Revolution Volume by Cylinder Shells

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To solve the volume of the region enclosed by the curves y=e^x and y=1/x rotated about the x-axis using the shells method, the upper curve is y=e^x and the lower curve is y=1/x. The height of the cylindrical shell is determined by the difference between these two functions, specifically e^x - 1/x. The radius of the shell is indeed x, leading to a surface area of 2πx(e^x - 1/x). The volume can be calculated by integrating this expression with respect to x over the interval from 1 to 2.
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Rotate about x-axis the region enclosed by y=e^x, y=1/x, x=1 and x=2. I can do the problem with the rings method but I don't how to even set up the integral to solve by the shells method. Help? Thanks
 
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The two curves cross well before x= 1 so you have a region in which y=ex is the upper curve and y= 1/x is the lower. That means that the length of your cylinder is exx- 1/x. The radius is x so you cylinder will have a surface area, for each x, of 2\pi x(e^x- 1/x). Multiply that by the "thickness", dx, to find the differential of volume.
 
Thanks, but since the region enclosed by the boundaries given is rotated about the x-axis, then, doesn't it mean that the radius of the cylinder is y? I'm a little confused. ??
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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