The Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » sci.math.* » sci.math

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: Area and volume of solid of revolution
Replies: 14   Last Post: Nov 28, 2005 11:13 PM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
Narcoleptic Insomniac

Posts: 797
From: Racine, WI
Registered: 3/10/05
Re: Area and volume of solid of revolution
Posted: Nov 27, 2005 6:58 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

On Nov 27, 2005 12:40 PM CT, RCA wrote:

> My question is - why does the area under the curve
> calculation ignore the "arc length" aspect.

Because we do not need to know the arc length of a curve
when finding the area under that curve.

> The arc length is used as the basis for the surface
> area of revolution


> ...while the arc length is approximated to the x
> component while calculating the volume (or the area
> under the curve). I dont understand how that is
> justified.

If by "arc length is approximated to the x component"
you mean integrating the formula

ds = [1 + (f'(x))^2]^(1/2) dx

...then no.

When finding the area under a curve we don't need to
calculate the arc length of this curve. The arc length
does not come into play when doing solids of revolution.

Maybe if we take a specific example things will clear up
a bit. Let's find the volume of a unit sphere and the
surface area of a unit sphere.

A circle of radius one centered at the origin is given
by the equation x^2 + y^2 = 1. The upper half of this
circle is given by y = [1 - x^2]^(1/2) and so we obtain

f(x) = [1 - x^2]^(1/2).

Now if we take the area under this curve and rotate it
about the x-axis we will obtain a sphere of radius one
centered at the origin.

Take a thin slice of this sphere perpendicular to the
xy-plane that goes through the point x = a. This thin
slice is a disk with radius f(a) and so the area of this
disk is pi * [f(a)]^2.

Thus, the volume for our sphere should be the integral
of pi * (1 - x^2) * dx from -1 to 1. Indeed, if we
evaluate this integral we get 4 * pi / 3, which is the
volume of a unit sphere.

It is important to see that we did not use arc length at
all to find this. All we did was take an area under a
curve, rotate this area, and then use calculus to find
our volume.

Now let's find the surface area of this unit sphere. If
we rotate the just the curve f(x) = [1 - x^2]^(1/2) about
the x-axis we will have a hollow sphere. Recall from the
previous post that the arc length of this curve is given
by the integral of

ds = [1 + (f'(x))^2]^(1/2) * dx

...and in this particular case we obtain...

ds = (1 - x^2)^(-1/2) * dx.

Now the infinitesmial area element is given by dS = 2 *
pi * f(x) * ds, and so the surface area is the integral
of dS from -1 to 1. In this case the integral becomes
2 * pi * dx which evaluates to 4 * pi.


Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2018. All Rights Reserved.