8.2 Area of the Region Bounded by a Curve and a Line

Method 1: Vertical Strips ($dx$)

1. Intersection: $x^2 = 4 \implies x = \pm 2$.
2. Upper and Lower: The region is bounded above by line $y=4$ and below by parabola $y=x^2$.
3. Integral: $A = \int_{-2}^{2} (y_{upper} - y_{lower}) dx = \int_{-2}^{2} (4 - x^2) dx$ Due to symmetry (even function), $A = 2 \int_{0}^{2} (4 - x^2) dx$. $A = 2 \left[ 4x - \frac{x^3}{3} \right]_{0}^{2}$ $A = 2 \left( 8 - \frac{8}{3} \right) = 2 \left( \frac{16}{3} \right) = \frac{32}{3} \text{ sq units}$

Method 2: Horizontal Strips ($dy$)

1. Limits: $y$ varies from 0 (vertex) to 4 (line).
2. Function: $y = x^2 \implies x = \pm \sqrt{y}$. Right branch ($x>0$): $x = \sqrt{y}$. Left branch: $x = -\sqrt{y}$. Width of strip = $x_{right} - x_{left} = \sqrt{y} - (-\sqrt{y}) = 2\sqrt{y}$.
3. Integral: $A = \int_{0}^{4} 2\sqrt{y} dy = 2 \left[ \frac{y^{3/2}}{3/2} \right]_{0}^{4}$ $A = 2 \cdot \frac{2}{3} [y^{3/2}]_{0}^{4} = \frac{4}{3} (4^{3/2})$ $A = \frac{4}{3} (8) = \frac{32}{3} \text{ sq units}$