Answer
$$\pi$$
Work Step by Step
\begin{align*}
\int_{0}^{1} \frac{4 r d r}{\sqrt{1-r^{4}}}&=\lim _{b \rightarrow 1}\int_{0}^{b} \frac{4 r d r}{\sqrt{1-r^{4}}}\\
&=\lim _{b \rightarrow 1}\left[2 \sin ^{-1}\left(r^{2}\right)\right]_{0}^{b}\\
&=\lim _{b \rightarrow 1^{-}}\left[2 \sin ^{-1}\left(b^{2}\right)-2 \sin ^{-1} 0\right]\\
&=2 \cdot \frac{\pi}{2}-0=\pi
\end{align*}