Find the volume of the frustrum of the right

circular cone generated by rotating the line y=x+1
about the x-axis between x = 0 and x = 3

To find the volume of the frustum of the right circular cone generated by rotating the line y = x + 1 about the x-axis between x = 0 and x = 3, we will use the method of cylindrical shells.

Here are the steps to calculate the volume:

Step 1: Determine the equations of the top and bottom circles of the frustum.
- The top circle is generated by rotating the point (3, 4) because the line y = x + 1 intersects the x-axis at (3, 4).
- The bottom circle is generated by rotating the point (0, 1) because the line y = x + 1 intersects the x-axis at (0, 1).
- The equations of the circles are:
- Top circle: (x - 3)^2 + (y - 4)^2 = r^2
- Bottom circle: (x - 0)^2 + (y - 1)^2 = R^2

Step 2: Determine the radii of the top and bottom circles.
- The radius of the top circle (r) is the distance between the point (3, 4) and the y-axis.
- The distance formula gives: r = sqrt((3 - 0)^2 + (4 - 0)^2) = 5.
- The radius of the bottom circle (R) is the distance between the point (0, 1) and the y-axis.
- The distance formula gives: R = sqrt((0 - 0)^2 + (1 - 0)^2) = 1.

Step 3: Determine the height of the frustum.
- The height of the frustum is the difference between the y-coordinate of the top point (4) and the y-coordinate of the bottom point (1).
- Therefore, the height is: h = 4 - 1 = 3.

Step 4: Calculate the volume of the frustum using the formula for the volume of a frustum of a cone.
- The formula for the volume of a frustum of a cone is: V = (1/3)πh(R^2 + Rr + r^2).
- Substituting the known values, we get: V = (1/3)π(3)(1^2 + 1(5) + 5^2).

Step 5: Simplify and calculate the volume.
- V = (1/3)π(3)(1 + 5 + 25) = (1/3)π(3)(31) = π(31) = 31π.

Therefore, the volume of the frustum of the right circular cone generated by rotating the line y = x + 1 about the x-axis between x = 0 and x = 3 is 31π cubic units.

I somehow missed your attempt at doing this. Either way, using the formula for volume of cones or doing the integration of circular areas (r=x+1) from x = 0 to x = 3 will work.