The problem involves setting up and evaluating the integral in spherical coordinates to find the volume of the solid enclosed by the cone $z = \sqrt{x^2 + y^2}$ and bounded above by the sphere $x^2 + y^2 + z^2 = 4$.

Steps to Solve:

1. Convert the Equations to Spherical Coordinates:

   • The cone equation $z = \sqrt{x^2 + y^2}$ converts to $\rho\cos\phi = \rho\sin\phi$, simplifying to $\tan\phi = 1$ or $\phi = \pi/4$.
   • The sphere equation $x^2 + y^2 + z^2 = 4$ converts to $\rho^2 = 4$ or $\rho = 2$.

2. Set Up the Integral: In spherical coordinates, the volume element is $dV = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta$. Thus, the integral becomes: $V = \int_{0}^{2\pi} \int_{0}^{\pi/4} \int_{0}^{2} \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta$

3. Evaluate the Integral:

   • Integrate $\rho^2$ with respect to $\rho$ from 0 to 2.
   • Then integrate $\sin\phi$ with respect to $\phi$ from 0 to $\pi/4$.
   • Finally, integrate with respect to $\theta$ from 0 to $2\pi$.

Let's perform these integrations.

Performing Integration:The volume of the solid enclosed by the cone $z = \sqrt{x^2 + y^2}$ and bounded above by the sphere $x^2 + y^2 + z^2 = 4$ is $\frac{8}{3} \pi (1 - \sqrt{2})$.

Would you like to explore this solution further or do you have any questions on this topic?

Here are some related questions you might consider:

1. How does the choice of limits impact the integral's value in spherical coordinates?
2. What would be the volume if the bounding sphere had a different radius?
3. How does the volume change if the cone's angle is different from $\pi/4$?
4. Can you solve the same problem using cylindrical coordinates?
5. What are the applications of integrating in spherical coordinates in physics and engineering?

Tip: When integrating in spherical coordinates, visualize the solid to accurately set the limits for $\rho$, $\phi$, and $\theta$.