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

 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

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

Yes.

> ...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
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.

Regards,
Kyle

Date Subject Author
11/26/05 RCA
11/26/05 narasimham
11/26/05 RCA
11/26/05 Jules
11/27/05 RCA
11/26/05 narasimham
11/27/05 Narcoleptic Insomniac
11/27/05 RCA
11/27/05 Narcoleptic Insomniac
11/27/05 RCA
11/27/05 Narcoleptic Insomniac
11/28/05 RCA
11/28/05 Dave Seaman
11/28/05 Narcoleptic Insomniac
11/28/05 RCA