To solve the given integral using polar coordinates, we will follow these steps:

Problem:

Evaluate
$\iint_R \sin(x^2 + y^2) \, dA$ where $R$ is the annular region defined by $4 \leq x^2 + y^2 \leq 25$.

Step 1: Convert the integral to polar coordinates

In polar coordinates:

• $x = r\cos\theta$, $y = r\sin\theta$
• $x^2 + y^2 = r^2$
• $dA = r \, dr \, d\theta$

The region $R$ becomes: $4 \leq r^2 \leq 25 \quad \Rightarrow \quad 2 \leq r \leq 5$ and $\theta$ ranges from $0$ to $2\pi$ (full circle).

Step 2: Rewrite the integral

The integral in polar coordinates is: $\int_0^{2\pi} \int_2^5 \sin(r^2) \cdot r \, dr \, d\theta$

Step 3: Separate the integral

The integrand does not depend on $\theta$, so the integral separates: $\int_0^{2\pi} \, d\theta \cdot \int_2^5 r \sin(r^2) \, dr$ The first integral evaluates to: $\int_0^{2\pi} \, d\theta = 2\pi$

Thus, the integral becomes: $2\pi \int_2^5 r \sin(r^2) \, dr$

Step 4: Simplify the radial integral

Use the substitution $u = r^2$:

• $du = 2r \, dr$
• When $r = 2$, $u = 4$
• When $r = 5$, $u = 25$

The integral transforms to: $\int_2^5 r \sin(r^2) \, dr = \frac{1}{2} \int_4^{25} \sin(u) \, du$

Step 5: Evaluate the integral

The integral of $\sin(u)$ is: $\int \sin(u) \, du = -\cos(u)$

Thus: $\frac{1}{2} \int_4^{25} \sin(u) \, du = \frac{1}{2} \left[ -\cos(u) \right]_4^{25}$ $= \frac{1}{2} \left[ -\cos(25) + \cos(4) \right]$

Step 6: Final expression

The full integral is: $2\pi \cdot \frac{1}{2} \left[ -\cos(25) + \cos(4) \right]$ $= \pi \left[ \cos(4) - \cos(25) \right]$

Final Answer:

$\iint_R \sin(x^2 + y^2) \, dA = \pi \left[ \cos(4) - \cos(25) \right]$

Would you like a detailed explanation of any step?

Expanding Questions:

1. How does the conversion from Cartesian to polar coordinates work in general?
2. Why is the Jacobian determinant $r$ included in the area element $dA$?
3. What are the implications of symmetry in polar coordinates when solving integrals?
4. Could this integral be solved using Cartesian coordinates, and if so, how?
5. How does substitution help simplify integrals involving trigonometric functions?

Tip:

When solving polar integrals, always verify the bounds and ensure that the Jacobian $r$ is included to account for the coordinate transformation.