To solve this problem using polar coordinates, we need to evaluate the double integral:

$\iint_R \sin(x^2 + y^2) \, dA$

where $R$ is the annular region $16 \leq x^2 + y^2 \leq 49$. Here's how we can solve this step-by-step:

Step 1: Change to polar coordinates

In polar coordinates: $x^2 + y^2 = r^2, \quad dA = r \, dr \, d\theta$

The integral becomes: $\iint_R \sin(x^2 + y^2) \, dA = \iint_R \sin(r^2) \, r \, dr \, d\theta$

The region $R$ corresponds to $r$ ranging from $\sqrt{16} = 4$ to $\sqrt{49} = 7$, and $\theta$ ranging from $0$ to $2\pi$.

Step 2: Separate the integral

The integral separates as: $\int_0^{2\pi} d\theta \int_4^7 \sin(r^2) \, r \, dr$

The $\theta$-integral is straightforward: $\int_0^{2\pi} d\theta = 2\pi$

Now, focus on the $r$-integral: $\int_4^7 \sin(r^2) \, r \, dr$

Step 3: Simplify the $r$-integral with substitution

Let $u = r^2$, so $du = 2r \, dr$, or $r \, dr = \frac{1}{2} du$.

When $r = 4$, $u = 16$.
When $r = 7$, $u = 49$.

The integral becomes: $\int_4^7 \sin(r^2) \, r \, dr = \frac{1}{2} \int_{16}^{49} \sin(u) \, du$

Step 4: Compute the $u$-integral

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

Thus: $\frac{1}{2} \int_{16}^{49} \sin(u) \, du = \frac{1}{2} \left[ -\cos(u) \right]_{16}^{49}$

Evaluate the bounds: $-\cos(49) + \cos(16)$

So the result is: $\frac{1}{2} \left( \cos(16) - \cos(49) \right)$

Step 5: Multiply by $\theta$-integral

Now multiply by $\int_0^{2\pi} d\theta = 2\pi$: $\text{Result} = 2\pi \cdot \frac{1}{2} \left( \cos(16) - \cos(49) \right)$

Simplify: $\text{Result} = \pi \left( \cos(16) - \cos(49) \right)$

Final Answer:

$\pi \left( \cos(16) - \cos(49) \right)$

Would you like a detailed explanation of any step or further simplification?

Related Questions:

1. How do polar coordinates simplify integration in circular regions?
2. What substitution rules work well for integrals like $\int \sin(r^2) \, r \, dr$?
3. Why is $dA = r \, dr \, d\theta$ in polar coordinates?
4. What is the geometric interpretation of this integral over the annular region?
5. Can numerical methods be used to approximate this result?

Tip:

When integrating over circular regions, always check if polar coordinates simplify the bounds and integrand before attempting Cartesian integration!