Answer
$\text{Area = $4 - \sqrt 7$}$
Work Step by Step
$\text{It is given that}$
\begin{align}
f(y) = \frac{y}{\sqrt {16-y^2}}; \ g(y) = 0; \ y = 3
\end{align}
$\text{The intersection of f(y) and g(y) is}$
\begin{align}
\frac{y}{\sqrt {16-y^2}} = 0 \\
y = 0 \Rrightarrow x = 0
\end{align}
$\text{Thus, the area of the given region is}$
\begin{align}
Area = \int_0^3 \frac{y}{\sqrt {16-y^2}} \ dy
\end{align}
$\text{Apply u - substitution $u = 16 - y^2; du = -2ydy$:}$
\begin{align}
Area = \int_{16}^7 -\frac{1}{2\sqrt u} du = \frac{1}{2} \int^{16}_7\frac{1}{\sqrt u} du = \\
= \left[ \sqrt u \right]_7^{16} = 4 - \sqrt 7
\end{align}
