**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 2π*r · h* and *volume* about 2π*r · h* · Δ *x* . This is the expression that we will treat in order to sum up the volumes of the cylinders.

**Examples**

**Example 1**

Use the method of cylindrical shells to calculate the volume of the solid enclosed when the curve *y = x*^{2} , 1 ≤ *x* ≤ 3, is rotated about the *y* -axis.

**Solution 1**

As usual, we think of the region under *y = x*^{2} and above the *x* -axis as composed of vertical segments or strips. The segment at position *x* has height *x*^{2}. Thus, in this instance, *h* = *x*^{2}, *r* = *x* , and the volume of the cylinder is 2π*x · x*^{2} · 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 ).

**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.

**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.

**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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