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Math Help - This question is so hard!! any help??

  1. #1
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    This question is so hard!! any help??

    Consider a function f(x) where f(x) \succeq 0 for a \preceq x \preceq b.

    Define the 'surface of revolution' to be the surface swept out by a complete revolution of the curve f(x), a \preceq x \preceq b, about the x-axis.

    This surface of revolution can be described using the parameterization r(u,v)=[v, f(v)cos u, f(v)sin u], a\preceq v \preceq b, 0 \preceq u \preceq 2\pi.

    Show that the area of this surface is given by A=2\pi \int_a^b f(v) \sqrt{1+(f'(v))^2}dv where f'(v)=\frac{df}{dv}
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  2. #2
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    Quote Originally Posted by Ash_underpar View Post
    Consider a function f(x) where f(x) \succeq 0 for a \preceq x \preceq b.

    Define the 'surface of revolution' to be the surface swept out by a complete revolution of the curve f(x), a \preceq x \preceq b, about the x-axis.

    This surface of revolution can be described using the parameterization r(u,v)=[v, f(v)cos u, f(v)sin u], a\preceq v \preceq b, 0 \preceq u \preceq 2\pi.

    Show that the area of this surface is given by A=2\pi \int_a^b f(v) \sqrt{1+(f'(v))^2}dv where f'(v)=\frac{df}{dv}
    Hi
    Please consider the following graph


    The curvilinear abscissa ds is
    ds=\sqrt{dv^2+dy^2}

    y=f(v)
    dy=f'(v) dv

    ds=\sqrt{dv^2+f'(v)^2dv^2}
    ds=\sqrt{1+(f'(v))^2}dv

    Rotation of curvilinear abscissa ds around x axis with an angle u leads to an elementary surface
    dA=ds f(v)du
    f(v) du is the equivalent of r d\theta, the radius being f(v) and the angle being u
    dA=f(v)\sqrt{1+(f'(v))^2}dv\;du

    Surface swept out by a complete revolution of the curve f(x) is
    A=\int_0^{2\pi} \int_a^b f(v) \sqrt{1+(f'(v))^2}dv \;du

    A=2\pi \int_a^b f(v) \sqrt{1+(f'(v))^2}dv
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