The problem asks to evaluate the double integral $\iint_{R} \sin(x^2 + y^2) \, dA$ over the region $R$ enclosed by the lines $y = x$, $y = 2x$, $x = 0$, and $x = \frac{1}{2}$.

Steps to Solve:

1. Understand the region $R$:

   • The region $R$ is bounded by:
     • $y = x$ (Line 1)
     • $y = 2x$ (Line 2)
     • $x = 0$ (Vertical line)
     • $x = \frac{1}{2}$ (Vertical line)

2. Set up the double integral:

   • The double integral can be expressed as: $\int_{x=0}^{\frac{1}{2}} \int_{y=x}^{2x} \sin(x^2 + y^2) \, dy \, dx$

3. Integrate with respect to $y$:

   • Consider the integral with respect to $y$ first: $\int_{y=x}^{2x} \sin(x^2 + y^2) \, dy$ To solve this, we perform substitution:
   • Let $u = x^2 + y^2$, then $du = 2y \, dy$.
   • Since the limits for $y$ are from $x$ to $2x$, we need to adjust the limits accordingly:
     • When $y = x$, $u = x^2 + x^2 = 2x^2$
     • When $y = 2x$, $u = x^2 + 4x^2 = 5x^2$
   • The integral becomes: $\frac{1}{2} \int_{2x^2}^{5x^2} \frac{\sin(u)}{\sqrt{u - x^2}} \, du$
   • This integral is non-trivial and might need a special function approach or numerical methods.

4. Integrate with respect to $x$:

   • After simplifying or approximating the $y$-integral, integrate with respect to $x$ from 0 to $\frac{1}{2}$.

Result:

The exact evaluation requires advanced methods. For practical purposes, the integral might be best evaluated numerically.

Would you like to proceed with a specific step or have a more detailed explanation for any part? Let me know!

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Relative Questions:

1. What is the method to solve double integrals over non-rectangular regions?
2. How can substitutions simplify the evaluation of integrals?
3. What are common challenges when integrating trigonometric functions?
4. How can numerical integration be applied to complex integrals?
5. What are some real-world applications of double integrals?

Tip: For complex integrals, substitution often simplifies the process by reducing the integral into a more manageable form.