Answer
The exact length of the polar curve:
$$ r=\theta^{2} \quad \text {for } \theta =0 \quad \text {to} \quad \theta=2 \pi $$
is equal to
$$
\begin{aligned} L &=\int_{a}^{b} \sqrt{r^{2}+(d r / d \theta)^{2}} d \theta \\
& =\frac{8}{3}\left[\left(\pi^{2}+1\right)^{3 / 2}-1\right]\end{aligned}
$$
Work Step by Step
The exact length of the polar curve:
$$ r=\theta^{2} \quad \text {for } \theta =0 \quad \text {to} \quad \theta=2 \pi $$
is equal to
$$
\begin{aligned} L &=\int_{a}^{b} \sqrt{r^{2}+(d r / d \theta)^{2}} d \theta \\ &=\int_{0}^{2 \pi} \sqrt{\left(\theta^{2}\right)^{2}+(2 \theta)^{2}} d \theta \\ &=\int_{0}^{2 \pi} \sqrt{\theta^{4}+4 \theta^{2}} d \theta \\ &=\int_{0}^{2 \pi} \sqrt{\theta^{2}\left(\theta^{2}+4\right)} d \theta \\
&= \int_{0}^{2 \pi} \theta \sqrt{\theta^{2}+4} d \theta \end{aligned}
$$
Now let $ u=\theta^{2}+4$, so that
$$
du=2 \theta d \theta \quad \quad [\theta d\theta=\frac{1}{2}du]
$$
and
$$
\begin{aligned}
\int_{0}^{2 \pi} \theta \sqrt{\theta^{2}+4} d \theta &=\int_{4}^{4 \pi^{2}+4} \frac{1}{2} \sqrt{u} d u \\
&=\frac{1}{2} \cdot \frac{2}{3}\left[u^{3 / 2}\right]_{4}^{4\left(\pi^{2}+1\right)} \\ & =\frac{1}{3}\left[4^{3 / 2}\left(\pi^{2}+1\right)^{3 / 2}-4^{3 / 2}\right] \\
& =\frac{8}{3}\left[\left(\pi^{2}+1\right)^{3 / 2}-1\right]
\end{aligned}.
$$
