Finding the Curl of a Vector Field

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Homework Help Overview

The discussion revolves around finding the curl of a vector field and its implications for calculating circulation around a curve. Participants are examining the mathematical properties of the curl and its integration over a specified area.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to compute the curl and its integral over a region, with some questioning the validity of their results and the integration limits. There is a discussion about the dependency of the curl on the variable y and its implications for the final answer.

Discussion Status

The conversation is ongoing, with participants providing hints and guidance on integration techniques. There is recognition of the complexity involved in the integral due to the non-constant nature of the curl, and some participants are exploring different integration boundaries.

Contextual Notes

There is mention of specific integration limits from y = -2 to y = 2, which appears to be a point of confusion for some participants. The original poster's understanding of the problem setup and the implications of the curl's dependency on y are also under scrutiny.

-EquinoX-
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Homework Statement



http://img5.imageshack.us/img5/8295/capturewmw.th.jpg

Homework Equations





The Attempt at a Solution



I tried to find the curl first and what i got is y - 3 and then I multiply that by the area of the circle which is 4pi.. am I doing something wrong?
 
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The circulation of a vector field \vec G around a curve C is given by
\operatorname{circ}_C(\vec G) = \iint_A \operatorname{curl}(\vec G) \, \mathrm d\vec a
Since the curl is not a constant on the disk, the integral is not as trivial as integrand * surface area.

You could have easily seen that your answer is wrong because 4pi(y - 2) still depends on y, while it should be a number.
 
well..how do I get around to solve this? I know the curl is y-3...
 
If you have to solve this question I assume you have learned how to integrate a function over some area.

I suggest integrating from y = -2 to y = 2 so that the x integral will be trivial (you only need to worry about the integration boundaries):

\operatorname{circ}_{C}(\vec G) \propto \int_{-2}^{2} \int_{\cdots}^{\cdots} (y - 2) \, dx \, dy
up to some proportionality factors... see the image below.

I hope that I have provided you with enough clues to solve the question now...
 

Attachments

  • circle-integration.png
    circle-integration.png
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so the curl is -4? I don't get it why it's -2 to 2
 

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