Question

The area in square units of the region bounded by the curve \(y = \sqrt{16-x^2}\) and lines \(x = 0,x = 4\) above the X-axis is

• A. \(16π\)
• B. \(12π\)
• C. \(8π\)
• D. \(4π\)

Hint:

The curve y = root of (16 - x squared) is the upper half of a circle of radius 4.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
\(y=\sqrt{16-x^2}\) is the upper half of the circle \(x^2+y^2=16\) of radius 4. The lines \(x=0\) and \(x=4\) cut out the first quadrant part.

Step 2: Area by integration
\[ A=\int_0^4\sqrt{16-x^2}\,dx=\left[\frac x2\sqrt{16-x^2}+8\sin^{-1}\frac x4\right]_0^4 \]
\[ =0+8\cdot\frac\pi2-0=4\pi \]

Step 3: Check by geometry
The area of the full circle is \(\pi r^2=16\pi\). One quarter is \(4\pi\). Both methods agree, so the answer is option (D). The options \(16\pi\), \(12\pi\) and \(8\pi\) correspond to the whole circle, three quarters and a half.

Final Answer:
The region is a quarter of a circle of radius 4, so its area is 4 pi, option (D). \[ \boxed{4\pi} \]