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# The Method of Cylindrical Shells Help

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## Introduction to The Method of Cylindrical Shells

Our philosophy will now change. When we divide our region up into vertical strips, we will now rotate each strip about the y -axis instead of the x -axis. Thus, instead of generating a disk with each strip, we will now generate a cylinder.

Look at Fig. 8.19. When a strip of height h and thickness Δ x , with distance r from the y -axis, is rotated about the y -axis, the resulting cylinder has surface area 2πr · h and volume about 2πr · h · Δ x . This is the expression that we will treat in order to sum up the volumes of the cylinders.

Fig. 8.19

#### Example 1

Use the method of cylindrical shells to calculate the volume of the solid enclosed when the curve y = x2 , 1 ≤ x ≤ 3, is rotated about the y -axis.

#### Solution 1

As usual, we think of the region under y = x2 and above the x -axis as composed of vertical segments or strips. The segment at position x has height x2. Thus, in this instance, h = x2, r = x , and the volume of the cylinder is 2πx · x2 · Δ x . As a result, the requested volume is

We easily calculate this to equal

#### Example 2

Use the method of cylindrical shells to calculate the volume enclosed when the curve y = x 2 , 0 ≤ x ≤ 3, is rotated about the x -axis ( Fig. 8.20 ).

Fig. 8.20

#### Solution 2

We reverse, in our analysis, the roles of the x - and y -axes. Of course y ranges from 0 to 9. For each position y in that range, there is a segment stretching from to x = 3. Thus it has length . Then the cylinder generated when this segment (thickened to a strip of width Δ y ) is rotated about the x -axis has volume

The aggregate volume is then

You Try It: Use the method of cylindrical shells to calculate the volume enclosed when the region 0 ≤ y ≤ sin x , 0 ≤ x ≤ π/2, is rotated about the y -axis.

## Different Axes

Sometimes it is convenient to rotate a curve about some line other than the coordinate axes. We now provide a couple of examples of that type of problem.

#### Example 1

Use the method of washers to calculate the volume of the solid enclosed when the curve , is rotated about the line y = −1. See Fig. 8.21.

Fig. 8.21

#### Solution 1

The key is to notice that, at position x , the segment to be rotated has height the distance from the point on the curve to the line y = −1. Thus the disk generated has area . The resulting aggregate volume is

You Try It: Calculate the volume inside the surface generated when is rotated about the line y = −3, 1 ≤ x ≤ 4.

#### Example 2

Calculate the volume of the solid enclosed when the area between the curves x = ( y − 2)2 + 1 and x = −( y − 2)2 + 9 is rotated about the line y = −2.

Fig. 8.22

#### Solution 2

Solving the equations simultaneously, we find that the points of intersection are (5, 0) and (5, 4). The region between the two curves is illustrated in Fig. 8.22.

At height y , the horizontal segment that is to be rotated stretches from (( y − 2)2 + 1, y ) to (−( y − 2)2 + 9, y ). Thus the cylindrical shell that is generated has radius y − 2, height 8 − 2( y − 2)2 , and thickness Δ y . It therefore generates the element of volume given by

2 · ( y − 2) · [8 − 2( y − 2)2 ] · Δ y.

The aggregate volume that we seek is therefore

You Try It: Calculate the volume enclosed when the curve y = cos x is rotated about the line y = 4, ≤ x ≤ 3 .

Find practice problems and solutions for these concepts at: Applications of the Integral Practice Test.

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