Question

Find the area enclosed by the curve \( y = x^2 \) and the line \( y = 16 \).

Hint:

Sketching the graphs is always a good first step. It helps you visualize which function is on top and what the limits of integration should be. Also, look for symmetries to simplify the integration process.

Answer 1

Step 1: Understanding the Concept:
The area enclosed between two curves is found by integrating the difference between the upper function and the lower function over the interval defined by their points of intersection.
Step 2: Key Formula or Approach:
1. Find the points of intersection by setting the two equations equal to each other.
2. Set up the definite integral: Area \( A = \int_{a}^{b} (y_{\text{upper}} - y_{\text{lower}}) \,dx \), where \( a \) and \( b \) are the x-coordinates of the intersection points.
3. Evaluate the integral.
Step 3: Detailed Explanation:
The given curves are \( y = x^2 \) (a parabola) and \( y = 16 \) (a horizontal line).
To find the points of intersection, we set \( x^2 = 16 \). This gives \( x = \pm 4 \). The points of intersection are (-4, 16) and (4, 16).
In the interval [-4, 4], the line \( y = 16 \) is above the parabola \( y = x^2 \). So, \( y_{\text{upper}} = 16 \) and \( y_{\text{lower}} = x^2 \).
The area is given by the integral: \[ A = \int_{-4}^{4} (16 - x^2) \,dx \] Since the integrand \( (16 - x^2) \) is an even function and the limits of integration are symmetric about 0, we can simplify the calculation: \[ A = 2 \int_{0}^{4} (16 - x^2) \,dx \] Now, we evaluate the integral: \[ A = 2 \left[ 16x - \frac{x^3}{3} \right]_{0}^{4} \] \[ A = 2 \left( \left(16(4) - \frac{4^3}{3}\right) - \left(16(0) - \frac{0^3}{3}\right) \right) \] \[ A = 2 \left( 64 - \frac{64}{3} - 0 \right) \] \[ A = 2 \left( \frac{192 - 64}{3} \right) = 2 \left( \frac{128}{3} \right) = \frac{256}{3} \] Step 4: Final Answer:
The area enclosed by the curve and the line is \( \frac{256}{3} \) square units.