Find the area of the ellipse (x²)/(a²) + (y²)/(b²) = 1

Find the area of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

We will evaluate the area using symmetry and a trigonometric substitution.

Step 1: Use symmetry

The ellipse is symmetric about both the $x$-axis and $y$-axis. We can calculate the area of the upper half (where $y \geq 0$) and multiply by 2. Or even better, calculate the area in the first quadrant and multiply by 4.

Step 2: Solve for $y$ in the first quadrant

From $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$:

\[\frac{y^2}{b^2} = 1 - \frac{x^2}{a^2} = \frac{a^2 - x^2}{a^2}\]

\[y^2 = \frac{b^2(a^2 - x^2)}{a^2}\]

For the upper half ($y \geq 0$):

\[y = \frac{b}{a}\sqrt{a^2 - x^2}\]

Step 3: Set up the integral

The area of the ellipse is:

\[A = 4 \times \text{(area in first quadrant)}\]

\[A = 4\int_0^a y \, dx = 4\int_0^a \frac{b}{a}\sqrt{a^2 - x^2} \, dx\]

\[= \frac{4b}{a}\int_0^a \sqrt{a^2 - x^2} \, dx\]

Step 4: Evaluate $\int_0^a \sqrt{a^2 - x^2} \, dx$

This is the area of a quarter circle of radius $a$. Using the substitution $x = a\sin\theta$, $dx = a\cos\theta \, d\theta$:

When $x = 0$: $\theta = 0$. When $x = a$: $\theta = \frac{\pi}{2}$

\[\int_0^a \sqrt{a^2 - x^2} \, dx = \int_0^{\pi/2} \sqrt{a^2 - a^2\sin^2\theta} \cdot a\cos\theta \, d\theta\]

\[= \int_0^{\pi/2} a\sqrt{1 - \sin^2\theta} \cdot a\cos\theta \, d\theta = \int_0^{\pi/2} a|\cos\theta| \cdot a\cos\theta \, d\theta\]

Since $\cos\theta \geq 0$ on $[0, \frac{\pi}{2}]$:

\[= a^2\int_0^{\pi/2} \cos^2\theta \, d\theta\]

Using $\cos^2\theta = \frac{1 + \cos 2\theta}{2}$:

\[= a^2\int_0^{\pi/2} \frac{1 + \cos 2\theta}{2} \, d\theta = \frac{a^2}{2}\left[\theta + \frac{\sin 2\theta}{2}\right]_0^{\pi/2}\]

\[= \frac{a^2}{2}\left[\left(\frac{\pi}{2} + 0\right) - (0 + 0)\right] = \frac{a^2}{2} \cdot \frac{\pi}{2} = \frac{\pi a^2}{4}\]

Step 5: Calculate the area of the ellipse

\[A = \frac{4b}{a} \cdot \frac{\pi a^2}{4} = \frac{4b\pi a^2}{4a} = \pi ab\]

\[\boxed{A = \pi ab \text{ square units}}\]

Verification: Special cases

Circle: If $a = b = r$ (circle of radius $r$):

\[A = \pi \cdot r \cdot r = \pi r^2 \quad \checkmark\]

This is the familiar formula for the area of a circle!

Numerical example: If $a = 3$ and $b = 2$:

\[A = \pi \cdot 3 \cdot 2 = 6\pi \approx 18.85 \text{ square units}\]

Alternative derivation using parametric equations

The ellipse can be parameterized as:

\[x = a\cos t, \quad y = b\sin t, \quad t \in [0, 2\pi]\]

Using Green's theorem or the shoelace formula:

\[A = \frac{1}{2}\left|\int_0^{2\pi} (x\frac{dy}{dt} - y\frac{dx}{dt}) \, dt\right|\]

With $\frac{dx}{dt} = -a\sin t$ and $\frac{dy}{dt} = b\cos t$:

\[A = \frac{1}{2}\int_0^{2\pi} (a\cos t \cdot b\cos t - b\sin t \cdot (-a\sin t)) \, dt\]

\[= \frac{1}{2}\int_0^{2\pi} (ab\cos^2 t + ab\sin^2 t) \, dt = \frac{ab}{2}\int_0^{2\pi} 1 \, dt = \frac{ab}{2} \cdot 2\pi = \pi ab\]

Summary

The area of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is:

\[\boxed{A = \pi ab \text{ square units}}\]

where $a$ is the semi-major (or semi-minor) axis along the $x$-direction and $b$ is the semi-major (or semi-minor) axis along the $y$-direction.

Key steps:

1. Use symmetry: calculate first quadrant area and multiply by 4
2. Express $y = \frac{b}{a}\sqrt{a^2 - x^2}$
3. Set up integral: $A = \frac{4b}{a}\int_0^a \sqrt{a^2 - x^2} \, dx$
4. Recognize quarter-circle: $\int_0^a \sqrt{a^2 - x^2} \, dx = \frac{\pi a^2}{4}$
5. Calculate: $A = \frac{4b}{a} \cdot \frac{\pi a^2}{4} = \pi ab$

Key insight:

The area of an ellipse is $\pi$ times the product of its semi-axes, generalizing the circle formula $\pi r^2 = \pi \cdot r \cdot r$.