Let's solve the problem step by step and determine the correct 3D graph.

Problem Description:

You are asked to find the volume of a solid obtained by rotating the region bounded by the curves $y = x + 1$, $y = 0$, $x = 0$, and $x = 5$ about the x-axis.

Step 1: Sketch the Region

The region is bounded by:

1. $y = x + 1$ is a line with slope 1 and y-intercept at 1.
2. $y = 0$ is the x-axis.
3. $x = 0$ is the y-axis.
4. $x = 5$ is the vertical line at $x = 5$.

So, the region is a right triangle with vertices at $(0, 1)$, $(0, 0)$, and $(5, 6)$.

Step 2: Set up the Volume Integral

The volume of the solid is calculated using the disk method since we are rotating around the x-axis.

1. The radius of each disk is given by $r(x) = y = x + 1$ (since the distance from the curve to the x-axis is the height of the region).
2. The area of a disk is $A(x) = \pi [r(x)]^2 = \pi (x + 1)^2$.
3. To find the volume, integrate the area from $x = 0$ to $x = 5$.

The volume integral is: $V = \int_0^5 \pi (x + 1)^2 \, dx$ Let's solve this integral.

Step 3: Solve the Integral

We will now compute: $V = \pi \int_0^5 (x^2 + 2x + 1) \, dx$

The integral becomes: $V = \pi \left[ \frac{x^3}{3} + x^2 + x \right]_0^5$ Substituting the limits: [ V = \pi \left[ \frac{5^3}{3} + 5^2 + 5 \right] - \pi \left[ \frac{0^3}{3} + 0^2 + 0 \right] ] [ V = \pi \left[ \frac{125}{3} + 25 + 5 \right] = \pi \left[ \frac{125}{3} + \frac{75}{3} + \frac{15}{3} \right] ] $V = \pi \left[ \frac{215}{3} \right] = \frac{215\pi}{3}$

Step 4: Match with the 3D Graph

The solid obtained from this rotation is a cone-like shape with a rounded base, which corresponds to Graph 2 in your diagram.

Final Answer:

• The correct 3D graph is Graph 2.
• The volume of the solid is $\frac{215\pi}{3}$ cubic units.

Would you like more details on any of these steps or on visualizing the solid? Here are some related questions:

1. How does the disk method apply to other curves?
2. What happens if you rotate around a different axis?
3. Could the washer method be used here instead?
4. How do you handle cases where the curve crosses the axis of rotation?
5. How would this problem change if it involved polar coordinates?

Tip: Always sketch the region and the axis of rotation to visualize the solid before setting up the integral.