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Y=x² limit [0,1] rotated about y=x find area of surface of revolution?

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6 years ago

```							If the function y= x2 is rotated about y = x , at some point (x, y) on the parabola, the distance from y=x is equal to(x-y)/ sqrt(2) = (x- x2) / sqrt (2) ; this will be the radius of surface of revolution at (x, y) on parabola.Take a differential surface with this radius and width ds along the surface. Surface area of this cylinder will be equal to:dA = 2 pi R ds = 2 (pi)  [(x- x2) / sqrt (2)] sqrt( 1+(y’)2) dx Hece area of surface of revolution is = sqrt(2) (pi) ∫ (x- x2) sqrt( 1+4x2) dx  (int. limit 0 to 1)= sqrt(2) (pi) [ ∫x sqrt( 1+4x2) dx – ∫ x2 sqrt( 1+4x2) dx (int. limit 0 to 1)Use indefinite integration formula for integration of form x sqrt ( a2+x2) , and x2 (a2+x2), and limit from 0 to 1.1/96 sqrt(4 x^2+1) (-24 x^3+32 x^2-3 x+8)+1/64 sinh^(-1)(2 x)  ( limit 0 to 1 )Approximately equal to .2420243429194321 using calculator.
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6 years ago
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