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Homework Help: Surface areas

  1. Jan 22, 2004 #1
    I know that the equation for the surface area of any solid of revolution around, say, the x-axis is
    [tex] SA = 2\pi\int_{a}^{b} y\sqrt{1 + (\frac{\,dy}{\,dx})^2} \,dx [/tex]

    What I need is the same formula except in parametric terms, like if the problem was given in terms of x(t) and y(t). Any takers?
  2. jcsd
  3. Jan 22, 2004 #2
    If it revolves about the x-axis on the closed interval [a,b], then
    [tex]SA = 2\pi\int_{a}^{b} y(t)\sqrt{[x'(t)]^2 + [y'(t)]^2} \,dt [/tex]

    For example. if the surface are of the solid generated by revolving the region enclosed by the curve with parametric equations x(t), y(t) from t = 0 to t = pi/2, then the upper limit, b = pi/2, lower limit a = 0.

    If it revolves about the y-axis on the closed interval [a,b], then
    [tex]SA = 2\pi\int_{a}^{b} x(t)\sqrt{[x'(t)]^2 + [y'(t)]^2} \,dt [/tex]
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