Calculate $\int_C{y^2\:ds}$ where $C$ is $x^2+y^2=9$ I need help calculating $\int_C{y^2\:ds}$ where $C$ is $x^2+y^2=9$.
So what I first did was convert $C$ into parametric and then I set it up like this:
$$\int_0^{2\pi}(3\sin t)^2\sqrt{(-3\sin t)^2+(3\cos t)^2}$$ and I got $27\pi$ as my final answer. Can someone verify this for me, both the integration process and how I set it up?

It's really simpler evaluate it in polar coordinates. $ds=3\,d\phi$, and integral is
$$
\int_0^{2\pi} 9\sin^2\phi\cdot 3\,d\phi = 27\pi
$$
