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The circle is centered at the origin with radius $r=3$. The standard equation is $x^2 + y^2 = 9$. Solving for $x$ in terms of $y$, we get $x = \sqrt{9 - y^2}$.
In standard CBSE Class 12 problems of this type, the shaded region typically represents the area bounded by the curve $x = \sqrt{9 - y^2}$ between two horizontal lines (e.g., $y=1$ and $y=3$) on both sides of the y-axis, or a specific segment. Given the options provided, the integral $\int_{1}^{3}\sqrt{9-y^{2}}dy$ represents the area in the first quadrant between $y=1$ and $y=3$.
Since the circle is symmetric about the y-axis, the total area of the shaded region spanning across both the first and second quadrants is twice the area calculated in the first quadrant. Thus, the area is $2 \int_{1}^{3}\sqrt{9-y^{2}}dy$.
Final Answer: B