Question

Using integration, find the area bounded by the circle whose centre is at origin and radius is 4 units.

Hint:

While integration is required for the proof, you can always check your answer using the basic geometric formula \( \text{Area} = \pi r^2 \). For \( r=4 \), \( \text{Area} = 16\pi \).

Answer 1

Step 1: Understanding the Concept:
The equation of a circle centered at the origin \( (0,0) \) with radius \( r \) is \( x^2 + y^2 = r^2 \). The total area can be found by integrating the function representing the upper semi-circle and multiplying by 4 (exploiting symmetry).
Step 2: Key Formula or Approach:
1. Circle Equation: \( x^2 + y^2 = 4^2 \implies y = \sqrt{16 - x^2} \).
2. Standard Integral: \( \int \sqrt{a^2 - x^2} dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C \).
Step 3: Detailed Explanation:
Total Area \( A = 4 \times (\text{Area of the first quadrant}) \).
The limits for the first quadrant are from \( x = 0 \) to \( x = 4 \).
\[ A = 4 \int_{0}^{4} \sqrt{16 - x^2} dx \] Applying the standard formula where \( a = 4 \):
\[ A = 4 \left[ \frac{x}{2}\sqrt{16 - x^2} + \frac{16}{2}\sin^{-1}\frac{x}{4} \right]_{0}^{4} \] \[ A = 4 \left[ \left( \frac{4}{2}\sqrt{16 - 16} + 8\sin^{-1}(1) \right) - \left( \frac{0}{2}\sqrt{16 - 0} + 8\sin^{-1}(0) \right) \right] \] \[ A = 4 [ (0 + 8 \times \frac{\pi}{2}) - (0 + 0) ] \] \[ A = 4 [ 4\pi ] = 16\pi \] Step 4: Final Answer:
The area bounded by the circle is \( 16\pi \) sq. units.