Question

The area of the region bounded by \(y=\sqrt{16-x^2}\) and x-axis is

• A. 8π square units
• B. 20π square units
• C. 16π square units
• D. 256π square units

Answer 1

The Correct Option is A

The given curve is \( y = \sqrt{16 - x^2} \). Since \( y \) must be non-negative (\( y \ge 0 \)), this equation represents the upper part of a curve.

To identify the shape of the curve, we can square both sides of the equation: \[ y^2 = (\sqrt{16 - x^2})^2 \] \[ y^2 = 16 - x^2 \] Rearranging the terms, we get: \[ x^2 + y^2 = 16 \]

This is the equation of a circle centered at the origin (0, 0) with radius \( r \) such that \( r^2 = 16 \). Therefore, the radius is \( r = \sqrt{16} = 4 \).

The equation \( y = \sqrt{16 - x^2} \) represents the upper semi-circle (\( y \ge 0 \)) of this circle. The region bounded by \( y = \sqrt{16 - x^2} \) and the x-axis (\( y = 0 \)) is the area of this upper semi-circle.

The area of a full circle with radius \( r \) is given by the formula \( A_{circle} = \pi r^2 \). For \( r = 4 \), the area of the full circle is \( A_{circle} = \pi (4)^2 = 16\pi \) square units.

The area of the region required (the upper semi-circle) is half the area of the full circle: \[ A_{region} = \frac{1}{2} A_{circle} = \frac{1}{2} (16\pi) = 8\pi \] square units.

The calculated area of the region bounded by \( y = \sqrt{16 - x^2} \) and the x-axis is \( 8\pi \) square units.

Answer 2

We are given the curve:

\[ y = \sqrt{16 - x^2} \]

Step 1: Identify the curve

This is the equation of a semi-circle (upper half) with radius 4 and centered at the origin.

Step 2: Limits of integration

Since it's a semi-circle over the x-axis, the bounds are from \( x = -4 \) to \( x = 4 \).

Step 3: Area under the curve

The area of a full circle is: \[ A = \pi r^2 = \pi \times 4^2 = 16\pi \] Since the curve is only the upper half of the circle (semi-circle), \[ \text{Required Area} = \frac{1}{2} \times 16\pi = 8\pi \]

Final Answer: \( \boxed{8\pi} \) square units