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 May 9th, 2015, 04:35 PM #1 Member   Joined: Apr 2014 From: australia Posts: 68 Thanks: 32 integral evaluation Hello, Find the work done by the force field F(x, y, z) = (–x^4 y^2 i) – (4 j) + (ln(x^4 – y^2) k) acting on a charged electric particle moving along the path given by the equation r(t) = (2 cos t i) + (2 sin t j) + (4 k); where the parameter t varies from pi /4 to 7pi /4. thanks in advance I have had a go at it and after evaluating the integral $\displaystyle I=int(F.dr)$ I get an answer = 8.sqrt(2) thanks in advance Thanks from Yury Stepanyants
May 9th, 2015, 05:03 PM   #2
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 Originally Posted by harley05 Hello, Find the work done by the force field F(x, y, z) = (–x^4 y^2 i) – (4 j) + (ln(x^4 – y^2) k) acting on a charged electric particle moving along the path given by the equation r(t) = (2 cos t i) + (2 sin t j) + (4 k); where the parameter t varies from pi /4 to 7pi /4. thanks in advance I have had a go at it and after evaluating the integral $\displaystyle I=int(F.dr)$ I get an answer = 8.sqrt(2) thanks in advance
You are correct. Nicely done!

Would you care to share your solution so others might see it?

-Dan

May 9th, 2015, 05:15 PM   #3
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Quote:
 Originally Posted by harley05 Hello, Find the work done by the force field F(x, y, z) = (–x^4 y^2 i) – (4 j) + (ln(x^4 – y^2) k) acting on a charged electric particle moving along the path given by the equation r(t) = (2 cos t i) + (2 sin t j) + (4 k); where the parameter t varies from pi /4 to 7pi /4. thanks in advance I have had a go at it and after evaluating the integral $\displaystyle I=int(F.dr)$ I get an answer = 8.sqrt(2) thanks in advance
Boy are you lucky there wasn't a function of $t$ attached to $k$ in the $r(t)$!

May 9th, 2015, 05:24 PM   #4
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Quote:
 Originally Posted by neelmodi Boy are you lucky there wasn't a function of $t$ attached to $k$ in the $r(t)$!
your not wrong there!

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