Find the area of the surface generated by revolving the given curve about the $y$-axis. \[ x=\sqrt{16-y^{2}},-3 \leq y \leq 3 \]

Step 1 ：We are given the curve \(x=\sqrt{16-y^{2}}\), and we are asked to find the area of the surface generated by revolving this curve about the y-axis from \(y=-3\) to \(y=3\).

Step 2 ：The formula for the surface area of a solid of revolution generated by revolving a curve \(y=f(x)\) from \(x=a\) to \(x=b\) about the y-axis is \(A = 2\pi \int_{a}^{b} x \sqrt{1 + (f'(x))^2} dx\).

Step 3 ：In this case, the curve is given in terms of y, so we need to find \(f'(x)\) in terms of y. The derivative of \(x=\sqrt{16-y^{2}}\) with respect to y is \(f'(y) = -y/\sqrt{16-y^{2}}\).

Step 4 ：We can then substitute \(f'(y)\) and x into the formula and integrate from \(y=-3\) to \(y=3\) to find the surface area.

Step 5 ：After performing the integration, we find that the surface area is \(48\pi\).

Step 6 ：Final Answer: The area of the surface generated by revolving the given curve about the y-axis is \(\boxed{48\pi}\).