Finding the Curl of a Vector Field

[-EquinoX-]

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?

 

[CompuChip]

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.

 

[-EquinoX-]

well..how do I get around to solve this? I know the curl is y-3...

 

[CompuChip]

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...

 

[-EquinoX-]

so the curl is -4? I don't get it why it's -2 to 2