Surface area of arcsec(x)

I have a problem that I am currently stuck on :O..
The problem is to find the surface area of arcsec(x) from x= -2 to x = - 2/sqrt(3) revolved around x=1 with respect to x , and y .

I managed to find the surface area of the function with respect to x.

But I am stuck on the y part.

Here's what I have so far,

x=sec(y).

Integral from arcsec(-2) to arcsec(-2/sqrt(3)) of 2pi (???) *sqrt(1+(tan(y) sec(y))^2 dy.
I can't figure out what my ??? part is,

since we revolve arcsec(x) to x =1 , the distance from x=1 to the function is easy to find in terms of x (1-x),

I have a feeling this might turn out to be multi integrals added together, but I need some clue to solve this problem.

Thanks for your time.

y = arcsec(x) from (-2,π/3) to (-2/√3, 5π/6)
x = secy = 1/cosy

If the curve is being rotated about the line x=1, the element of area between y and y+dy is
circumf * dy
= 2π(x-1) dy
= 2π(cos(y) - 1) dy

Integrating this gives
2π[sin(y) - y] evaluated from 5π/6 to π/3
= 2π[((√3)/2 - π/3) - (1/2 - 5π/6)]
= 2π[(√3 - 1)/2 + π/6]