To solve the given integral in spherical coordinates, we have the integral:

$\int_{0}^{2\pi} \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \int_{0}^{3} \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta.$

Step 1: Integrate with respect to $\rho$

The integral with respect to $\rho$ is: $\int_{0}^{3} \rho^2 \, d\rho.$ Compute this integral: $\int_{0}^{3} \rho^2 \, d\rho = \left[ \frac{\rho^3}{3} \right]_{0}^{3} = \frac{3^3}{3} - \frac{0^3}{3} = \frac{27}{3} = 9.$

Step 2: Integrate with respect to $\phi$

Now the integral becomes: $9 \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \sin\phi \, d\phi.$ Compute this integral: $\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \sin\phi \, d\phi = \left[ -\cos\phi \right]_{\frac{\pi}{6}}^{\frac{\pi}{4}}.$ Evaluating at the bounds: $-\cos\left( \frac{\pi}{4} \right) + \cos\left( \frac{\pi}{6} \right) = -\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}.$

Therefore, the integral with respect to $\phi$ is: $9 \left( \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \right).$

Step 3: Integrate with respect to $\theta$

The remaining integral is: $\int_{0}^{2\pi} 9 \left( \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \right) d\theta.$ Since the expression is independent of $\theta$, this integral becomes: $9 \left( \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \right) \cdot \int_{0}^{2\pi} d\theta = 9 \left( \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \right) \cdot 2\pi.$

Final Answer

The evaluated integral is: $18\pi \left( \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \right).$

Would you like a more detailed explanation or have any questions about the steps?

Here are five related questions to expand your knowledge:

1. How do you convert Cartesian coordinates to spherical coordinates?
2. What are the limits of integration in spherical coordinates for a full sphere?
3. How do you set up triple integrals in cylindrical coordinates?
4. What is the Jacobian when converting to spherical coordinates?
5. How do you interpret the geometric meaning of this integral in 3D space?

Tip: When working with spherical coordinates, always remember to include the Jacobian term $\rho^2 \sin\phi$ in the integral to account for the coordinate transformation.