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) $2\int_{0}^{3}\sqrt{9-x^{2}}dx$
Prev Next

AI Tutor Explanation

Powered by Gemini

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

AI generated content. Review strictly for academic accuracy.

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.

More from this Chapter

LA
The area of the region bounded by the line \(y=mx (m>0)\), the curve \(x^{2}+y^{2}=4\) and the \(x\)-axis in the first quadrant is \(\frac{\pi}{2}\) units. Using integration, find the value of m.
MCQ_SINGLE
The area of the region enclosed between the curve \(y=x|x|\), x-axis, \(x=-2\) and \(x=2\) is:
MCQ_SINGLE
The area of the region bounded by the curve \(y^{2}=x\) between \(x=0\) and \(x=1\) is:
SA
Sketch the graph of $y=|x+3|$ and find the area of the region enclosed by the curve, x-axis, between $x=-6$ and $x=0$, using integration.
LA
Find the area of the region bounded by the curve $4x^{2}+y^{2}=36$ using integration.
View All Questions