It looks like you've shared a screenshot of a math problem. The task involves converting an integral into polar coordinates.

The given integral is:

$\int \int_D e^{-x^2 - y^2} \, dA$

where $D$ is the region bounded by the semicircle $x = \sqrt{36 - y^2}$ and the y-axis.

To help you convert this into polar coordinates, let's first recall the key transformations:

1. $x = r \cos \theta$
2. $y = r \sin \theta$
3. The differential area element $dA = r \, dr \, d\theta$.

Next, let's focus on the bounds. The equation $x = \sqrt{36 - y^2}$ represents a semicircle of radius 6 (since $\sqrt{36} = 6$) centered at the origin along the x-axis. So, the region described by this inequality can be described in polar coordinates as:

• $r$ ranges from 0 to 6 (because the radius of the semicircle is 6).
• $\theta$ ranges from 0 to $\pi$ (since it’s the upper half-plane, above the x-axis).

Now we can rewrite the integral:

$\int_0^\pi \int_0^6 e^{-r^2} r \, dr \, d\theta$

To solve this, we need to evaluate the inner integral with respect to $r$, and then the outer integral with respect to $\theta$.

Let me know if you'd like to proceed with this or need any further steps to clarify. Would you like me to continue or explain in more detail?