How to calculate volume of solid generated by revolving?

(Shell Method, about the line x = k, i.e., a line parallel to the y-axis) The volume of the solid generated by revolving about the line x = k the region between the graphs of continuous functions y = F(x) and y = f (x), F(x) ≥ f (x), a ≤ x ≤ b, k not between a and b, is

How to find the volume of a solid formed by shell method?

Find the volume of the solid formed by rotating the region bounded by and about using the Shell Method. Solution Since our shells are parallel to the axis of rotation, we must consider the radius and height functions in terms of . The radius of a sample shell will be and the height of a sample shell will be .

When the region is rotated this thin slice is formed?

When the region is rotated, this thin slice forms a cylindrical shell, as pictured in part (b) of the figure. The previous section approximated a solid with lots of thin disks (or washers); we now approximate a solid with many thin cylindrical shells.

How do you find the area of a solid disk?

In the case that we get a solid disk the area is, A = π(radius)2 A = π (radius) 2 where the radius will depend upon the function and the axis of rotation. In the case that we get a ring the area is,

What is the axis of rotation for x = 2?

The axis of rotation, x = 2, is a line parallel

What is the difference between the axis of rotation and inner radius?

So, we know that the distance from the axis of rotation to the x x -axis is 4 and the distance from the x x -axis to the inner ring is x x. The inner radius must then be the difference between these two.

How do you find the cross section of a solid object?

One of the easier methods for getting the cross-sectional area is to cut the object perpendicular to the axis of rotation. Doing this the cross section will be either a solid disk if the object is solid (as our above example is) or a ring if we’ve hollowed out a portion of the solid (we will see this eventually).