Overview and Background

Each concentric shell width is 'dx'

As displayed on the model, the equation shown is y=-x^2+a (some arbitrary constant).

A student may be asked a question like "What is the volume of the figure created when y=-x^2+9 is revolved around the y axis?"

Knowing that each 'slice' of width dx and height y gives a cross-sectional area that can then be multiplied by 2pi × x (the average radius of slice) to get the volume of that individual shell, the student can use integration to find the entire volume of the rotational solid:

2pi × the integral of yxdx = 2pi × int((-x^2+9)(x)dx = int((-x^3+9x)dx) = -x^4/4 + 9/2x^2
evaluated from x = 0 to x = 3: 2pi × (-20.25 + 40.5) = 2pi × 20.25
= 40.5pi

Lesson Plan and Activity

1) Review conceptual understanding of integrating with respect to x: the area under the curve is broken up into very thin, rectangular slices of width 'dx'. The area of each of these rectangles is height × width, where the height is the value of the function at x' and the width is dx. Therefore, A=y(x')dx

2) Remind students of geometric formula for finding the volume of a solid cylinder, V=pi×r^2×h. Lead the students to realize that to find the volume of a hollow cylinder or "shell" the formula can be modified to be V=2pi×(r1^2-r2^2)×h.

3) Illustrate a function being revolved around the y axis, restate the thin rectangular slices and lead students to draw similarities between these slices, the revolution, and the cylindrical shell equation.

4) Piece by piece, show that 2pi is a constant and can be left alone, r1^2-r2^2 = xdx, and h = y(x'). That addresses a single shell, but they know how to integrate to add up all the dx's. This leaves you with V = 2pi × int(yxdx) from x=0 to whatever upper bound is appropriate for the problem.

5) Pass around the model and allow students to stack and unstack, revolve, and understand how the 2 dimensional function translates to a revolved solid that can be broken up into concentric shells.

Solids of Rotation - Shell Method