Using Green's theorem to evaluate

In summary, LCKurtz is trying to find the limits for x and θ for a circle, but is having trouble. He finds another method and integrates the equations to get the correct answer.
  • #1
DryRun
Gold Member
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4
Homework Statement
Using Green's theorem, evaluate:
http://s2.ipicture.ru/uploads/20120117/6p57O2HO.jpg

The attempt at a solution
[tex]\frac{\partial P}{\partial y}=3x+2y[/tex]
[tex]\frac{\partial Q}{\partial x}=2y+10x[/tex]
[tex]\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=7x[/tex]
To do the integration, I'm using the parametric equations of the circle; x= cosθ and y=sinθ
[tex]\int\int 7\cos \theta \,.rdrd\theta[/tex]
The curve C is a circle with centre (1,-2) and radius r=1
I've drawn the graph in my copybook.

Description of circle:
For θ fixed, r varies from r=-2 to r=-3
θ varies from -∏/2 to -∏/2

However, i can't get the correct answer as I'm quite sure that I've got the wrong limits. Usually, i deal with circles with centre (0,0) so it's easier to find the limits for r and θ.
 
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  • #2
Any help?
 
  • #3
sharks said:
Homework Statement
Using Green's theorem, evaluate:
http://s2.ipicture.ru/uploads/20120117/6p57O2HO.jpg

The attempt at a solution
[tex]\frac{\partial P}{\partial y}=3x+2y[/tex]
[tex]\frac{\partial Q}{\partial x}=2y+10x[/tex]
[tex]\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=7x[/tex]
To do the integration, I'm using the parametric equations of the circle; x= cosθ and y=sinθ
[tex]\int\int 7\cos \theta \,.rdrd\theta[/tex]
The curve C is a circle with centre (1,-2) and radius r=1
I've drawn the graph in my copybook.

Description of circle:
For θ fixed, r varies from r=-2 to r=-3
θ varies from -∏/2 to -∏/2

However, i can't get the correct answer as I'm quite sure that I've got the wrong limits. Usually, i deal with circles with centre (0,0) so it's easier to find the limits for r and θ.

Your equations ##x=r\cos\theta,\ y=r\sin\theta## give a circle at the origin, not centered at ##(1,-2)##. First figure out how to fix that (think about translation). Then, standard convention in polar coordinates uses positive r, so it wouldn't go from -2 to -3. And of course if you use the same upper and lower limits for ##\theta##, you will get 0. Think about if ##
\theta## starts at ##-\frac \pi 2## and increases by ##2\pi##, where will it end?
 
  • #4
Hi LCKurtz

Here is the Cartesian graph of path C:

http://s1.ipicture.ru/uploads/20120118/SQzbOw3U.gif

An attempt at translation of axes:

Let x = X+h
Let y=Y+k
Then, X=rcosθ-h and Y=rsinθ-k

Here, x=1 and y=-2.
I'm not sure. Is this correct?

I'm going to try another method:
From the general equation for polar coordinates, [itex]x^2+y^2=r^2[/itex], at (1,-2), r=√5
This would mean that r varies from r=0 to r=√5
Then, at (1,-2), x=1 and y=-2
Substituting into x=rcosθ and y=rsinθ
1=√5cosθ, θ= 1.107
and -2=√5sinθ, θ=-1.107
Thus θ varies from θ=-1.107 to θ=1.107
Again, I'm not sure. Maybe there's a simpler and more accurate method?

I did some more research and i believe this method is correct:
The new centre of circle is: [(x-1),(y+2)] and the radius stays same, r=1.
For [itex]x=r\cos\theta,\ y=r\sin\theta[/itex], these become [itex](x-1)=r\cos\theta,\ (y+2)=r\sin\theta[/itex]
So, [itex]x=r\cos\theta+1,\ y=r\sin\theta-2[/itex]
Since, r=1, the parametric equations of the circle become: [itex]x=\cos\theta+1,\ y=\sin\theta-2[/itex]

Therefore, the integral becomes:
[tex]\int\int 7(\cos\theta+1) \,.rdrd\theta=\int\int 7\cos\theta+7 \,.rdrd\theta[/tex]
The description of the circle:
For θ fixed, r varies from r=0 to r=1.
θ varies from θ=0 to θ=2∏.

The integration gives: 7∏, which is the correct answer. So, i assume that i finally found the right method.
 
Last edited:
  • #5
sharks said:
So, [itex]x=r\cos\theta+1,\ y=r\sin\theta-2[/itex]
Since, r=1, the parametric equations of the circle become: [itex]x=\cos\theta+1,\ y=\sin\theta-2[/itex]

Therefore, the integral becomes:
[tex]\int\int 7(\cos\theta+1) \,.rdrd\theta=\int\int 7\cos\theta+7 \,.rdrd\theta[/tex]
The description of the circle:
For θ fixed, r varies from r=0 to r=1.
θ varies from θ=0 to θ=2∏.

The integration gives: 7∏, which is the correct answer. So, i assume that i finally found the right method.

Yes! Good work, and you learned something in the process. :biggrin:
 
  • #6
Thanks for your guidance, LCKurtz. It's much appreciated.:smile:
 

FAQ: Using Green's theorem to evaluate

What is Green's theorem?

Green's theorem is a mathematical theorem that relates the line integral of a two-dimensional vector field over a closed curve to a double integral over the region enclosed by the curve.

How is Green's theorem used in science?

Green's theorem is often used in physics and engineering to calculate various physical quantities, such as work and flux, in a two-dimensional setting. It can also be applied in fluid dynamics, electromagnetism, and other fields where vector calculus is used.

Can Green's theorem be applied to any shape?

Green's theorem can only be applied to closed curves and regions. This means that the shape must have a defined boundary that completely encloses a two-dimensional region.

What are the benefits of using Green's theorem?

Green's theorem can often simplify complex integrals and allow for easier calculation of physical quantities. It also provides a connection between line integrals and double integrals, making it a powerful tool in certain applications.

Are there any limitations to using Green's theorem?

Green's theorem is limited to two-dimensional vector fields and regions. It also assumes that the region is simply connected, meaning that there are no holes or gaps in the enclosed area. Additionally, it may not always be the most efficient method for evaluating integrals, and other techniques may be more suitable in certain situations.

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