If we consider a small strip width dxdx, it will have radius y(x)y(x) as it is revolved about the x axis, and thus circuference 2 pi y2πy.
The arc length dsds of the tip of strip dxdx is:
ds = sqrt ( 1 + (y')^2) dx
With y' = x(1 + 2 ln x)#
and so the surface area of the element is
dS = 2 pi y ds
= 2 pi x^2 ln x sqrt ( 1 + ( x(1 + 2 ln x))^2) dx
For x in [1,3], the surface area S is therefore:
S = 2 pi int_1^3 \ x^2 ln x sqrt ( 1 + ( x(1 + 2 ln x))^2) dx
However because y < 0 for x in [1,3], which would generate a negative radius, we need to be sure to place a negative number on the integration.
The surface area in total is therefore
S = 2 pi ( int_1^3 \ x^2 ln x sqrt ( 1 + ( x(1 + 2 ln x))^2) dx - int_0^1 \ x^2 ln x sqrt ( 1 + ( x(1 + 2 ln x))^2) dx )
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