To solve this problem, we need to reverse the order of integration for the double integral:

$\int_0^{\frac{1}{4}} \int_{\sqrt{y}}^{\frac{1}{2}} \cos(4 \pi x^3) \, dx \, dy$

Step 1: Set up the region in the $xy$-plane

The limits of the integral are:

• For $y$: $0 \leq y \leq \frac{1}{4}$
• For $x$: $\sqrt{y} \leq x \leq \frac{1}{2}$

We need to rewrite these bounds in terms of $x$ and $y$ to change the order of integration.

1. From $x$: $\sqrt{y} \leq x \Rightarrow y \leq x^2$.
2. For $y$: $y \leq \frac{1}{4}$ and for $x$: $x \leq \frac{1}{2}$.

The region in terms of $x$ and $y$ is thus: $0 \leq x \leq \frac{1}{2}, \quad 0 \leq y \leq x^2$

Step 2: Rewrite the integral with reversed limits

Now, we can rewrite the integral with $x$ as the outer integral and $y$ as the inner integral:

$\int_0^{\frac{1}{2}} \int_0^{x^2} \cos(4 \pi x^3) \, dy \, dx$

Step 3: Integrate with respect to $y$

Since $\cos(4 \pi x^3)$ does not depend on $y$, we can treat it as a constant with respect to $y$:

$\int_0^{\frac{1}{2}} \left( \int_0^{x^2} \cos(4 \pi x^3) \, dy \right) dx = \int_0^{\frac{1}{2}} \cos(4 \pi x^3) \cdot y \Big|_0^{x^2} \, dx$

Evaluating the inner integral:

$= \int_0^{\frac{1}{2}} \cos(4 \pi x^3) \cdot x^2 \, dx$

Step 4: Integrate with respect to $x$

Now, we are left with the integral:

$\int_0^{\frac{1}{2}} x^2 \cos(4 \pi x^3) \, dx$

To evaluate this, let’s make the substitution $u = 4 \pi x^3$, which will simplify the integral.

1. Differentiating, we get $du = 12 \pi x^2 \, dx$, or $\frac{du}{12 \pi} = x^2 \, dx$.
2. When $x = 0$, $u = 0$.
3. When $x = \frac{1}{2}$, $u = 4 \pi \left(\frac{1}{2}\right)^3 = \frac{\pi}{2}$.

Thus, the integral becomes:

$\int_0^{\frac{\pi}{2}} \cos(u) \cdot \frac{1}{12 \pi} \, du = \frac{1}{12 \pi} \int_0^{\frac{\pi}{2}} \cos(u) \, du$

Step 5: Evaluate the integral

$= \frac{1}{12 \pi} \sin(u) \Big|_0^{\frac{\pi}{2}} = \frac{1}{12 \pi} \left( \sin\left(\frac{\pi}{2}\right) - \sin(0) \right) = \frac{1}{12 \pi} \cdot 1 = \frac{1}{12 \pi}$

Final Answer

$\int_0^{\frac{1}{4}} \int_{\sqrt{y}}^{\frac{1}{2}} \cos(4 \pi x^3) \, dx \, dy = \frac{1}{12 \pi}$

Would you like further details or have any questions?

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Related questions:

1. How would you reverse the integration order for a different type of region?
2. Can you evaluate an integral involving a different trigonometric function, like $\sin(x^2)$?
3. What are some common substitution techniques for integrals involving $x^2$ or $x^3$?
4. How does changing the limits affect the difficulty of integrating a cosine or sine function?
5. How would you approach this problem if it had an additional exponential factor?

Tip: When reversing the order of integration, always sketch the region in the $xy$-plane for clarity.