Answer
The answer is below.
Work Step by Step
Type I region,
\[
\begin{aligned}
\iint_{R} y d A &=\int_{0}^{5} \int_{5-x}^{\sqrt{25-x^{2}}} y d y d x \\
&=\int_{0}^{5}\left[\frac{y^{2}}{2}\right]_{5-x}^{\sqrt{25-x^{2}}} d x \\
&=\frac{1}{2} \int_{0}^{5}\left(\left(25-x^{2}\right)-(5-x)^{2}\right) d x \\
&=\frac{1}{2} \int_{0}^{5}\left(-2 x^{2}+10x \right) d x \\
&=\frac{125}{6}
\end{aligned}
\]
Type II region,
\[
\begin{aligned}
\iint_{R} y d A &=\int_{0}^{5} \int_{5-y}^{\sqrt{25-y^{2}}} y d x d y \\
&=\int_{0}^{5} y(\sqrt{25-y^{2}}-(5-y)) d y \\
&=\int_{0}^{5} y \sqrt{25-y^{2}} d y-\int_{0}^{5} y(5-y) d y
\end{aligned}
\]
Replacing $y=\sin \theta$ on the first integral gives
\[
\frac{125}{6}
\]
