Question

The area bounded by the circle \(x^2 + y^2 = 16\) and the lines \(x = 0\) and \(x = 2\) is

• A. \( \left[ 4\sqrt{3} + \frac{8\pi}{3} \right] \) sq. units
• B. \( \frac{1}{2}\left[ 4\sqrt{3} + \frac{8\pi}{3} \right] \) sq. units
• C. \( \left[ 4\sqrt{3} - \frac{8\pi}{3} \right] \) sq. units
• D. \( \frac{1}{2}\left[ 4\sqrt{3} - \frac{8\pi}{3} \right] \) sq. units

Hint:

For regions bounded by circles and vertical lines, express the curve as \(y\) in terms of \(x\) and use symmetry whenever possible to simplify calculations.

Answer 1

The Correct Option is A

Step 1: Understand the region.
The given circle is \[ x^2 + y^2 = 16 \Rightarrow y = \sqrt{16 - x^2} \] The region is bounded between \(x = 0\) and \(x = 2\) in the first quadrant.
Step 2: Write the area integral.
Area bounded by the curve and the x-axis is \[ A = 2 \int_{0}^{2} \sqrt{16 - x^2}\,dx \] The factor 2 accounts for symmetry about the x-axis.
Step 3: Evaluate the integral.
Using the standard result \[ \int \sqrt{a^2 - x^2}\,dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) \] with \(a = 4\), we get \[ A = 2\left[ \frac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}\left(\frac{x}{4}\right) \right]_{0}^{2} \] Step 4: Substitute limits.
\[ A = 2\left[ \sqrt{12} + 8\sin^{-1}\left(\frac{1}{2}\right) \right] = 2\left[ 2\sqrt{3} + \frac{4\pi}{3} \right] \] \[ A = 4\sqrt{3} + \frac{8\pi}{3} \]