Consider the small element width dx as shown, being revolved around the line x = -1
The cross sectional area (csa) of the washer of width dx is the csa of the outer circle minus the csa of the inner
ie #dA = pi (1 + x + dx)^2 - pi (1 + x)^2#
simplify the algebra with #u = 1+ x# so we have
#dA = pi ((u+dx)^2 - u^2) = pi (2u dx + dx^2)#
#= pi (2 (1+x) dx + dx^2) #
we can already see that #(dA)/dx|_{dx to 0} = pi (2 (1+x) color{red}{+ dx}) #
so we can ignore #mathcal(O)(dx^2)# in the original expression
thus
#dA = 2 pi (1+x) dx #
the volume of that small element is
#dV = (y_2 - y_1) dA#
# = (y_2 - y_1) * 2 pi (1+x) dx #
# = 2 pi (1+x) (sqrt x -x^2) dx #
#\implies V = 2 pi int_0^1 dx qquad (1+x) (sqrt x -x^2) #
#= 2 pi int_0^1 dx qquad (sqrt x -x^2 + x^{3/2} - x^3) #
#= 2 pi (2/3 x^{3/2} -x^3/3 + 2/5x^{5/2} - x^4/4)_0^1 #
#= 2 pi (2/3 - 1/3 +2/5 - 1/4)#
#= 29/30 pi#