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May 9th, 2015, 05:35 PM   #1
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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
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May 9th, 2015, 06:03 PM   #2
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Quote:
Originally Posted by harley05 View Post
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
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May 9th, 2015, 06:15 PM   #3
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Quote:
Originally Posted by harley05 View Post
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 $r(t)$!
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May 9th, 2015, 06:24 PM   #4
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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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