1. The problem asks to evaluate the volume element in spherical coordinates at $r=2$, $\theta=30^\circ$, with $dr=d\theta=d\phi=1$. 2. The volume element in spherical coordinates is given by the formula: $$dV = r^2 \sin\theta \, dr \, d\theta \, d\phi$$ where $r$ is the radius, $\theta$ is the polar angle (measured from the positive z-axis), and $\phi$ is the azimuthal angle. 3. Convert $\theta=30^\circ$ to radians since trigonometric functions use radians: $$\theta = 30^\circ = \frac{\pi}{6}$$ 4. Substitute the values into the volume element formula: $$dV = (2)^2 \sin\left(\frac{\pi}{6}\right) \times 1 \times 1 \times 1$$ 5. Calculate each part: $$2^2 = 4$$ $$\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$$ 6. Multiply all together: $$dV = 4 \times \frac{1}{2} = 2$$ 7. Therefore, the volume element at the given coordinates is: $$\boxed{2}$$