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Compute $\int_C (8-\sqrt{x^2 +y^2}) ~ds$ where $C$ is the circle $x^2 + y^2 =4$.

Answer: $24\pi$

How is the answer $24\pi$? I converted the integral into a double integral of polar coordinates and got $\frac{80}{3}\pi$ as my answer. Can someone please help me?

I converted the integral into $\int_0^{2\pi}\int_0^2 (8-r)r~drd\theta$. Is this correct?

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    $\begingroup$ If you're integrating over a circle, then that's a one-dimensional -- not double -- integral. To see where you went wrong you will have to share more details of your calculation. $\endgroup$ – Eckhard Apr 11 '13 at 17:49
  • $\begingroup$ Are you integrating around the circle or over the disk? $ds$ looks like a bit of arc length. $\endgroup$ – Ross Millikan Apr 11 '13 at 17:53
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    $\begingroup$ Since $r$ is fixed you just integrate over $\theta$. Just fill in $r=2$ and you'll get the desired result. $\endgroup$ – Matt L. Apr 11 '13 at 17:55
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\begin{equation*} \int_C (8-\sqrt{x^2+y^2})=\int_C (8-2)\, ds=6\int_C ds=6\cdot 4\pi=24\pi. \end{equation*}

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Integrating around the circle, the integrand is constant at $6$ and the arc length is $4\pi$. The product is $24\pi$

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