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I have a problem understanding the definition of the surface area integral of a revolution . I know that we can obtain its volume using the disc method . The method defines the width of the disc as infinitely small (dx) , it is then pretty obvious that the integral is defined as the area made by each point multiplied by the width , which is dx .I found however by the definition of the surface area integral that the width is not defined as dx . Can somebody please clarify why it doesn't work the same way with surface areas ? ! thanks

Chris Godsil
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doumham
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  • I would say - instead of dx to take $ \text{ds} = \sqrt{1+\text{y'}^2},\text{dx}$ – georg Dec 19 '16 at 15:37
  • thanks for your answer but this is actually my question, why would we take ds instead of dx . I have understood it already but I won't mind if someone can clarifies it more . :) – doumham Dec 19 '16 at 16:21
  • Sorry. Apparently I misunderstood question. – georg Dec 19 '16 at 18:35

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