#### Answer

$\frac{1}{3}(\pi^2+4)^{3/2} -\frac{8}{3}.$

#### Work Step by Step

Since $r=f(\theta)= \theta^2$, then $f'(\theta)=2\theta$.
The length is given by
\begin{align*} \text {Length }&=\int_{0}^{\pi} \sqrt{f(\theta)^{2}+f^{\prime}(\theta)^{2}} d \theta\\ &=\int_{0}^{\pi}\sqrt{\theta^4+4\theta^2} d \theta\\
&=\int_{0}^{\pi}\theta\sqrt{\theta^2+4} d \theta \\
&=\frac{1}{2}\int_{0}^{\pi}2\theta(\theta^2+4)^{1/2} d \theta\\
&=\frac{2}{6}(\theta^2+4)^{3/2} |_{0}^{\pi}\\
&=\frac{1}{3}(\pi^2+4)^{3/2} -\frac{8}{3}.
\end{align*}