Find the volume of the solid obtained by rotating the region in the first quadrant bounded by the given curve about the $y$-axis. \[ y=9-(x-5)^{2} \text {. } \]

Step 1 ：Rewrite the given equation in terms of \(x\): \[y = 9 - (x - 5)^2\] to get \[x = 5 + \sqrt{9 - y}\] and \[x = 5 - \sqrt{9 - y}\]

Step 2 ：Since we are considering the region in the first quadrant, we only need to consider the positive root. So, \(f(y) = 5 + \sqrt{9 - y} - (5 - \sqrt{9 - y}) = 2\sqrt{9 - y}\)

Step 3 ：The limits of integration are the y-values where the curve intersects the y-axis. Setting \(x = 0\) in the original equation, we get \(y = 9 - (0 - 5)^2 = 9 - 25 = -16\) and \(y = 9 - (10 - 5)^2 = 9 - 25 = -16\). So, \(a = -16\) and \(b = 9\)

Step 4 ：Substitute these values into the formula for the volume of a solid of revolution: \[V = \pi \int_{-16}^{9} [2\sqrt{9 - y}]^2 dy\] to get \[V = \pi \int_{-16}^{9} 4(9 - y) dy\]

Step 5 ：Simplify the integral to get \[V = 4\pi \int_{-16}^{9} (9 - y) dy\]

Step 6 ：Evaluate the integral to get \[V = 4\pi [9y - \frac{1}{2}y^2]_{-16}^{9}\]

Step 7 ：Substitute the limits of integration to get \[V = 4\pi [(9*9 - \frac{1}{2}*9^2) - (9*-16 - \frac{1}{2}*(-16)^2)]\]

Step 8 ：Simplify the expression to get \[V = 4\pi [(81 - 40.5) - (-144 - 128)]\]

Step 9 ：Further simplify the expression to get \[V = 4\pi [40.5 + 272]\]

Step 10 ：Simplify the expression to get \[V = 4\pi [312.5]\]

Step 11 ：Finally, simplify the expression to get \[V = 1250\pi\]

Step 12 ：So, the volume of the solid obtained by rotating the region in the first quadrant bounded by the given curve about the y-axis is \(\boxed{1250\pi}\) cubic units