The problem asks to identify and sketch the graph of the equation $z = \sqrt{x^2 + y^2} + 1$ in three-space.

1. Problem Description

The problem asks to identify and sketch the graph of the equation $z = \sqrt{x^2 + y^2} + 1$ in three-space.

2. Solution Steps

The equation is $z = \sqrt{x^2 + y^2} + 1$.

Let's analyze this equation to determine the type of surface it represents. We notice that the right-hand side involves $\sqrt{x^2 + y^2}$, which is the radial distance in the $xy$-plane. Let $r = \sqrt{x^2 + y^2}$. Then the equation becomes $z = r + 1$.

Since $r \ge 0$, we have $z \ge 1$.

We can also rewrite the equation as $z-1 = \sqrt{x^2 + y^2}$.

Squaring both sides, we get $(z-1)^2 = x^2 + y^2$.

This is the equation of a cone whose vertex is at $(0, 0, 1)$. The variable $z$ must be greater than or equal to 1 because of the square root in the original equation.

When $z = 1$, $x^2 + y^2 = 0$, so $x = 0$ and $y = 0$. This is the vertex of the cone.

When $z = 2$, $x^2 + y^2 = (2-1)^2 = 1$, which is a circle of radius 1 centered at $(0, 0, 2)$.

When $z = 3$, $x^2 + y^2 = (3-1)^2 = 4$, which is a circle of radius 2 centered at $(0, 0, 3)$.

Therefore, the graph is a cone opening upwards, with vertex at $(0, 0, 1)$. This cone is the same as the cone $z = \sqrt{x^2 + y^2}$ shifted up by 1 unit in the positive z-direction.

3. Final Answer

The graph is a cone with vertex at (0, 0, 1), opening upwards.

The equation represents a cone.