Given the parametric equations

`x(t) =k*(cos(t)+t*sin(t))`

`y(t) =k*(sin(t)-t*cos(t))`

the graph is attached below for `-pi <=t<=pi` . For the domain `0<=t<=pi` only the upper half of the graph is present (the part above x axis).

To find the surface are of the figure `y = y(x)` rotated about the x axis one need to find the the value of the integral

`S = int_0^pi 2*pi*y(t)*ds`

where `ds =sqrt((dx/dt)^2+ (dy/dt)^2)*dt`

Since

`(dx)/dt =k*(-sin(t)+sin(t)+t*cos(t)) =k*t*cos(t)`

`(dy)/dt = k*(cos(t) -cos(t) +t*sin(t)) =k*t*sin(t)`

the length of arc `ds` is

`ds = k*t*dt`

Thus the surface of the figure obtained by rotation is

`S =int_0^pi 2*pi*(sin(t) -t*cos(t))*k^2*t*dt`

We have to evaluate two integrals

`I_1 = int_0^pi t*sin(t)*dt = -t*cos(t) (0->pi) +int_0^pi cos(t)*dt =`

`=[-tcos(t) +sin(t)](0 -> pi) =-pi*cos(pi)+sin(pi) =pi +1`

`I_2 =int_0^pi t^2*cos(t)*dt = t^2*sin(t) (0->pi) -2int_0^pi(t*sin(t))*dt = pi^2*sin(pi) -0^2*sin(0) -2(pi+1) =-2*pi-2`

Above both integrals `I_1` and `I_2` have been computed integrating by parts.

Hence

`S =2*pi*k^2(I_1 -I_2) =2*pi*k^2(pi+1+2pi+2) =`

`=2*pi*k^2*(3*pi+3)=6*pi*k^2(pi+1) `

Answer: the area of the surface is `S =6*pi*k^2(pi+1)`

Images:

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