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Math Help - Differential Area of a Circular Hoop

  1. #1
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    Differential Area of a Circular Hoop

    Hello,

    OK. So suppose you have a circular hoop of differential thickness dx that has a radius r. I have been told in multiple courses that the differential area of such a hoop is dA=2πr*dx. It is hard for me to see why this is.

    Can somebody please explain how this derivation comes about? A proof or explanation would be appreciated.
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  2. #2
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    Re: Differential Area of a Circular Hoop

    Quote Originally Posted by thekonstant View Post
    Hello,

    OK. So suppose you have a circular hoop of differential thickness dx that has a radius r. I have been told in multiple courses that the differential area of such a hoop is dA=2πr*dx. It is hard for me to see why this is.

    Can somebody please explain how this derivation comes about? A proof or explanation would be appreciated.
    There are two ways to look at this. If you were to "cut" the hoop and straighten it out, it would be a rectangle of width dx and length equal to the circumference of the circle, 2\pi r. The area is, then, 2\pi r dx.

    Or: think of the hoop as the area between two circles, one of radius r and the other of radius r+ dx. The larger circle has area \pi (r+ dx)^2= \pi(r^2+ 2rdx+ dx^2) and the smaller area \pi r^2. The difference is \pi (2rdx+ dx^2) but since dx is a "differential", we ignore its square leaving just 2\pi r dx.
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  3. #3
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    Re: Differential Area of a Circular Hoop

    Fantastic Response. I tended to think of the hoop the second way you mentioned, but couldn't quite get past the squared differential.

    Thanks for clearing things up.
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