1. **State the problem:** Find the area of an ellipse with semi-major axis $a$ and semi-minor axis $b$ without using polar coordinates. 2. **Recall the formula for an ellipse:** The ellipse centered at the origin with axes along $x$ and $y$ is given by $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.$$ We want to find the area enclosed by this curve. 3. **Express $y$ in terms of $x$:** From the ellipse equation, solve for $y$: $$y = \pm b \sqrt{1 - \frac{x^2}{a^2}}.$$ The positive root represents the upper half of the ellipse. 4. **Set up the integral for the area:** The total area is twice the area under the upper half from $x = -a$ to $x = a$: $$\text{Area} = 2 \int_{-a}^a b \sqrt{1 - \frac{x^2}{a^2}} \, dx.$$ 5. **Simplify the integral:** Substitute $x = a \sin \theta$, so $dx = a \cos \theta \, d\theta$. When $x = -a$, $\theta = -\frac{\pi}{2}$; when $x = a$, $\theta = \frac{\pi}{2}$. 6. **Rewrite the integral in terms of $\theta$:** $$\text{Area} = 2b \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{1 - \sin^2 \theta} \cdot a \cos \theta \, d\theta = 2ab \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2 \theta \, d\theta.$$ 7. **Use the identity for $\cos^2 \theta$:** $$\cos^2 \theta = \frac{1 + \cos 2\theta}{2}.$$ 8. **Evaluate the integral:** $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2 \theta \, d\theta = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1 + \cos 2\theta}{2} \, d\theta = \frac{1}{2} \left[ \theta + \frac{\sin 2\theta}{2} \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \frac{1}{2} \left( \pi + 0 - (-\pi + 0) \right) = \frac{1}{2} (2\pi) = \pi.$$ 9. **Substitute back to find the area:** $$\text{Area} = 2ab \cdot \pi = \pi ab.$$ **Final answer:** The area of the ellipse is $$\boxed{\pi ab}.$$