The given integral is:

$\int_0^1 \int_{3y}^3 \sin(x^2) \, dx \, dy$

To solve this double integral, we first need to switch the bounds of integration and then evaluate the resulting integral in the order $dy\, dx$.

Step 1: Understanding the region of integration

The bounds of the given integral are as follows:

• $y$ ranges from 0 to 1: $0 \leq y \leq 1$
• For a fixed $y$, $x$ ranges from $3y$ to 3: $3y \leq x \leq 3$

This describes a region in the $xy$-plane, and our goal is to switch the order of integration to evaluate the integral more easily.

Step 2: Determine the new bounds after switching

In terms of the variable $x$, notice that:

• $x$ goes from 0 to 3: $0 \leq x \leq 3$
• For a fixed $x$, $y$ ranges from $0$ to $x/3$: $0 \leq y \leq x/3$

Step 3: Rewrite the double integral

After switching the bounds of integration, we have:

$\int_0^3 \int_0^{x/3} \sin(x^2) \, dy \, dx$

Step 4: Evaluate the inner integral

The inner integral is with respect to $y$, and since $\sin(x^2)$ is independent of $y$, it can be factored out:

$\int_0^{x/3} \sin(x^2) \, dy = \sin(x^2) \cdot \left( \frac{x}{3} - 0 \right) = \frac{x}{3} \sin(x^2)$

Step 5: Evaluate the outer integral

Now, we need to evaluate the remaining integral with respect to $x$:

$\int_0^3 \frac{x}{3} \sin(x^2) \, dx$

To solve this integral, we use the substitution:

• Let $u = x^2$, so that $du = 2x \, dx$, or $\frac{du}{2} = x \, dx$.

This transforms the integral into:

$\frac{1}{3} \int_0^9 \frac{\sin(u)}{2} \, du = \frac{1}{6} \int_0^9 \sin(u) \, du$

Step 6: Solve the integral

The integral of $\sin(u)$ is $-\cos(u)$, so we get:

$\frac{1}{6} \left[ -\cos(u) \right]_0^9 = \frac{1}{6} \left( -\cos(9) + \cos(0) \right)$

Since $\cos(0) = 1$, the result is:

$\frac{1}{6} \left( -\cos(9) + 1 \right)$

Final Answer:

The value of the double integral is:

$\frac{1}{6} \left( 1 - \cos(9) \right)$

Would you like more details or have any questions?

Here are five related questions:

1. How do you decide when to switch the bounds of integration in a double integral?
2. What is the geometric interpretation of switching integration order in double integrals?
3. What are common substitution techniques for evaluating integrals involving trigonometric functions?
4. Can all double integrals be simplified by switching the order of integration?
5. How does the sine function affect the behavior of an integral?

Tip: Substitution is a powerful tool for solving integrals, especially when dealing with functions that involve powers or trigonometric components.