The problem asks for the integral that calculates the area of the rose curve given by the polar equation $\rho = 4 \cos 3\varphi$.

To find the area $A$ of one petal of a rose curve given by $\rho = a \cos(n\varphi)$ or $\rho = a \sin(n\varphi)$, the formula is:

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

For the given equation $\rho = 4 \cos 3\varphi$:

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

Thus, the area integral becomes:

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

Since this is a three-petaled rose (because $n = 3$ is odd), the limits of integration for one petal range from $\varphi = -\frac{\pi}{6}$ to $\varphi = \frac{\pi}{6}$, or equivalently, you could integrate from $0$ to $\frac{\pi}{3}$ for one full petal.

The total area of all three petals would then be:

$A_{\text{total}} = 3 \times 8 \int_{0}^{\frac{\pi}{3}} \cos^2 3\varphi \, d\varphi = 24 \int_{0}^{\frac{\pi}{3}} \cos^2 3\varphi \, d\varphi$

So the correct answer among the options would be:

$\boxed{24 \int_{0}^{\frac{\pi}{3}} \cos^2 3\varphi \, d\varphi}$

This corresponds to the third option in the provided choices.

Would you like more details or have any further questions?

Here are some related questions you might find helpful:

1. How do you determine the number of petals in a rose curve given by $\rho = a \cos(n\varphi)$?
2. What is the general formula for the area of a polar curve?
3. How does the range of integration change for different numbers of petals in a rose curve?
4. What is the difference between the integrals for $\cos$ and $\sin$ functions in polar coordinates?
5. How can trigonometric identities simplify the integration of $\cos^2$ functions?

Tip: When integrating trigonometric functions squared, use the identity $\cos^2 x = \frac{1 + \cos(2x)}{2}$ to simplify the process.