1. **State the problem:** We are given the parametric equations of an ellipse: $$x = a \cos t, \quad y = b \sin t, \quad 0 \leq t \leq 2\pi$$ We need to find the area enclosed by this ellipse. 2. **Recall the formula for the area of an ellipse:** The area $A$ of an ellipse with semi-major axis $a$ and semi-minor axis $b$ is given by: $$A = \pi a b$$ 3. **Explanation:** The ellipse described by the parametric equations has semi-major axis length $a$ along the x-axis and semi-minor axis length $b$ along the y-axis. 4. **Derivation using parametric form (optional):** The area enclosed by a parametric curve $x(t), y(t)$ for $t$ in $[\alpha, \beta]$ is: $$A = \int_{\alpha}^{\beta} y(t) x'(t) dt$$ Here, $$x'(t) = \frac{d}{dt}(a \cos t) = -a \sin t$$ So, $$A = \int_0^{2\pi} b \sin t (-a \sin t) dt = -ab \int_0^{2\pi} \sin^2 t dt$$ 5. **Simplify the integral:** Using the identity $\sin^2 t = \frac{1 - \cos 2t}{2}$, $$A = -ab \int_0^{2\pi} \frac{1 - \cos 2t}{2} dt = -\frac{ab}{2} \int_0^{2\pi} (1 - \cos 2t) dt$$ 6. **Evaluate the integral:** $$\int_0^{2\pi} 1 dt = 2\pi$$ $$\int_0^{2\pi} \cos 2t dt = 0$$ Therefore, $$A = -\frac{ab}{2} (2\pi - 0) = -\frac{ab}{2} \times 2\pi = -ab \pi$$ 7. **Interpret the negative sign:** The negative sign appears because $x'(t)$ is negative. The area must be positive, so we take the absolute value: $$A = ab \pi$$ **Final answer:** $$\boxed{\pi a b}$$