Find the volume of a solid of revolution using the shell method
Today we’ll look at an alternate way to find volumes of a solid of revolution: the shell method. This uses cylindrical shells. As we get more proficient with the shell method, we will compare and contrast it to the disk method to see which is best for a particular problem. http://mathdemos.org/mathdemos/shellmethod/gallery/gallery.html Cake washer Cake shell
Imagine peeling off a shell layer of volume, much like a soup can label. The shape would be a rectangle, with a height of h and a base of the circumference, 2 π r . The thickness of the volume would be w =∆ x or w =∆ y , however things were oriented. Each layer you peeled off would have a radius indicated as p from the axis of the rotation. This radius would change a little each time as you peeled off a new layer. The height might also change as you peeled off new layers. Volume of shell = 2 π (radius)(height)(thickness) For the orientation shown to the left, ∆V = 2 π p(y)h(y)∆y Notice that we are building up parallel to the x-axis (axis of rotation) not perpendicular, as we do with disk method. Summing up these volumes leads to integration: h 2 π r Thickness
Ex 1 p. 468 Using the shell method to find volume (see if you can see some advantages!) Find the volume of the solid of revolution formed by revolving the region bounded by and the x-axis (0 ≤ x ≤ 2) about the y-axis. h=y p=x The radius, p = x. The height h = y(x). We are building a shell layer ∆x thick. Because x ranges from an inside radius of 0 to outside radius of 2, the volume is What were some of the advantages? If we use the shell method, we need a radius in terms of x and a height in terms of x.
Ex 2 p.469 Using the shell method to find volume Find the volume of the solid of revolution formed by revolving the region bounded by the graph of and the y-axis (0 ≤ y ≤ 1) about the x-axis With the shell method, we need to find a radius and a height. The radius p is y, the height is x(y), the thickness is ∆y, and we go from a radius with a low y of 0 to a high y of 1. p=y h = x(y) Notice that we rotated about the x-axis, the representative layer was parallel to the x-axis, and everything inside the integral was about Y!
For the disk method, the representative rectangle is ALWAYS PERPENDICULAR to the axis of rotation. If that allows you to determine radii with the correct variable inside the integration, and you don’t have to split things up a bunch to get the correct volume, then disk method will be great. Here is the secret of when to use the disk method and when to use the shell method! For the shell method, the representative rectangle is ALWAYS PARALLEL to the axis of rotation. If that allows you to determine radii with the correct variable inside the integration, and you don’t have to split things up a bunch to get the correct volume, then shell method will be the one to choose. For some situations you can choose. Others you have only one method that will work.
Let’s turn back to Ex 4 p460 and look at the disk method: Part of this region will revolve and be washers, and part will be disks. So the region must be split up at the level where y=1. I am revolving about y-axis, so y is being sliced up as dy. I will need functions in terms of y for the radii. In the upper part, the inner radius is determined by and the outer radius is still x = 1 X = 1 Ex 3 p.470 Shell method preferable Find the volume of the solid formed by revolving the region bounded by the graphs of y = x 2 +1, y = 0, x = 0, and x = 1 about the y-axis.
Ex 3 p.470 Now let’s look at it with the shell method By way of reminder, we are revolving the region bounded by the graphs of y = x 2 +1, y = 0, x = 0, and x = 1 about the y-axis. We need to identify the radius from axis, p = x. The height would be y(x), and the thickness would be Δ x. Going from a radius with low x of 0 and the highest x of 1, the integral becomes p = x h= y(x) Same answer, only one integral so LESS WORK!
Remember the shells formed by your cake layers. Visualize! 7.3a p. 472/ 1-19 odd
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