# Cylindrical Shell Method

1. Sep 3, 2006

### shaneewert

Using the method of cylindrical shells, find the volume generated by rotating the region the region bounded by the given curves about the specified axis.

y=(x-1)^(1/2), y=0, x=5; about y = 3

Please tell me how to set up the integral! Any help is MUCH appreciated.

So far I have Integral from 0 to 2 of (3-(y^2+1))*y dy I know that isn't right, because I am not getting the right answer! The book says that it is 24pi.

Thank you.

2. Sep 3, 2006

### Galileo

First, draw a picture. Then cut the region into horizontal slabs of height dy (or $\Delta y$ if you prefer). Imagine a slab at height y being rotated around the line y=3. What would be the volume the obtained cylinder?
Next, add the contributions from all cylinders, i.e. integrate.

3. Sep 4, 2006

### Swapnil

I think that using vertical cylindrical shells would be most effective for this problem.

I would use vertical cylindrical shell whose outer radius r2 is $$3+\sqrt{x-1}$$ while its inner radius r1 is 3 and whose height is dx.

Now the infinitesimal volume of any arbitrary shell would be

$$dV = \pi\cdot dx\cdot \big[ (3+\sqrt{x-1})^2 - 3^2) \big]$$

Now since x would have to go from 1 to 5 to cover the required region, we just integrate the above expression from 1 to 5 and we get:

$$V = \pi\cdot\int_0^5 \big[ (3+\sqrt{x-1})^2 - 3^2) \big] dx$$

This should give you the right answer.

Last edited: Sep 4, 2006