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Can we get a negative value for areas of surfaces of revolution?

  1. Jan 29, 2008 #1
    the question is..

    Find the areas of the surfaces generated by revolvin the curves about the indicated axes.

    x = (1/3)y^(3/2) - y^(1/2), 1≦y≦3; revolved about y-axis

    so i use the general formula "S = Integral 2π (radius)(dS)"
    and the radiu in this case is x which is (1/3)y^(3/2) - y^(1/2) but yea..
    i just keep getting negative value when i subsitute everything in the equation...

    Last edited: Jan 29, 2008
  2. jcsd
  3. Jan 29, 2008 #2


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    Yes, the surface area is given by [itex]2\pi \int x dS[/itex]. Here dS is the differential of arc length, [itex]\sqrt{1+ (dx/dy)^2} dy[/itex]
    Since [itex]x= (1/3)y^(3/2)- y^(1/2)[/itex], [itex]dx/dy= (1/2)y^{1/2}- (1/2)y^{-1/2}[/itex] and [itex](dx/dy)^2= (1/4)(y- 2+ y^{-1})[/itex]. Then [itex](dx/dy)^2+ 1= (1/4)(y+ 2+ y^{-1})= (1/4)(y^{1/2}+ y^{-1/2})^2[/itex] so [itex]dS= \sqrt{(1/4)(y+ 2+ y^{-1})}= (1/2)(y^{1/2}+ y^{-1/2})dy[/itex].

    The surface area is
    [tex]\pi\int_1^3 (y^{1/2}+ y^{-1/2}) dy[/tex]
    That should be easy to integrate and you definitely won't get a negative value!

    To answer you original question, no "area" is never negative.
  4. Jan 29, 2008 #3
    Yes i got to that part too but what happen to the radius (x)?
    cuz after i multiply the radius with the dS and subsitute value 3 and 1 into y-value i got
    Last edited: Jan 29, 2008
  5. Jan 30, 2008 #4


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    You are right- I forgot to multiply by [itex]x= (1/3)y^{3/2}- y^{1/2}[/itex]. That would make the integral
    [tex]\pi \int_1^3 ((1/3)y^2- (2/3)y- 1)dy[/tex]
  6. Jan 30, 2008 #5
    yea..and after i do anti-derivative i got

    π [ (1/9)y^3 - (2/6)y^2 -y ]

    which gives me a -16/9...

    i just dun understand how come my answer is negative..

    btw thanks for helping!
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