Determining the Surface Area of a Solid of Revolution
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Key Questions

If a surface is obtained by rotating about the xaxis from
#t=a# to#b# the curve of the parametric equation#{(x=x(t)),(y=y(t)):}# ,then its surface area A can be found by
#A=2pi int_a^by(t)sqrt{x'(t)+y'(t)}dt# If the same curve is rotated about the yaxis, then
#A=2pi int_a^b x(t)sqrt{x'(t)+y'(t)}dt#
I hope that this was helpful.

The formula ti evaluate the surface area of a function
#y(x)# rotated around the#x# axis is:# S=2piinty(x)dl # where
#dl=sqrt(dx^2+dy^2)# . If we express all in the#x# variable running in some interval#[a,b]# we have then:# S=int_a^by(x)sqrt(1+(dy/dx)^2)dx # In our case is very easy to verify that
#y(x)=2/x# and the interval on which#x# runs is#[x(1),x(5)]=[2,50]# .So in order to evaluate this surface we have to solve the integral:
# S=int_2^(50)2/xsqrt(1+4/x^4)dx # From now on the solution is quite long, so I will show you just the basic steps:
 Multiply numerator and denominator by
#x^2#
#=> S=int_2^(50)2(sqrt(4+x^4))/x^3=int_2^(50)4(sqrt(1+x^4/4))/x^3#  Perform the change of variable
#x^2/2=sinh(z)#
#=> int_(z_1)^(z_2)(cosh^2z)/(sinh^2z)dz=int_(z_1)^(z_2)(coth^2z)dz# where
#z_1=sinh^1(2)# and#z_2=sinh^1(1250)#  Find the antiderivative of
#coth^2(z)=1+(1/sinhz)^2#  You shold end up whith the expression:
# S=(zcothz)_(z_1)^(z_2) # which finally gives:
# S~=6.498 #  Multiply numerator and denominator by