$$\oint \limits_C (7y+x+2)dx+(5+y+2x)dy$$ where the curve C is the circle: $(x-a)^2 +(y-b)^2=25$. This integral calls for Green's Theorem.

$$\iint \limits_D-5dydx$$ I beleive the region D is best represented in polar. Is it correct to write the region like this $r=2a\cos(\theta)+2b\sin(\theta)$? Then the bounds are as such $\{(r,\theta) \mid 0<r<2a\cos(\theta)+2b\sin(\theta), 0<\theta<2\pi\}$

$$\text{So,}\oint \limits_C (7y+x+2)dx+(5+y+2x)dy=\int_0^{2\pi}\int_0^{2a\cos(\theta)+2b\sin(\theta)}-5r\ drd\theta$$

Is this correct, or am I missing something?


You can evaluate this double integral geometrically: It is the signed volume of a cylinder with height $-5$ and base of radius $5$, so has volume $-5(\pi(5)^2) =-125\pi$.

Alternatively, you could evaluate the integral in polar as you propose. Your proposed parameterization of the circle is incorrect however. You can write any point in this shifted circle as $(x,y) = (a+rcos(\theta),b+r\sin(\theta))$. Can you take it from here?

  • $\begingroup$ Isn't the circle with center (a,b) given by the equation $r=2a\cos(\theta)+2b\sin(\theta)$ $\endgroup$ – John Dec 11 '13 at 22:40
  • $\begingroup$ Something has to be wrong because you do not even have an $r$ in there. $\endgroup$ – Steven Gubkin Dec 11 '13 at 22:42
  • $\begingroup$ I see where your going at, its clear that the radius here is $\sqrt{a^2+b^2}$. So there seems to be a dependency of the center on the radius. Is there a more general polar equation for the circle? $\endgroup$ – John Dec 11 '13 at 22:45
  • $\begingroup$ I think that you are really pretty lost here. It might be better to sit down with someone who can work with you face to face. The integral you seek is just $$\int_0^{2\pi}\int_0^5 -5rdrd\theta$$. $\endgroup$ – Steven Gubkin Dec 11 '13 at 22:48
  • $\begingroup$ If you assume that r=5, you are saying that the circle, region D, is centered at zero--which is not true. In this case, it doesn't make a difference because D is bounded by C if r=5. I was asking a general question about circles and polar equations. $\endgroup$ – John Dec 11 '13 at 22:53

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