Determining the Surface Area of a Solid of Revolution

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M8-9: surface are of surfaces of revolution

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Key Questions

  • If a surface is obtained by rotating about the x-axis 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 y-axis, 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=(z-cothz)|_(z_1)^(z_2) #

    which finally gives:

    # S~=6.498 #

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