Double integration may be applied with the element of summation comprising some notional approximate rectangle (becoming exact in the limit as the length of the sides approaches 0) of sides dx and r d theta where r is the radius of the circle being swept to form the surface, such that, in this case, r = x sqrt(x - 1) and theta is the angle through which the radius is swept. It will be necessary to sweep from theta = 0 to theta = 2 pi, over the interval x = 2 to x = 3.
Denoting the required area by A, retaining r for the moment to illustrate the point, the integral may be set up as
A = int_(theta = 0)^(theta = 2pi) int_(x = 2)^(x = 3) r dx d theta
that is
A = int_(theta = 0)^(theta = 2pi) int_(x = 2)^(x = 3) xsqrt(x-1) dx d theta
Taking the inner integral first,
int_2^3 xsqrt(x-1) dx
this may be solved using the substitution
u(x) = x - 1
so that
(du)/dx = 1
so that
int du = int dx
Noting
u = x - 1
implies
x = 1 + u
so that
xsqrt(x-1) = (1 + u) sqrt(u) = sqrt(u) + u^(3/2)
and, for the limits,
u(2) = 1
and
u(3) = 2
the required (inner) integral is
int_1^2 (u^(1/2) + u^(3/2)) du
which, by the sum rule is
int_1^2 u^(1/2) du + int_1^2 u^(3/2) du
= (2/3)[u^(3/2)]_1^2 + (2/5)[u^(5/2)]_1^2
= (2/3)(2^(3/2)-1) + (2/5)(2^(5/2) -1)
~~ 3.0817 (to 4 decimal places, courtesy of Gnu Octave but defer that collapsing to an approximate value until the outer integral is evaluated ... )
Denoting the first (inner) integral by K (as it evaluates to a constant), the outer integral may now be evaluated
A = int_0^(2pi) K d theta
= K[theta]_0^(2pi)
= 2 pi K
So the overall double integral evaluates to
A = 2 pi ((2/3)(2^(3/2)-1) + (2/5)(2^(5/2) -1))
~~19.363 square units (to be specified from the units of x)
(to 3 decimal places, again courtesy of Gnu Octave)
