To find the area enclosed by the polar curve $r = 4(1 + \cos\theta)$, we use the formula for the area enclosed by a polar curve:

$A = \frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 \, d\theta$

Step 1: Simplify the equation

The given equation is $r = 4(1 + \cos\theta)$. The square of $r$ is:

$r^2 = [4(1 + \cos\theta)]^2 = 16(1 + \cos\theta)^2$

Step 2: Expand $(1 + \cos\theta)^2$

$(1 + \cos\theta)^2 = 1 + 2\cos\theta + \cos^2\theta$

Substitute $\cos^2\theta = \frac{1 + \cos(2\theta)}{2}$:

$(1 + \cos\theta)^2 = 1 + 2\cos\theta + \frac{1 + \cos(2\theta)}{2} = \frac{3}{2} + 2\cos\theta + \frac{\cos(2\theta)}{2}$

Thus: $r^2 = 16\left(\frac{3}{2} + 2\cos\theta + \frac{\cos(2\theta)}{2}\right) = 24 + 32\cos\theta + 8\cos(2\theta)$

Step 3: Determine integration limits

Since $r = 4(1 + \cos\theta)$ is a limaçon, the curve is symmetric about the polar axis. The full curve is traced as $\theta$ goes from $0$ to $2\pi$.

Step 4: Set up the integral

$A = \frac{1}{2} \int_{0}^{2\pi} \left(24 + 32\cos\theta + 8\cos(2\theta)\right) d\theta$

Step 5: Solve the integral

Break it into parts:

1. $\int_{0}^{2\pi} 24 \, d\theta = 24\theta \Big|_{0}^{2\pi} = 24(2\pi) - 24(0) = 48\pi$
2. $\int_{0}^{2\pi} 32\cos\theta \, d\theta = 32[\sin\theta]_{0}^{2\pi} = 32(\sin(2\pi) - \sin(0)) = 0$
3. $\int_{0}^{2\pi} 8\cos(2\theta) \, d\theta = 8\left[\frac{\sin(2\theta)}{2}\right]_{0}^{2\pi} = 8 \cdot 0 = 0$

Step 6: Add results

$A = \frac{1}{2}(48\pi + 0 + 0) = \frac{48\pi}{2} = 24\pi$

Final Answer:

The area enclosed by the curve is: $\boxed{24\pi}$

Do you want further details or have any questions?

---

Here are 5 related questions:

1. How is the area formula for polar coordinates derived?
2. What happens if the curve is not symmetric?
3. How do you determine the limits of integration for a polar curve?
4. What role does trigonometric identity play in polar integrals?
5. Can this method handle curves with overlapping regions?

Tip: Always sketch the polar curve to better understand the limits of integration and symmetry!