Volume of the Solid involving Hyperbolic functions

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Homework Statement



The area bounded by y=2 coshx, the x-axis, the y-axis, and the line x=4 is revolved about the x-axis. Find the volume of the solid generated.


Homework Equations



I sliced the area along the axis of revolution. That is the strip is dx. So the equation necessary is pi(R^2-r^2) from the limits of integration (0,4)

The Attempt at a Solution


pi\int((2cosh(x))^2-16) from 0 to 4
I am uncertain if my logic is sound. Any feedback would be appreciated.
 
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I think it would be best to use disks to solve this problem. If I were to use shells the slicing would be awkward. Can anyone give me some pointers?
 
Why not just integrate pi*(2*cosh(x))^2 dx from 0 to 4

Your vertical strip is just between y=2*cosh(x) and y=0. So pi(R^2-0). x=4 is just the boundary condition, your limits are from x=0 to x=4
 
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Thank you. The thought occurred to me that I could just integrate that. And the logic behind that seems sound by my accounts. I just needed some additional feedback.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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