
The cylindrical shell method finds the volume of a solid created when a 2D region is revolved around an axis by summing up the volumes of thin cylindrical shells.
Each shell is like a thin soup can label:
Height = value of the function
Radius = distance from axis
Thickness = dx or dy (depending on integration direction)
The shell method is especially convenient when:
Use vertical slices, thickness dx:
V=$$2\pi\int_{a}^{b}(radian)(height)dx$$
Where:
radius = x (distance from y-axis)
height = f(x) or difference of functions
So:
V=$$2\pi\int_{a}^{b}x[f(x)]dx$$
Use horizontal slices, thickness dy:
V=$$2\pi\int_{a}^{b}y[g(y)]dy$$
Same idea:
radius = distance from axis
height = function expressed in terms of y
A thin shell has:
Radius = r
Height = h
Thickness = Δx
Circumference = 2πr
Volume of thin shell:
Volume=$$2\pi\cdot{h}\cdot\Delta{x}$$
Sum → integral:
V=$$\int2\pi{r}hdx$$
Find the volume generated when the region under
$$y=x^2$$
from x=0 to x=2 is rotated around the y-axis.
Shell Method Setup:
radius = x
height = $$x^2$$
thickness = dx
V=$$2\pi\int_{0}^{2}x(x^2)dx=2\pi\int_{0}^{2}x^3dx$$
=$$2\pi\left[\frac{x^4}{4}\right]_{0}^{2}=2\pi\cdot\frac{16}{4}=8\pi$$
Region between
$$y=4x , y=x^2$$
rotated around the y-axis.
Height of shell:
height = $$4x-x^2$$
Radius:
r = $$x$$
Intersection:
$$4x=x^2 \longrightarrow x^2-4x=0$$ then $$x(x-4)=0 \longrightarrow x=0 , x=4$$
Integral:
V=$$2\pi\int_{0}^{4}x(4x-x^2)dx$$
The AP exam often gives situations where shell method is much better:
→ washers would require solving for inverse functions.
→ shell radius becomes |3 – x|, still easy.
→ shells avoid rewriting in terms of y.
If rotating around x=k:
radius: |x – k|
height: f(x)−g(x)
Formula:
V=$$2\pi\int_{a}^{b}|x-k|\cdot(f(x)-g(x))dx$$
If the region lies on one side, drop the absolute value.
Use horizontal shells:
radius: |y – k|
height: g(y)−h(y)
V=$$2\pi\int_{a}^{b}|y-k|\cdot(g(y)-h(y))dy$$
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