Question

Using integration, find the area bounded by the circle \( x^2 + y^2 = 16 \) in the first quadrant.

Hint:

To calculate areas of regions bounded by circles, use trigonometric substitution and the symmetry of the problem to simplify the integral.

Answer 1

Step 1: Understanding the equation of the circle.
The equation of the circle is \( x^2 + y^2 = 16 \). This represents a circle with radius 4 centered at the origin. Step 2: Expressing the equation for integration.
To find the area in the first quadrant, we solve for \( y \): \[ y = \sqrt{16 - x^2} \] The area in the first quadrant can be found by integrating from 0 to 4 (the radius of the circle): \[ A = \int_0^4 \sqrt{16 - x^2} \, dx \] Step 3: Using a standard trigonometric substitution.
We can use the substitution \( x = 4 \sin \theta \), which gives \( dx = 4 \cos \theta \, d\theta \). The limits of integration change as follows: - When \( x = 0 \), \( \theta = 0 \) - When \( x = 4 \), \( \theta = \frac{\pi}{2} \) Thus, the integral becomes: \[ A = \int_0^{\frac{\pi}{2}} \sqrt{16 - (4 \sin \theta)^2} \times 4 \cos \theta \, d\theta \] Simplifying the expression: \[ A = \int_0^{\frac{\pi}{2}} 16 \cos^2 \theta \, d\theta \] Step 4: Solving the integral.
Using the identity \( \cos^2 \theta = \frac{1 + \cos(2\theta)}{2} \), we get: \[ A = 16 \int_0^{\frac{\pi}{2}} \frac{1 + \cos(2\theta)}{2} \, d\theta \] \[ A = 8 \int_0^{\frac{\pi}{2}} (1 + \cos(2\theta)) \, d\theta \] The integral of \( 1 \) is \( \theta \), and the integral of \( \cos(2\theta) \) is \( \frac{\sin(2\theta)}{2} \). Thus: \[ A = 8 \left[ \theta + \frac{\sin(2\theta)}{2} \right]_0^{\frac{\pi}{2}} \] Step 5: Conclusion.
Evaluating the limits, we get: \[ A = 8 \left[ \frac{\pi}{2} + \frac{\sin(\pi)}{2} - 0 \right] = 8 \times \frac{\pi}{2} = 4\pi \] Thus, the area bounded by the circle in the first quadrant is \( 4\pi \).