Answer
$$\frac{ A }{2} \sqrt{1+ A ^{2}}+\frac{1}{2} \ln ( A+\sqrt{1 + A^{2}})$$
Work Step by Step
Since $r= \theta $ , then $f'(\theta )= 1$ and hence:
\begin{aligned}
\text { Length } &=\int_{0}^{A} \sqrt{f(\theta)^{2}+f^{\prime}(\theta)^{2}} d \theta \\
&=\int_{0}^{A} \sqrt{1+ \theta^{2}} d \theta
\end{aligned}
Use the formula $$ \int \sqrt{a^{2}+u^{2}} d u=\frac{u}{2} \sqrt{a^{2}+u^{2}}+\frac{a^{2}}{2} \ln (u+\sqrt{a^{2}+u^{2}})+C$$
Then
\begin{aligned}
\text { Length } &=\int_{0}^{A} \sqrt{1+ \theta^{2}} d \theta \\
&= \frac{ \theta }{2} \sqrt{1+ \theta ^{2}}+\frac{1}{2} \ln ( \theta +\sqrt{1 + \theta^{2}})\bigg|_{0}^{A}\\
&= \frac{ A }{2} \sqrt{1+ A ^{2}}+\frac{1}{2} \ln ( A+\sqrt{1 + A^{2}})
\end{aligned}
