given curve:(adsbygoogle = window.adsbygoogle || []).push({});

x=cost

y=sin(2t)

z=cos(2t)

0<=t<=2pi

vectorian field:

F=(exp(x^3),3z/(y^2+z^2),-3y/(y^2+z^2))

so the ingral SF*dr (on the curve)

at the direction of t from 0 to 2pi is?

if i assign the parametrion to the field i get 12pi from integrals on y and z

but i still need to proove that the integral there:

https://www.physicsforums.com/showthread.php?t=43059

is zero

and i think its impossible..

i think using stocks is impossible because building sutaible surface to the curve is imposible

any other sugestions?

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Work integral on hard field

Can you offer guidance or do you also need help?

Draft saved
Draft deleted

Loading...

Similar Threads - integral hard field | Date |
---|---|

Hard Integral Question | Jan 13, 2016 |

Hard Integration | May 4, 2015 |

Hard integrals in 2D | Oct 17, 2014 |

Hard Integrals | May 9, 2013 |

A hard integration. | Apr 10, 2012 |

**Physics Forums - The Fusion of Science and Community**