If R is revolved about the y-axis, find the volume of the solid of revolution (a) by the disk/washer method, and (b) by the shell method. y x y 2x y x3 For problems 3 - 4, let R be the region bounded by the given curves. You can also get some more practice with the washer method here. Draggable points let you control the limits of integration, the axis of revolution, and the position of the line that will become the sample shell. disk/washer method, and (b) by the shell method. Exampleįind the volume obtained by rotating the area bounded by \(y=x\) and \(y= \sqrt\)! Hopefully this has helped you with the washer method, but if there’s still a topic you’d like to learn about take a look at some of my other lessons and problem solutions about integrals. So let’s jump into an example and I’ll explain the difference as we go. Exactly as you would expect from the name, a washer is just a disk with a hole taken out of its center. Now it clearly seems to be a closed region so you ought to be able to convert it to a filled region by first making a boundary and then filling it, but it fails. So if you try to ask for the Volume it will give zero. You can think of the main difference between these two methods being that the washer method deals with a solid with a piece of it taken out. begingroup MichaelE2 Right, but that is just the shell, a 2D region embedded in 3D space. The washer method for finding the volume of a solid is very similar to the disk method with one small added complexity. It is similar to the disk method and washer method because it involves solids of revolution, but the process in.
Once you have the disk method down, the next step would be to find the volume of a solid using the washer method.
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