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6.4 Volumes by Shells

Last time: method of disk/washer

Stack volumes of thin disks.

A cone shown with horizontal disk slices and a corresponding 2D graph with horizontal rectangles perpendicular to the y-axis.

disk/washer: rectangles perpendicular to axis of revolution

Today: method of cylindrical shells

A cone composed of nested vertical cylindrical shells, labeled as thin shell onion layers.

Stacking infinitely-many shells to form, in this case, a cone.

thin shell (think of onion layers)

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Example

Use shells to find volume of solid obtained by revolving the region bounded by \( y = 2 - x \), \( y = 0 \), \( x = 0 \), revolved around y-axis.

set up rectangle parallel to axis of revolution

2D graph of y=2-x from x=0 to 2, showing a vertical representative rectangle parallel to the y-axis.
3D visualization of the cone being formed by revolving the triangular region, highlighting a single cylindrical shell.

Calculate volume of one shell, then integrate to accumulate all shells

2D diagram showing the radius as the x-distance and height as 2-x for a vertical rectangle of thickness dx.
Diagram of a single cylindrical shell with radius, height equal to rectangle height, and thickness dx.

radius = how far rectangle is from axis of revolution

here, radius = \( x \)

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Volume of a Shell: Unwrapping the Method

To find the volume of a cylindrical shell, we imagine "unwrapping" it into a rectangular slab.

  • Height: \( h = 2 - x \)
  • Thickness: \( dx \)
  • Length: This corresponds to the circumference of the shell.
    \( \text{Length} = \text{circumference} = 2\pi \cdot \text{radius} = 2\pi \cdot x \)
A 3D rectangular slab labeled with height 2-x, thickness dx, and length equal to circumference 2 pi x.

Volume of one shell:

\[ \text{volume} = (2\pi x)(2 - x)(dx) = 2\pi x(2 - x) dx \]

This represents one very thin rectangle/shell. We want to accumulate all such shells.

The integration bounds are from the start at \( x = 0 \) to the end at \( x = 2 \).

Volume of the Resulting Cone

\[ \int_{0}^{2} 2\pi x(2 - x) dx = \dots = \frac{8\pi}{3} \]

Note: \( 0 \) is the left end of region and \( 2 \) is the right end of region. The integrand represents one shell.

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Example: Volume by Shell Method

Consider the region bounded by \( y = e^{-x^2} \), \( x = 0 \), \( y = 0 \), and \( x = 2 \), revolved around the \( y \)-axis.

When using the shell method, we use a representative rectangle parallel to the axis of revolution.

  • The height of the rectangle is given by the function: \( \text{height} = e^{-x^2} \)
  • The thickness is \( dx \).
  • The radius of the shell is \( x \).
Coordinate graph showing the curve y=e^(-x^2) and a vertical differential element dx at distance x.

Unwrapping the shell results in a slab with:

  • \( \text{Length} = 2\pi \cdot \text{radius} = 2\pi \cdot x \)
3D diagram of a bell-shaped solid with an internal cylindrical shell being extracted and unwrapped.

Volume of one shell:

\[ 2\pi x e^{-x^2} dx \]

Volume of Object

Sum all thin shell volumes by integration:

\[ \int_{0}^{2} 2\pi x e^{-x^2} dx \]

Using substitution where \( u = -x^2 \):

\[ \dots = \pi(1 - e^{-4}) \]
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Example

Region bounded by y = x3, x = 1, y = 0 revolved around y = 1.

Coordinate graph showing the region bounded by y=x^3, x=1, and y=0, with axis of revolution y=1.

The region is bounded by the curve \( y = x^3 \), the vertical line \( x = 1 \), and the x-axis \( y = 0 \). The axis of revolution is the horizontal line \( y = 1 \).

3D sketch of the resulting solid of revolution, labeled as hollowed out.

hollowed out

shell: rectangle

parallel to axis of rev

Diagram illustrating the radius (1-y) and height y of a horizontal shell relative to the axis y=1.

radius = distance of rectangle from axis of rev.

\[ \text{radius} = 1 - y \]

thickness = \( dy \)

integrate in terms of \( y \)

NO \( x \) in any part

Diagram showing the horizontal shell height as the difference between x=1 and the curve x=y^(1/3).

"height" = \( 1 - y^{1/3} \)

  • right curve: \( x = 1 \)
  • left curve: \( y = x^3 \) but turned into \( x = y^{1/3} \)
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volume of one shell = \( 2\pi (1-y)(1-y^{1/3}) dy \)

radius
"height"
thickness

Accumulate ALL by integration from \( y = 0 \) to \( y = 1 \)

(bottom of region)(top of region)
\[ \int_{0}^{1} 2\pi (1-y)(1-y^{1/3}) dy = \dots = \boxed{\frac{5\pi}{14}} \]