In the previous chapter, we explored methods to compute indefinite and definite integrals. In this chapter, we apply integral calculus to determine the geometric area bounded by simple curves, lines, and parabolas.
Let $y = f(x)$ be a continuous function bounded on the interval $[a, b]$. The area $A$ enclosed between the curve, the $x$-axis, and the vertical ordinate lines $x = a$ and $x = b$ is given by the definite integral expression:
$$A = \int_{a}^{b} f(x) \, dx = \int_{a}^{b} y \, dx$$Similarly, if the curve is defined as $x = g(y)$ and is bounded vertically by the horizontal abscissa lines $y = c$ and $y = d$, the enclosed area is computed as:
$$A = \int_{c}^{d} g(y) \, dy = \int_{c}^{d} x \, dy$$| Geometric Profile | Mathematical Equation Base | Active Integration Formulation | Evaluation System Limits |
|---|---|---|---|
| Standard Circle | $x^2 + y^2 = a^2$ | $y = \sqrt{a^2 - x^2}$ | Total Area $= 4 \int_{0}^{a} \sqrt{a^2 - x^2} \, dx = \pi a^2$ |
| Standard Ellipse | $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ | $y = \frac{b}{a}\sqrt{a^2 - x^2}$ | Total Area $= 4 \frac{b}{a} \int_{0}^{a} \sqrt{a^2 - x^2} \, dx = \pi a b$ |
| Vertical Parabola | $y^2 = 4ax$ | $y = 2\sqrt{a}\sqrt{x}$ | Area evaluated symmetrically along the coordinate axes. |
Mathematical Problem Statement: Find the area enclosed by the circle $x^2 + y^2 = a^2$.
Step-by-Step Algorithmic Solution Trace:
- Express the equation in terms of $y$ for the first quadrant setup: $y = \sqrt{a^2 - x^2}$.
- The region is symmetric across all four quadrants, so integrate from $x = 0$ to $x = a$ and scale by $4$:$$A = 4 \int_{0}^{a} \sqrt{a^2 - x^2} \, dx$$
- Apply the standard integration formula $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$:$$A = 4 \left[ \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) \right]_{0}^{a}$$
- Substitute upper limit $a$ and lower limit $0$:$$A = 4 \left[ \left(0 + \frac{a^2}{2}\sin^{-1}(1)\right) - (0 + 0) \right] = 4 \left( \frac{a^2}{2} \cdot \frac{\pi}{2} \right) = \pi a^2$$
Final Evaluation Output: Total area $= \pi a^2$ square units.