Finding Shell Radius: Why (3)^1/2 - (3-y^2)^1/2?

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Homework Statement


30b1j4i.jpg

This is the main question.So I'm stuck at the b part
2nqcjfp.jpg

It says that the shell radius is (3)^1/2 - (3-y^2)^1/2 but i would only say (3-y^2)^1/2.Why do we say otherwise?

Homework Equations

The Attempt at a Solution


i would say the shell radius is (3-y^2)^1/2 but it's wrong
 
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On my device at least your post is sideways so practically unreadable. I probably have nothing helpful to say on the problem, but possibly helpful is to tell you that people who could help won't if they can't read your post, and if it is difficult in such a careless way can be irritated and not want to take any trouble themselves.
 
elfinitty said:

Homework Statement


[ IMG]http://i62.tinypic.com/30b1j4i.jpg[/PLAIN]
This is the main question.So I'm stuck at the b part
[ IMG]http://i61.tinypic.com/2nqcjfp.jpg[/PLAIN]
It says that the shell radius is (3)^1/2 - (3-y^2)^1/2 but i would only say (3-y^2)^1/2.Why do we say otherwise?

Homework Equations



The Attempt at a Solution


i would say the shell radius is (3-y^2)^1/2 but it's wrong
It was rather difficult to read. Moreover, it would have been helpful for you to have accompanied the images with more text explaining what you were trying to do here.

For using the shell method and revolving about the x-axis:
I would say the radius of the shell is y, the width is ##\displaystyle \ \sqrt{3}-\sqrt{3-y^2} \ ##, and the thickness is dy .
 
Thread closed as posting homework questions as images is not allowed per Physics Forums rules. elfinitty, if you want further help, please re-post in a new thread with the question as readable text. The Physics Forums LaTeX features can help you to format equations.
 
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