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

In summary, the question is asking for the surface area of a curve revolvin about the y-axis. However, the surface area always has a negative value.
  • #1
joseph9496
5
0
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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  • #2
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.
 
  • #3
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:
  • #4
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]
 
  • #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!
 

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

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

Yes, it is possible to get a negative value for areas of surfaces of revolution. This can happen when the function used to generate the surface has negative values or when the surface is located below the x-axis.

2. How does a negative value for areas of surfaces of revolution affect calculations?

A negative value for areas of surfaces of revolution does not affect calculations as long as it is accounted for. It simply indicates that the surface is located below the x-axis, and the absolute value of the area should be considered.

3. Can negative areas of surfaces of revolution have physical meaning?

Yes, negative areas of surfaces of revolution can have physical meaning in certain scenarios. For example, in physics, a negative area can represent the area under a graph that represents work done in the opposite direction of motion.

4. Are there any limitations to getting a negative value for areas of surfaces of revolution?

There are no inherent limitations to getting a negative value for areas of surfaces of revolution. It ultimately depends on the function used and the location of the surface.

5. How can we avoid getting a negative value for areas of surfaces of revolution?

To avoid getting a negative value for areas of surfaces of revolution, the function used to generate the surface should have only positive values. Additionally, the boundaries of the surface should be carefully chosen to ensure that the surface is located above the x-axis.

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