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

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Discussion Overview

The discussion revolves around the calculation of the surface area generated by revolving a specific curve about the y-axis. Participants explore the application of the surface area formula and the implications of obtaining a negative value in their calculations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a curve defined by the equation x = (1/3)y^(3/2) - y^(1/2) and seeks to find the surface area when revolved about the y-axis, using the formula S = Integral 2π (radius)(dS).
  • Another participant confirms the correct form of the surface area formula and provides the expression for dS, indicating that the surface area should not yield a negative value.
  • A participant questions the treatment of the radius (x) in their calculations, expressing confusion over obtaining a negative result after substituting values into the integral.
  • One participant acknowledges an oversight in their calculations regarding the multiplication by the radius and presents a revised integral expression.
  • Another participant reiterates their negative result after performing the anti-derivative, expressing confusion about how a surface area could be negative.

Areas of Agreement / Disagreement

Participants generally agree that the surface area should not be negative, but there is disagreement regarding the calculations leading to the negative result. The discussion remains unresolved as participants attempt to clarify the source of the negative value.

Contextual Notes

There are limitations in the discussion, including potential misunderstandings of the integral setup and the treatment of the radius in the surface area formula. The calculations depend on the correct application of mathematical principles, which have not been fully resolved.

joseph9496
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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...

thanks.
 
Last edited:
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joseph9496 said:
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 then general formula "S = 2π INTEGRAL (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...

thanks.

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.
 
HallsofIvy said:
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!

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
-16/9..
 
Last edited:
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]
 
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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