8.1 Area Under Simple Curves

The circle is symmetric about both x and y axes. Total Area = $4 \times$ Area of the region in the first quadrant.

1. Equation: $y^2 = a^2 - x^2 \implies y = \sqrt{a^2 - x^2}$ (for 1st quadrant, $y>0$).
2. Limits: $x$ varies from $0$ to $a$.
3. Integral Setup: $A_1 = \int_{0}^{a} y dx = \int_{0}^{a} \sqrt{a^2 - x^2} dx$
4. Standard Integral Formula: $\int \sqrt{a^2 - x^2} dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$
5. Evaluate: $A_1 = \left[ \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) \right]_{0}^{a}$ Upper Limit ($a$): $\frac{a}{2}(0) + \frac{a^2}{2}\sin^{-1}(1) = \frac{a^2}{2} \cdot \frac{\pi}{2} = \frac{\pi a^2}{4}$. Lower Limit ($0$): $0 + 0 = 0$. $A_1 = \frac{\pi a^2}{4}$
6. Total Area: $4 A_1 = 4 \left( \frac{\pi a^2}{4} \right) = \pi a^2$. verified.