Question

1) What is the arc length of C? 2) The surface of revolution on the x axis?

c: x= cost + ln(csct-cot t), y=sint (`pi/6` `<=` t `<=` `pi/3` )

Answer 1

Arc length: `log(sqrt3)`
Surface area of revolution: `pi(sqrt(3)-1)`
Bonus...
Volume of solid of revolution: `pi/24[3sqrt(3)-1]`

Explanation:

Arc length

The arc length of a curve given parametrically is
`s=int_(t_1)^(t_2) sqrt((dx/dt)^2+(dy/dt)^2)` dt
Derivation here: http://tutorial.math.lamar.edu/Classes/CalcII/ParaArcLength.aspx

In our case, `x=cost+ln(csct-cott)`, `y=sint`, so `dx/dt=-sint+ (-csctcott+csc^2t)/(csct-cott)` and `dy/dt=cost`.
Use the quotient rule and the definitions of `csc` and `cot` in terms of `sin` and `cos` to deduce that `d/dt csct=-csctcott` and `d/dtcott=-csc^2t`.

The expression for `dx/dt` simplifies substantially:
`dx/dt=-sint+ (csc^2t-csctcott)/(csct-cott)`
`dx/dt=-sint+csct(csct-cott)/(csct-cott)`
`dx/dt=-sint+csct`

So arc length `s=int_(pi/6)^(pi/3)sqrt((-sint+csct)^2+(cost)^2)dt`
Multiply out within the square root:
`s=int_(pi/6)^(pi/3)sqrt(sin^2t-2+csc^2t+cos^2t)dt`
Recall the identity `sin^2t+cos^2t=1`
`s=int_(pi/6)^(pi/3)sqrt(-1+csc^2t)dt`
Recall also the similar identity `cot^2t=-1+csc^2t`, obtained by dividing through the above identity by `sin^2t`
`s=int_(pi/6)^(pi/3)sqrt(cot^2t)dt`
`s=int_(pi/6)^(pi/3)cottdt`

To integrate `cot`, recall its definition as `cott=cost/sint` and notice that the numerator is the derivative of the denominator. So
`int cott dt=log|sint|+C`, the natural logarithm. Thus

`s=[log|sint|]_(pi/6)^(pi/3)`

Note that evaluation of integrals of a logarithm is only valid when both limits are on the same half of the logarithm curve, i.e. if the point that the log blows up to minus infinity at is not between the two points. Happily, this is the case here; both limits are the same side of the vertical asymptote at `x=0`.

`s=log(sqrt3/2)-log(1/2)`
Recall that by definition `log(A/B)=logA-logB`
`s=log((sqrt3/2)/(1/2))`
`s=log(sqrt3)`

This is the simplest form that this answer can take. It's nice, isn't it? We started out with such an obscure and messy function and gradually simplified it down to this expression. Someone somewhere invested a lot of effort in designing this question!

Surface of revolution

I'm going to assume that you are asking for the calculation of the surface area of the surface of revolution here, which is of a piece with the first part of the question. We already know from the question how to specify the surface.

The surface area for a rotation about the `x`-axis of a parametric curve is given by
`S=int_(t_0)^(t_1)2piysqrt((dx/dt)^2+(dy/dt)^2)dt`
Derivation here: http://tutorial.math.lamar.edu/Classes/CalcII/ParaSurfaceArea.aspx
Note that this is a pretty similar formula to the arc length formula we evaluated above; we already know that
`sqrt((dx/dt)^2+(dy/dt)^2)=cott`, so
`S=int_(pi/6)^(pi/3)2pisintcottdt`
Cancel factor of `sint`
`S=2piint_(pi/6)^(pi/3)costdt`
`S=2pi[sint]_(pi/6)^(pi/3)`
`S=2pi[sqrt3/2-1/2]`
`S=pi(sqrt(3)-1)`

Volume of revolution

Extra bonus answer! You didn't ask this, but it's the other linked quantity, and often is asked in such questions - the volume enclosed by the surface of revolution, the solid of revolution. It probably isn't asked for here as the question is designed to have you perform the elegant reduction `sqrt((dx/dt)^2+(dy/dt)^2)=cott` that we did above, and the volume formula is slightly different in that respect.

The volume of a solid of revolution about the `x`-axis is
`V=int_(t_0)^(t_1)piy^2dx/dtdt`
So
`V=piint_(pi/6)^(pi/3)sin^2t(-sint+csct)dt`
`V=piint_(pi/6)^(pi/3)-sin^3t+sintdt`
Recall again the identity `sin^2t+cos^2t=1`
`V=piint_(pi/6)^(pi/3)-sint(1-cos^2t)+sintdt`
Multiply out and cancel
`V=piint_(pi/6)^(pi/3)-sint+sintcos^2t+sintdt`
`V=piint_(pi/6)^(pi/3)sintcos^2tdt`

Notice that this integrand is of the form produced by a chain rule differentiation: `d/dt f(g(t))=g'(t)f'(g(t))`. In this case `g(t)=cost` and `f(g)=g^3`. We need a minus sign also.

`V=pi[-1/3cos^3t]_(pi/6)^(pi/3)`
`V=pi[-1/3(1/2)^3-(-1/3(sqrt3/2)^3)]`
`V=pi[-1/3*1/8+1/3*(3sqrt(3))/8]`
`V=pi/24[3sqrt(3)-1]`