Surface area of solid of revolution with undefined derivative

[Swallow]

Hello there.
Suppose I have a function:
y=3x[tex]^{2/3}[/tex]-1

I want to find the surface area of the solid formed when the part of the curve between x=0 and x=8 is revolved about the x-axis.The curve crosses the x-axis at a point (1/3)[tex]^{3/2}[/tex]
The derivative of the function is:
y'=2x[tex]^{-1/3}[/tex] which is undefined at x=0, so can i still find the surface area by using the limits 0 to 8 in the formula for surface area i.e.

S.A.= Integral of( 2*pi* f(x)*[tex]\sqrt{1+f'(x)^{2}}[/tex]) (with limits 0 to 8).

moreover, if I can find the surface area do i have to break the integral at the point at which the function changes form negative to positive?

 

[Calrik]

This probably sounds like a stupid question but given you are talking about a differential which is a sphere's surface why is it undefined at x=0.

Surely it is just dimensionless?

Sorry probably a mindless question between 0 and 8 though isn't it just a lower limit not undefined?

Might help might not to think of how the integral of a circles area or or a spheres volume relates to the derivative of a surface.

 

[Swallow]

No Calrik, the diffrential isn't a sphere's surface, the final result (Obtained after integrating) gives you the surface area of the solid formed when that particular curve is revolved about the x-axis, AND that surface area has a definite value (it certainly isn't undefined).
Edit:
Oh and no problem, the only stupid question is the one that you don't ask. :)

 

[Calrik]

No Calrik, the diffrential isn't a sphere's surface, the final result (Obtained after integrating) gives you the surface area of the solid formed when that particular curve is revolved about the x-axis, AND that surface area has a definite value (it certainly isn't undefined)

Ok sorry, brain fart didn't quite understand what you were driving at.

At least I bumped your thread.

Carry on.

I think by topographical definition if you break the surface the calculus is non linear. That said you can stretch it to from a proof I guess.

Ie if we don't break it it defines the limits of what we can do to find an answer that is non "imaginary" if you see what I mean.

I'm probably not helping here am I, I'll get me coat. :Biggrin:

 

[Swallow]

*BUMP*
Anyone?
Its a pretty straightforward question.

 

[HallsofIvy]

Yes, it is not uncommon to be able to integrate a function across some point where the integrand does not exist- it's an "improper" integral- take the limit as x approaches that point from each side. Here, since the point, x= 0, is one end of the range of integration, you just take the limit as x approaches 0 from above.

 

[SammyS]

Hello there.
Suppose I have a function:
y=3x[tex]^{2/3}[/tex]-1

I want to find the surface area of the solid formed when the part of the curve between x=0 and x=8 is revolved about the x-axis.
The curve crosses the x-axis at a point (1/3)[tex]^{3/2}[/tex]
The derivative of the function is:
y'=2x[tex]^{-1/3}[/tex] which is undefined at x=0, so can i still find the surface area by using the limits 0 to 8 in the formula for surface area i.e.

S.A.= Integral of( 2*pi* f(x)*[tex]\sqrt{1+f'(x)^{2}}[/tex]) (with limits 0 to 8).

moreover, if I can find the surface area do i have to break the integral at the point at which the function changes form negative to positive?

The function in your integral for surface area must be non-negative. One way to handle this is to break up the integral. Also, as you pointed out, [tex]\displaystyle y'=2x^{-1/3}[/tex] is undefined at x=0, making the integral from 0 to 8 an improper integral.

S.A.= [tex]\displaystyle \lim_{a\to 0^+}\int_a^8 \left( 2\pi |f(x)|\sqrt{1+f'(x)^{2}}\right)\,dx[/tex]

[tex]=\lim_{a\to 0^+}\int_a^{(1/3)^{3/2}} \left( -2\pi f(x)\sqrt{1+f'(x)^{2}}\right)\,dx \ \ +\,\int_{(1/3)^{3/2}}^8 \left( 2\pi f(x)\sqrt{1+f'(x)^{2}}\right)\,dx[/tex]
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As an alternative, you could integrate over y rather than x. That would eliminate having an improper integral., although you would still have to break up the integral.

In general:

S.A.= [tex]\displaystyle \int_C 2\pi |y|\ ds[/tex]

If [tex]\displaystyle x=g(y)[/tex], then S.A.= [tex]\displaystyle \int_{y_1}^{y_2} 2\pi|y|\sqrt{1+g'(y)}\ dy[/tex]

For x ≥ 0, and y=3x[tex]^{2/3}[/tex]-1, we have x=[tex]\displaystyle x=g(y)=\left({{y+1}\over{3}}\right)^{3/2}[/tex]

So S.A.= [tex]\displaystyle -\int_{-1}^{0} 2\pi y\sqrt{1+g'(y)}\ dy\ \ +\int_{0}^{11} 2\pi y\sqrt{1+g'(y)}\ dy[/tex]

Find [tex]\displaystyle g'(y)={{d}\over{dy}}g(y),[/tex] then do the integration.

Added in edit:
I think you'll find that this integration is easier than the one over x.