# In each of the following problem where applicable you will be expected to: a. Draw a sketch of the curve showing its start and stop points and its direction. b. For integrals involving ds (arc length or surface area) find a simplified expression (single fraction with positive exponents) for the radical expression. c. If applicable, draw a box around the simplified integral that needs to be solved. d. For integrals that require one of the methods of chapter 7, find the antiderivative in factored form, before evaluating the integral. e. Solve the integral. 1) Find the surface area if x=k(cos(t)+tsin(t)), y=k(sin(t))-tcos(t)),0 `<=t<=pi,k>0 `is revolved about the x-axis.2) Find the arc lenght of y= ln`(e^x-2)/(e^x+2)`1<=x<=2``

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)`