The Method of Cylindrical Shells Help
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.
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.
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
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 ).
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.
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.
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.
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.
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.
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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