
so negative pi should be the answer?
anonymous
  • anonymous
and cos (0) is 1 oh it cancels out!
anonymous
  • anonymous
no that still doesn't seem right... :(
anonymous
  • anonymous
i meant 0 for middle integral
anonymous
  • anonymous
answer is \(\pi\) ?
anonymous
  • anonymous
the first integral would be zero too and then we have left \[[-tcost+sint]_0^{\pi}=\pi\]
anonymous
  • anonymous
YAY!
anonymous
  • anonymous
We are DONE.....;-D

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