The question asks for the integral used to calculate the area of a rose curve, specifically for the polar equation $\rho = 4 \cos(3\varphi)$.

Step-by-Step Solution:

The general formula for the area enclosed by a polar curve $\rho = f(\varphi)$ is:

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

In this case, $\rho = 4 \cos(3\varphi)$, so:

$\rho^2 = (4 \cos(3\varphi))^2 = 16 \cos^2(3\varphi)$

Thus, the area becomes:

$A = \frac{1}{2} \int_{\alpha}^{\beta} 16 \cos^2(3\varphi) \, d\varphi = 8 \int_{\alpha}^{\beta} \cos^2(3\varphi) \, d\varphi$

For a rose curve of the form $\rho = 4 \cos(3\varphi)$, the total area is calculated over one petal, and because the curve has symmetry, the limits of integration for one petal range from $\varphi = -\frac{\pi}{6}$ to $\frac{\pi}{6}$. The entire area is given by multiplying the area of one petal by the number of petals, which for this curve (with $3 \varphi$) is 6 petals.

The correct integral that gives the total area is:

$8 \int_0^{2\pi} \cos^2(3\varphi) \, d\varphi$

Therefore, the correct option is:

• $8 \int_0^{2\pi} \cos^2(3\varphi) \, d\varphi$

This matches the fourth option.

Let me know if you need further details or explanations. Here are 5 additional questions to explore this concept:

1. How do you derive the formula for the area enclosed by a polar curve?
2. What is the significance of the limits of integration for different polar curves?
3. How does the number of petals affect the total area of a rose curve?
4. Can this method be applied to curves with sine functions, such as $\rho = a \sin(k\varphi)$?
5. How do you simplify integrals involving trigonometric functions like $\cos^2(\theta)$?

Tip: For integrals involving $\cos^2(\theta)$, remember to use the identity $\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$ to simplify calculations.