To find the volume of the solid obtained by rotating the region bounded by the curves $y = 3 \cos(x^2)$, $y = 0$, $x = 0$, and $x = \sqrt{\frac{\pi}{3}}$ about the $x$-axis, we can use the method of disks.

Step 1: Set up the volume integral

When rotating a region about the $x$-axis, the volume $V$ is given by:

$V = \pi \int_{a}^{b} \left[ f(x) \right]^2 \, dx$

Here:

• $f(x) = 3 \cos(x^2)$
• The bounds are $x = 0$ to $x = \sqrt{\frac{\pi}{3}}$.

So the volume integral becomes:

$V = \pi \int_{0}^{\sqrt{\frac{\pi}{3}}} \left[ 3 \cos(x^2) \right]^2 \, dx$

Step 2: Simplify the integrand

Simplify the integrand:

$V = \pi \int_{0}^{\sqrt{\frac{\pi}{3}}} 9 \cos^2(x^2) \, dx$

Factor out the constant:

$V = 9\pi \int_{0}^{\sqrt{\frac{\pi}{3}}} \cos^2(x^2) \, dx$

Step 3: Use an appropriate substitution or numerical methods

The integral $\int \cos^2(x^2) \, dx$ does not have an elementary antiderivative. Therefore, this integral is typically evaluated using numerical methods.

However, we can express the integral using the half-angle identity for cosine:

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

For our integral:

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

Thus, the integral becomes:

$V = 9\pi \int_{0}^{\sqrt{\frac{\pi}{3}}} \frac{1 + \cos(2x^2)}{2} \, dx$

Simplifying:

$V = \frac{9\pi}{2} \int_{0}^{\sqrt{\frac{\pi}{3}}} \left[ 1 + \cos(2x^2) \right] \, dx$

This integral can be split into two parts:

$V = \frac{9\pi}{2} \left[ \int_{0}^{\sqrt{\frac{\pi}{3}}} 1 \, dx + \int_{0}^{\sqrt{\frac{\pi}{3}}} \cos(2x^2) \, dx \right]$

The first integral is straightforward:

$\int_{0}^{\sqrt{\frac{\pi}{3}}} 1 \, dx = \sqrt{\frac{\pi}{3}}$

The second integral $\int_{0}^{\sqrt{\frac{\pi}{3}}} \cos(2x^2) \, dx$ does not have a simple closed form and is usually evaluated numerically.

Let's denote this integral by $I$:

$I = \int_{0}^{\sqrt{\frac{\pi}{3}}} \cos(2x^2) \, dx$

Thus, the volume $V$ becomes:

$V = \frac{9\pi}{2} \left[ \sqrt{\frac{\pi}{3}} + I \right]$

Step 4: Compute the numerical value

The integral $I$ can be computed numerically, but we can provide a simplified expression here:

$V = \frac{9\pi}{2} \left[ \sqrt{\frac{\pi}{3}} + I \right]$

Since $I$ is a definite integral, it is a fixed number that can be approximated using numerical integration methods such as Simpson's rule, trapezoidal rule, or using a calculator or software.

Final Answer:

The volume of the solid obtained by rotating the region bounded by $y = 3 \cos(x^2)$, $y = 0$, $x = 0$, and $x = \sqrt{\frac{\pi}{3}}$ about the $x$-axis is given by:

$V = \frac{9\pi}{2} \left[ \sqrt{\frac{\pi}{3}} + I \right]$

where $I$ is the integral $\int_{0}^{\sqrt{\frac{\pi}{3}}} \cos(2x^2) \, dx$, which needs to be evaluated numerically for an exact value.

If you need the exact numerical value, I can help compute it. Let me know!

Here are 5 related questions:

1. How do you approach integrals that do not have elementary antiderivatives?
2. What is the half-angle identity, and how is it used in integration?
3. How can you evaluate integrals numerically?
4. What are some alternative methods for finding volumes of revolution?
5. How does the choice of axis of rotation affect the integral setup?

Tip: When dealing with complex integrals that do not have elementary solutions, consider using numerical methods or computer algebra systems to approximate the result.