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Surface area of a rotated curve

  1. Nov 28, 2006 #1
    Here's the question:
    Find the area of the surface obtained by rotating the curve

    [​IMG]

    from x=0 to x=9 about the x-axis.

    I'm supposed to parametrize the curve, using rcos(theta) as x and rsin(theta) as y, at least I think I am.

    That would give f(x,y) = 3rcos^3(theta),rsin(theta)

    Then find the partial derivatives with respect to r and theta and find their cross product. Then find the magnitude of the cross product and integrate with limits int[0-2pi] int[0-9].

    Is this right? I can't find any information on the internet to do it this way and the book isn't much help either.

    Thanks.
     
  2. jcsd
  3. Nov 29, 2006 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    No. That is a circle in the xy-plane. Your curve, in the xy-plane, is y= 3x3, not at all a circle. When you "rotate" around the x-axis, every point, with coordinates (x0, y) in the xy-plane, traces out a circle of points in the (x0, y, z) plane.
    Since the radius of the circle is y= 3x^3, that would have parametric equations [itex]x= x_0, y= 3x_0^3cos(\theta ), z= 3x_0^3 sin(\theta)[/itex].

    Using the parametric equations I gave, with x0 and [itex]\theta[/itex] as parameters, that should work. However, it looks awfully complicated to me. The area of the curved surface of a cylinder with radius y and height ds is [itex]2\pi y ds[/itex]. For y= 3x3,
    [tex]ds= \sqrt{1+ (9x^2)^2}dx= \sqrt{1+ 81x^4}dx[/tex].
    The integral for surface area would be given by
    [tex]2\pi \int_0^9(3x^3)\sqrt{1+81x^4}dx[/tex]
     
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