Find the area bounded by the lemniscate: rho² = a² cos 2

Find the area bounded by the lemniscate: $\rho^2 = a^2 \cos 2\varphi$.

We will use the polar-coordinates area formula together with the symmetry of the two lobes.

Step 1: Understand the lemniscate

The lemniscate $\rho^2 = a^2 \cos 2\varphi$ is a figure-eight shaped curve. For the curve to be defined, we need $\cos 2\varphi \geq 0$. This occurs when:

\[-\frac{\pi}{4} \leq \varphi \leq \frac{\pi}{4} \quad \text{or} \quad \frac{3\pi}{4} \leq \varphi \leq \frac{5\pi}{4}\]

The lemniscate has two symmetric lobes.

Step 2: Formula for area in polar coordinates

The area enclosed by a polar curve $\rho = f(\varphi)$ from $\varphi = \alpha$ to $\varphi = \beta$ is:

\[A = \frac{1}{2}\int_{\alpha}^{\beta} \rho^2 \, d\varphi\]

Step 3: Use symmetry

The lemniscate is symmetric about both axes. We can calculate the area of the right lobe (where $-\frac{\pi}{4} \leq \varphi \leq \frac{\pi}{4}$) and multiply by 2.

For the right lobe:

\[A_{\text{right}} = \frac{1}{2}\int_{-\pi/4}^{\pi/4} \rho^2 \, d\varphi = \frac{1}{2}\int_{-\pi/4}^{\pi/4} a^2\cos 2\varphi \, d\varphi\]

Step 4: Evaluate the integral

\[A_{\text{right}} = \frac{a^2}{2}\int_{-\pi/4}^{\pi/4} \cos 2\varphi \, d\varphi\]

\[= \frac{a^2}{2}\left[\frac{\sin 2\varphi}{2}\right]_{-\pi/4}^{\pi/4}\]

\[= \frac{a^2}{4}[\sin 2\varphi]_{-\pi/4}^{\pi/4}\]

At $\varphi = \frac{\pi}{4}$: $\sin\left(2 \cdot \frac{\pi}{4}\right) = \sin\frac{\pi}{2} = 1$

At $\varphi = -\frac{\pi}{4}$: $\sin\left(2 \cdot \left(-\frac{\pi}{4}\right)\right) = \sin\left(-\frac{\pi}{2}\right) = -1$

Therefore:

\[A_{\text{right}} = \frac{a^2}{4}[1 - (-1)] = \frac{a^2}{4} \cdot 2 = \frac{a^2}{2}\]

Step 5: Total area

The total area of the lemniscate (both lobes) is:

\[A = 2 \cdot A_{\text{right}} = 2 \cdot \frac{a^2}{2} = a^2\]

\[\boxed{A = a^2 \text{ square units}}\]

Alternative approach: Direct integration

We can also integrate over both lobes directly:

\[A = \frac{1}{2}\int_{-\pi/4}^{\pi/4} a^2\cos 2\varphi \, d\varphi + \frac{1}{2}\int_{3\pi/4}^{5\pi/4} a^2\cos 2\varphi \, d\varphi\]

By symmetry, both integrals give $\frac{a^2}{2}$, so:

\[A = \frac{a^2}{2} + \frac{a^2}{2} = a^2\]

Verification using substitution

Let $u = 2\varphi$, so $du = 2d\varphi$ or $d\varphi = \frac{du}{2}$.

When $\varphi = -\frac{\pi}{4}$: $u = -\frac{\pi}{2}$. When $\varphi = \frac{\pi}{4}$: $u = \frac{\pi}{2}$

\[A_{\text{right}} = \frac{a^2}{2}\int_{-\pi/2}^{\pi/2} \cos u \cdot \frac{du}{2} = \frac{a^2}{4}[\sin u]_{-\pi/2}^{\pi/2}\]

\[= \frac{a^2}{4}[1 - (-1)] = \frac{a^2}{2} \quad \checkmark\]

Numerical example

For $a = 2$:

\[A = 2^2 = 4 \text{ square units}\]

Summary

The total area enclosed by the lemniscate $\rho^2 = a^2 \cos 2\varphi$ is:

\[\boxed{A = a^2 \text{ square units}}\]

Key steps:

1. Identify the curve: lemniscate with two symmetric lobes
2. Use polar area formula: $A = \frac{1}{2}\int \rho^2 \, d\varphi$
3. Calculate one lobe: $A_{\text{right}} = \frac{1}{2}\int_{-\pi/4}^{\pi/4} a^2\cos 2\varphi \, d\varphi = \frac{a^2}{2}$
4. Use symmetry: Total area = $2 \times \frac{a^2}{2} = a^2$

Key insight:

The lemniscate is a beautiful curve with a simple area formula. Despite its complex shape (figure-eight), the total enclosed area is simply $a^2$, where $a$ is the parameter in the equation.