1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Greens Theorem

  1. Apr 28, 2012 #1
    1. The problem statement, all variables and given/known data
    Let C be the boundary of the region bounded by the curves y=[itex]x^{2}[/itex] and y=x. Assuming C is oriented counter clockwise, Use green's theorem to evaluate the following line integrals (a) [itex]\oint(6xy-y^2)dx[/itex] and (b) [itex]\oint(6xy-y^2)dy[/itex]


    2. Relevant equations



    3. The attempt at a solution

    [itex]\int^{0}_{1} 6x^2 - x^2[/itex]
    [itex]\int^{0}_{1} 5x^2 [/itex] = -[itex]\frac{5}{3}[/itex]
    and
    [itex]\int^{1}_{0} 6x^3 - x^4[/itex] = [itex]\frac{6}{4} - \frac{1}{5} = \frac{13}{10}[/itex]

    so [itex]\oint[/itex] = -[itex]\frac{11}{30}[/itex]

    but
    [itex]\int\int_{R}[/itex] [itex]\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}[/itex])dxdy
    M=6xy-[itex]y^{2}[/itex] and N=0
    [itex]\frac{\partial M}{\partial y} 6x-2y[/itex]

    [itex]\int\int_{R}[/itex](6x-2y)dxdy

    [itex]\int^{1}_{0} [ \int^{x}_{y=x^2} (6x-2y)dy] dx [/itex]

    [itex]\int^{1}_{0} 5x^2 - 6x^3 - x^4 dx [/itex]

    = [itex]\frac{-1}{30}[/itex]
    anyone got any idea what im doing wrong here!stumped
     
  2. jcsd
  3. Apr 28, 2012 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Poorly written, but assuming you are doing the circuit integral -11/30 is correct.

    Check your sign on that ##x^4## term in the second to last line. And don't you want ##-M_y## for Green's theorem?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Greens Theorem
  1. Green's Theorem (Replies: 8)

  2. Greens theorem (Replies: 10)

  3. Green's Theorem (Replies: 4)

  4. Green's Theorem (Replies: 2)

Loading...