Class CBSE Class 12 Mathematics Applications of Integrals Q #1711
COMPETENCY BASED
APPLY
1 Marks 2026 AISSCE(Board Exam) MCQ SINGLE
The area of the shaded region of the circle given below is equal to :
(A) $\int_{1}^{3}\sqrt{9-y^{2}}dy$
(B) $2\int_{1}^{3}\sqrt{9-y^{2}}dy$
(C) $\int_{0}^{3}\sqrt{9-x^{2}}dx$
(D) $2\int_{0}^{3}\sqrt{9-x^{2}}dx$
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Detailed Solution

Step 1: Identify the Equation of the Circle

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 $x = \sqrt{9 - y^2}$.

Step 2: Analyze the Shaded Region

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 $y=3$.

Step 3: Determine the Symmetry

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

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student must translate a geometric representation into a definite integral using the concept of area under a curve.
Knowledge Dimension: PROCEDURAL
Justification: The student is required to execute the procedure of setting up a definite integral based on the geometric properties of a circle.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. This question tests the student's ability to visualize and model geometric areas using calculus, which is a core competency in the Application of Integrals chapter.

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