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Greens theorem-is this along the right line

  1. Jul 18, 2012 #1
    1. The problem statement, all variables and given/known data
    Cis the boundary of the region given by curves [itex]y = x^{2}[/itex] and y=x use Greens theorem to evaluate the following line integrals.
    a) [itex]\oint(6xy-y^2 )dx[/itex]
    b) [itex]\oint(6xy-y^2 )dy[/itex]


    2. Relevant equations



    3. The attempt at a solution

    so greens theorem states [itex]\oint Mdx + Ndy = \int\int \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} dxdy[/itex]

    so for (a) M=6xy-[itex]y^2[/itex] N=0
    The 2 curves intersect at (1,1)

    so [itex]\oint (6xy-y^2)dx = \int^{1}_{0} \int^{x^2}_{x} (6x-2y) dydx[/itex]
    =[itex]\int^{1}_{0} -x^4 + 6x^3 -7x^2 dx[/itex]
    =[itex]\frac{-31}{30}[/itex]

    and for (b) N=6xy-[itex]y^2[/itex] M=0
    The 2 curves intersect at (1,1)

    so [itex]\oint (6xy-y^2)dy = \int^{1}_{0} \int^{x^2}_{x} (6y-2y)dydx[/itex]
    =[itex]\int^{1}_{0} 2x^4 - 2x^2 dx[/itex]
    =[itex]\frac{-4}{15}[/itex]

    am i doing this right?
     
  2. jcsd
  3. Jul 18, 2012 #2

    sharks

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    Gold Member

    Always sketch the graph, to know the boundary of the region enclosed by the two curves. Check the attachment.

    First, describe the region enclosed by the two curves (let's call it, region R).

    For x fixed, y varies from ##y = x^2## to ##y= x##
    x varies from x = 0 to x = 1

    Let's consider part (a). You have correctly recognized terms M and N. Now, you need to find:
    $$\iint_R \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} dxdy=\iint_R 0 - \frac{\partial M}{\partial y} dxdy=-\iint_R \frac{\partial M}{\partial y} dxdy$$
    Set the limits:$$-\int_0^1 \int_{x^2}^x \frac{\partial M}{\partial y} dydx$$
    You have to find: $$\frac{\partial M}{\partial y}=6x-2y$$
    Thus, you get:$$-\int_0^1 \int_{x^2}^x (6x-2y) dydx=-\frac{11}{30}$$
    Now, you should be able to do part (b).
     

    Attached Files:

    Last edited: Jul 18, 2012
  4. Jul 18, 2012 #3
    Thanks sharks,
    i think that cleared up a few things.
    When i am doing part B do the limits change?as in...

    [itex]\int_0^1 \int_{y}^{\sqrt{y}} (6y) dxdy [/itex]

    or do i keep the same limits?
     
  5. Jul 18, 2012 #4

    sharks

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    Gold Member

    Since the region of integration remains unchanged, you can use the same limits as in part (a).

    Alternatively, you could also have defined the limits as:
    $$\int_0^1 \int_{x^2}^x (6y) dydx$$but the answer should be the same.
     
  6. Jul 18, 2012 #5
    Thanks sharks,I did it out both ways and got the same answer [itex]\frac{2}{5}[/itex]

    I have a similar question...

    [itex]\oint (x^2 y)dx + (y+xy^2) dy [/itex] between the curves [itex]y=x^2 [/itex] and [itex] x=y^2 [/itex]

    i sketched out the region and it very similar to the above question.

    so from green theorem i get [itex]\int^{1}_{0} \int^{\sqrt{x}}_{x^2} y^2 - x^2 dydx[/itex] which when i do it out i get 0. but from my sketch its not 0. am i taking the wrong limits?i've done it out 2 times. and got the same answer
     
  7. Jul 19, 2012 #6
    [itex] \int^{1}_{0} \int^{\sqrt{x}}_{x^2} (y^2 - x^2) dydx [/itex]

    [itex] \int^{1}_{0} (\frac{1}{3} y^3 - x^2 y)^\sqrt{x}_{x^2} dx [/itex]

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

    = [itex] (\frac{2}{15}x^{\frac{5}{2}} - \frac{2}{7}x^{\frac{7}{2}} - \frac{1}{21}x^{7} + \frac{1}{5}x^{5})^{1}_{0} [/itex]

    which works as zero...this cant be right?
     
  8. Jul 19, 2012 #7

    sharks

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    Gold Member

    The answer is indeed 0. See the two attachments.
     

    Attached Files:

    Last edited: Jul 19, 2012
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