Help with solid of revolution volume question

student93
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Homework Statement



The problem is attached in this post.

Homework Equations



The problem is attached in this post.

The Attempt at a Solution



I used washer method and set my outer radius as 2+2+√(x-1) and my inner radius as 2. I set my upper limit as 5 and my lower limit as 2.

I set my integral as π∫((4+√(x-1))^2-(4) dx, from 2 to 5 = 485π/6 which is incorrect, the actual answer is 157π/6.
 

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student93 said:
I used washer method and set my outer radius as 2+2+√(x-1)

Hi student93!

Why do you set the outer radius to 2+2+√(x-1)? Outer radius is simply 2+√(x-1), do you see why?
 
Could you please explain why that's the value of the outer radius? Also is the value of my inner radius correct?
 
student93 said:
Could you please explain why that's the value of the outer radius?
Draw a graph and select a disk, it becomes clear why the outer radius is that.
Also is the value of my inner radius correct?
Yes. :)
 
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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