Answer
$$\frac{20}{3} \sqrt{5}-\frac{32}{5} \sqrt{3}$$
Work Step by Step
Given
$$ \int_{0}^{2} r \sqrt{5-\sqrt{4-r^{2}}} d r$$
Let
$$ u= 5-\sqrt{4-r^{2}} \ \ \ \Rightarrow \ \ \ du =\frac{rdr}{\sqrt{4-r^{2}}} \ \ \Rightarrow rdr = (5-u)du$$
At $$ r= 0 \to u= 3, \ \ r= 2\to u= 5$$
Then
\begin{aligned}
\int_{0}^{2} r \sqrt{5-\sqrt{4-r^{2}}} d r&=\int_{3}^{5} \sqrt{u}(5-u) d u\\
&=\int_{3}^{5}\left(5 u^{1 / 2}-u^{3 / 2}\right) d u\\
&=\left.\left(\frac{10}{3} u^{3 / 2}-\frac{2}{5} u^{5 / 2}\right)\right|_{3} ^{5}\\
&=\left(\frac{50}{3} \sqrt{5}-10 \sqrt{5}\right)-\left(10 \sqrt{3}-\frac{18}{5} \sqrt{3}\right)\\
&=\frac{20}{3} \sqrt{5}-\frac{32}{5} \sqrt{3}\end{aligned}
