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GREEN'S THRM Calc III

  1. Apr 27, 2013 #1
    1. The problem statement, all variables and given/known data

    The problem is attached with my work so far.

    Consider the vector Field F, where C is the perimeter of the circle of radius 3 centered at (2,5)


    2. Relevant equations

    ∫ Mdx+Ndy----->∫∫ (∂N/∂x)-(∂M/∂y) da

    3. The attempt at a solution

    Attached is my work thus far. I am having trouble figuring out how to answer this question. I know that I cannot find a potential function f: ∇f=F because ∂M/∂y ≠ ∂N/∂x so i am assuming I could use greens thrm or another method in which i would just do ∫F(r(t)) (dot) (dr/dt)*(dt) (attached second picture) but i am unsure if I parametrized the circle correctly.

    Any help would be greatly appreciated. Also I am unable to find a similar question in my text book, if anyone can point me in the right direction to a similar problem that would also be great.

    thank you
     

    Attached Files:

    Last edited by a moderator: Apr 27, 2013
  2. jcsd
  3. Apr 27, 2013 #2

    SammyS

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    It would help if you would give a complete statement of the problem.

    It's kinda/sorta given in one of your attachments.

    ##\displaystyle \oint_{C} (N\, \mathrm{d}x + M\, \mathrm{d}y) = \iint_{D} \left(\frac{\partial M}{\partial x} - \frac{\partial N}{\partial y}\right)\, \mathrm{d}x\, \mathrm{d}y \ ## is for the case where path C is taken counter-clockwise. It line integral for a clockwise path is the negative of this.


    Doing the area integral should be easy. Define the region of integration.

    You have the equation of the circle: [itex]\displaystyle \ (x-2)^2+(y-5)^2=9 \ .\ [/itex] Solve that for x or y, depending upon which integral you want to integrate over first.
     
  4. Apr 27, 2013 #3

    Thank you for the reply, I greatly appreciate it.

    Maybe I misunderstood you, the question is stated word for word in the first sentence of the first attachment.

    As for solving the area integral. I am trying to understand if I can put everything in terms of one variable t. So i do not have to solve y=5+(9-(x-2)^(2))^(1/2) or y=5-(9-(x-2)^(2))^(1/2)
    or can i just list the end points as -1<x<5 and 2<y<8 or is there is a different method. as i stated, if you know of any similar questions that has a circle that is not centered at the orgin, i would greatly appreciate it.

    thank you
     
  5. Apr 27, 2013 #4

    okay so if i use the first definition ∫Ndx+Mdy in terms of t i get

    ∫ ((cost)^(2))*(sint)^(2)) from 0 to 2pi
     
  6. Apr 27, 2013 #5

    SammyS

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    You seemed unsure as to whether in path of integration was clockwise or was counter-clockwise.

    If you solve for y, as you are doing, then that indicates that you integrate with respect to y first, then with respect to x.

    If you merely do the following ##\displaystyle \ \int_{-1}^{5} \int_{2}^{8} \left(\frac{\partial M}{\partial x} - \frac{\partial N}{\partial y}\right)\, \mathrm{d}y\, \mathrm{d}x\,,\ ## then you would be integrating over the entire rectangle -1<x<5 and 2<y<8 .

    You need the limits of integration for y to go from the lower semi-circle to upper semi-circle.
     
  7. Apr 27, 2013 #6

    SammyS

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    For one thing, you do not show what your variable of integration is --- you have no dt .

    For another, that integral is counter-clockwise.

    Most importantly: How do you get that from the original problem ?

    The area integral, given by Green's Theorem looks much easier to work with.
     
  8. Apr 28, 2013 #7
    That is what I was asking, if i can parametrize the entire thing in terms of one variable t, as we are taught in the text book. so if we have the equation

    (x-2)^(2) + (y-5)^(2)=9 and we know that x=rcost and y=rsint as a asked can we put
    (3cost-2)^(2) +(3sint-5)^(2)=9

    i do not think you are going to use x and y, i believe you are going to use t OR theta and convert into polar integral form. this problem should have taken less than 5 min to solve, and if the equation for y=5+(9-(x-2)^(2))^(1/2) or y=5-(9-(x-2)^(2))^(1/2) that would take a lot of time and i know that that is not right.

    thank you for the help.
     
    Last edited by a moderator: Apr 29, 2013
  9. Apr 29, 2013 #8

    SammyS

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    Yes, this could likely be done in five minutes ... if you do the area integral.

    That's the integral on the right-hand side of ## \displaystyle \oint_{C} (N\, \mathrm{d}x + M\, \mathrm{d}y) = \iint_{D} \left(\frac{\partial M}{\partial x} - \frac{\partial N}{\partial y}\right)\, \mathrm{d}x\, \mathrm{d}y \ . ##

    The only issue you seemed to have with that was determining the limits of integration for the inner integral.

    More on that later .


    Since you seemed to be determined to do the problem directly, by doing line integral on the left-hand side of the above, I indicated that what you had done in attempting that
    ∫ ((cost)^(2))*(sint)^(2))dt from 0 to 2pi​
    was not correct. What I didn't say was that it appears to be completely missing the vector valued function, F(x,y) .

    If you want to persue this method, let me know and I'll try to help.


    However, my recomendation is to evaluate the area integral ## \displaystyle \ \int_{-1}^{5} \int_\text{Lower}^\text{Upper} \left(\frac{\partial M}{\partial x} - \frac{\partial N}{\partial y}\right)\, \mathrm{d}y\, \mathrm{d}x\,,\ ##

    where "Lower" means the lower half of the circle and "Upper" means the upper half of the circle. You have already solved for equations corresponding to Upper and Lower portions of the circle.

    You have done double and triple integral previously, haven't you?
     
  10. Apr 29, 2013 #9

    Thank you for the help, in terms of y I do not know how to do it w/o making it complicated, but...

    ∫ <(2((2+cosθ)(5+3cosθ)^(2)) , (3((2+cosθ)^(2)*(5+3cosθ))> (dot) <-3sinθ,3cosθ> dθ from 0 to 2pi


    again thank you for the help.
     
  11. Apr 29, 2013 #10

    SammyS

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    The equation, which you gave for the lower half of the circle is
    ## y = 5-\sqrt{9-(x-2)^2} ##​
    For the upper half we have
    ## y = 5+\sqrt{9-(x-2)^2} ##​

    Those are the limits of integration for y.

    The integrand is (∂N/∂x - ∂M/∂y) = 2xy .
     
  12. Apr 29, 2013 #11

    This is extremely helpful.

    If i have trouble converting the equations into x=rcosθ and y=rsinθ i know now by what you have stated i can still set up and solve (however messy is may be) in terms of x and y and still get the same solution. Ill post a picture later of the two solved integrals.

    thank you for help.
     
  13. Apr 29, 2013 #12

    SammyS

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    You can do the area integral (double integral) in polar coordinates.

    1.
    Shift your coordinate system so that you translate the point (2,5) to the origin. Then use standard polar coordinates.

    Don't forget to alter the integrand appropriately.


    2.
    Use "Modified polar coordinates".

    Let ##\ x=2+r\cos(\theta)\ ## and ##\ y=5+r\sin(\theta)\ ##
     
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