To convert the given integral from Cartesian coordinates $(x, y)$ to polar coordinates $(r, \theta)$, let's go step by step:

Problem Setup:

1. The region $C$ is the right-hand half of the disc centered at the origin with radius 2. In polar coordinates:

   • The radius $r$ varies from $0$ to $2$ ($r \in [0, 2]$).
   • The angle $\theta$ sweeps from $0$ to $\pi$ ($\theta \in [0, \pi]$) because it is the right-hand side of the disc.

2. The integrand $e^{-(x^2 + y^2)}$ in Cartesian coordinates transforms to $e^{-r^2}$ in polar coordinates (since $x^2 + y^2 = r^2$).

3. The differential area element $dA$ in polar coordinates becomes $r \, dr \, d\theta$.

Integral in Polar Coordinates:

The double integral becomes: $\iint_C e^{-(x^2 + y^2)} \, dA = \int_0^\pi \int_0^2 e^{-r^2} \cdot r \, dr \, d\theta$

Expression:

$\iint_C e^{-(x^2 + y^2)} \, dA = \int_0^\pi \int_0^2 r e^{-r^2} \, dr \, d\theta$

Simplifying the Integral:

1. The inner integral with respect to $r$: $\int_0^2 r e^{-r^2} \, dr$ Let $u = r^2$, so $du = 2r \, dr$. The limits for $u$ change as follows:

   • When $r = 0$, $u = 0$.
   • When $r = 2$, $u = 4$.

   Substituting, the integral becomes: $\int_0^2 r e^{-r^2} \, dr = \frac{1}{2} \int_0^4 e^{-u} \, du = \frac{1}{2} \left[ -e^{-u} \right]_0^4 = \frac{1}{2} \left( 1 - e^{-4} \right)$

2. The outer integral with respect to $\theta$: $\int_0^\pi \, d\theta = \pi$

3. Combining results: $\iint_C e^{-(x^2 + y^2)} \, dA = \pi \cdot \frac{1}{2} \left( 1 - e^{-4} \right) = \frac{\pi}{2} \left( 1 - e^{-4} \right)$

Final Answer:

The integral evaluates to: $\boxed{\frac{\pi}{2} \left( 1 - e^{-4} \right)}$

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Would you like further clarification or details on any step? Here are related questions you might consider:

1. How is the region of integration defined in polar coordinates for other shapes?
2. What is the general formula for changing Cartesian integrals to polar coordinates?
3. Can this integral be evaluated numerically for specific applications?
4. How does the exponential decay $e^{-r^2}$ affect the integration?
5. What would change if the region were the left-hand side of the disc instead?

Tip: Always check the bounds of integration carefully when transitioning between Cartesian and polar coordinates!