To solve this problem, we aim to compute the surface area of the portion $\sigma$ of the cone $z = \sqrt{x^2 + y^2}$ located below the plane $z = 2$. Using the surface area formula with the $xy$-approach, we proceed as follows:

Surface Area Formula

The surface area of a surface $z = f(x, y)$ is given by: $\text{Surface Area} = \iint_R \sqrt{1 + \left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2} \, dA,$ where $R$ is the region of projection in the $xy$-plane.

Here, $z = \sqrt{x^2 + y^2}$, and the plane $z = 2$ intersects the cone when: $\sqrt{x^2 + y^2} = 2 \implies x^2 + y^2 = 4.$ Thus, $R$ is the disk $x^2 + y^2 \leq 4$ in the $xy$-plane.

Step 1: Compute Partial Derivatives

The partial derivatives of $z = \sqrt{x^2 + y^2}$ are: $\frac{\partial z}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}}, \quad \frac{\partial z}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}}.$ Their squares are: $\left( \frac{\partial z}{\partial x} \right)^2 = \frac{x^2}{x^2 + y^2}, \quad \left( \frac{\partial z}{\partial y} \right)^2 = \frac{y^2}{x^2 + y^2}.$ Adding these: $\left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2 = \frac{x^2 + y^2}{x^2 + y^2} = 1.$

Step 2: Surface Area Integrand

The integrand becomes: $\sqrt{1 + \left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2} = \sqrt{1 + 1} = \sqrt{2}.$

Thus, the surface area is: $\text{Surface Area} = \sqrt{2} \iint_R 1 \, dA.$

Step 3: Compute $\iint_R 1 \, dA$

The region $R$ is a disk of radius 2. In polar coordinates: $x = r\cos\theta, \quad y = r\sin\theta, \quad dA = r \, dr \, d\theta,$ with $0 \leq r \leq 2$ and $0 \leq \theta \leq 2\pi$. The integral becomes: $\iint_R 1 \, dA = \int_0^{2\pi} \int_0^2 r \, dr \, d\theta.$ Compute the inner integral: $\int_0^2 r \, dr = \left[\frac{r^2}{2}\right]_0^2 = \frac{2^2}{2} - 0 = 2.$ Now compute the outer integral: $\int_0^{2\pi} 2 \, d\theta = 2 \cdot 2\pi = 4\pi.$

Step 4: Final Surface Area

Multiply by $\sqrt{2}$: $\text{Surface Area} = \sqrt{2} \cdot 4\pi = 4\pi\sqrt{2}.$

Final Answer:

The surface area of $\sigma$ is: $\boxed{4\pi\sqrt{2}}$

Would you like further clarification or additional steps detailed?

Related Questions:

1. How does the partial derivative computation contribute to the integrand simplification?
2. What would change if the cone equation were modified, e.g., $z = 2\sqrt{x^2 + y^2}$?
3. How can this result be generalized for cones with different apex angles?
4. Can the polar coordinates method be extended to more complex surfaces?
5. What is the physical interpretation of the surface area in practical applications?

Tip:

Always visualize the geometry of the surface and projection region when setting up integrals for surface area calculations!