two Q's: finding center of mass of a bounded region, and finding length of curve

Quote:

2.) sketch and calculate the length of the polar curve y=theta^2 from theta=0 to theta=4pi.

Shouldn't that be $r={\theta}^{2}$ ?.

Anyway, arc length in polar is given by $L=\int_{\alpha}^{\beta}\sqrt{r^{2}+\left(\frac{dr} {d{\theta}}\right)^{2}}d{\theta}$

$\int_{0}^{4\pi}\sqrt{{\theta}^{4}+4{\theta}^{2}}d{ \theta}$

$\int_{0}^{4\pi}\sqrt{{\theta}^{2}({\theta}^{2}+4)} d{\theta}$

$=\int_{0}^{4\pi}{\theta}\sqrt{{\theta}^{2}+4}d{\th eta}$

You could use trig sub to solve this by using the sub:

${\theta}=2tan(x), \;\ d{\theta}=2sec^{2}(x)dx$