1. **State the problem:** Convert the double integral $$\int_0^6 \int_{-\sqrt{6x - x^2}}^{\sqrt{6x - x^2}} y \, dy \, dx$$ to polar coordinates. 2. **Understand the region:** The limits for $y$ are from $-\sqrt{6x - x^2}$ to $\sqrt{6x - x^2}$, which describes a semicircle centered at $(3,0)$ with radius 3 because $6x - x^2 = 9 - (x-3)^2$. 3. **Polar coordinates:** Recall that $x = r \cos \theta$, $y = r \sin \theta$, and the Jacobian determinant for the area element is $dx \, dy = r \, dr \, d\theta$. 4. **Determine angular limits:** Since $x$ goes from 0 to 6 and $y$ covers the full vertical semicircle, the angle $\theta$ ranges from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$. 5. **Determine radial limits:** The boundary circle is $r = 6 \cos \theta$ because the circle is shifted along the $x$-axis. 6. **Rewrite the integral:** Substitute $y = r \sin \theta$ and $dx \, dy = r \, dr \, d\theta$: $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_0^{6 \cos \theta} (r \sin \theta) (r) \, dr \, d\theta = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_0^{6 \cos \theta} r^2 \sin \theta \, dr \, d\theta$$ 7. **Compare with options:** This matches option (d): $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_0^{6 \cos \theta} r^2 \sin \theta \, dr \, d\theta$$ **Final answer:** Option (d) is the correct polar coordinate form of the given integral.