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




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 about the xaxis we will obtain a sphere of radius one centered at the origin.
Take a thin slice of this sphere perpendicular to the xyplane 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 xaxis 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



