Question

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. \( 2 \int_{0}^{3} \sqrt{9 - x^2} \, dx \)

Hint:

Integrating along the y-axis is often much cleaner when the region is bounded symmetrically by horizontal boundaries. The formula used is \(\text{Area} = \int_{c}^{d} (x_{\text{right}} - x_{\text{left}}) \, dy\).

Answer 1

The Correct Option is B

Concept: The problem asks for the area of a region bounded by a circle \( x^2 + y^2 = 9 \) and a horizontal straight line \( y = 1 \). The shaded region lies above the line \( y = 1 \) inside the circle up to its highest point \( y = 3 \). We can set up the integration with respect to the y-axis to find the area directly.

Step 1: Identify the boundaries of the region.
The circle equation is given by: \[ x^2 + y^2 = 9 \] This is a circle centered at the origin \( (0,0) \) with radius \( r = \sqrt{9} = 3 \). The lower boundary of the shaded region is the line \( y = 1 \). The upper limit of the shaded region along the y-axis is the peak of the circle, where \( x = 0 \), giving \( y = 3 \). Thus, the limits of integration for \( y \) are from \( y = 1 \) to \( y = 3 \).

Step 2: Express \( x \) as a function of \( y \).
From the circle equation, solve for \( x \): \[ x^2 = 9 - y^2 \quad \Rightarrow \quad x = \pm \sqrt{9 - y^2} \] The right half of the circle corresponds to \( x = +\sqrt{9 - y^2} \) and the left half corresponds to \( x = -\sqrt{9 - y^2} \).

Step 3: Set up the area integral.
By symmetry across the y-axis, the total area is twice the area contained in the first quadrant between \( y = 1 \) and \( y = 3 \): \[ \text{Area} = 2 \int_{1}^{3} x \, dy = 2 \int_{1}^{3} \sqrt{9 - y^2} \, dy \] This expression directly matches option (B).