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?