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LuisRodg

Real Member

Registered: 2007-10-23

Posts: 322

Surface area of a solid of rotation.

Today in my Calculus II class we went over how to find the surface area of a solid of rotation. The professor explained it in terms of:

Which Im not sure I really understand whats going on so Im going to have to do some extra reading etc which is fine. However, before doing surface area we were doing volume which I understood perfectly. You get the solid and you take a slice which is a circle itself so to get the volume you integrate the area function of the cross-section:

So for that formula we are using the area of a circle to find the volume of the solid by integrating. So right now if we are trying to find the surface area, why cant you use the circumference of a circle and integrate that? Such as:

While in the lecture I made the comment during the class and he told me that I cant do it like that and he gave no explanation...

Could anyone help me clarify this issue?

Last edited by LuisRodg (2008-02-04 05:05:40)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Surface area of a solid of rotation.

s is the arc length of the curve. As x increases by dx and y increases by dy, s increases by ds. In the limit as dx → 0, ds becomes more and more like a straight-line increment – i.e. ds → √[(dx)[sup]2[/sup]+(dy)[sup]2[/sup]] (by Pythagoras).

No, you can’t use the circumference of a circle as approximation, because not all curves have the curvature of a circle.

Last edited by JaneFairfax (2008-02-04 12:02:01)

LuisRodg

Real Member

Registered: 2007-10-23

Posts: 322

Re: Surface area of a solid of rotation.

What do you mean that not all curves have the same curvature of a circle? When you rotate a curve about the x-axis (or y) and you take a cross-section of it. It will always be a circle. Wouldnt it?

Theres what I mean. Take whatever curve, rotate it and when you take a cross-section, it will always be a circle for which you can take the circumference. Add up all the circumferences of all the circles and you have the surface area of the solid.

Can you point out where im wrong?

EDIT: Also, pardon my 1337 Paint skillz.

Last edited by LuisRodg (2008-02-04 13:00:12)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Surface area of a solid of rotation.

I am talking about the concept of curvature,

http://en.wikipedia.org/wiki/Curvature

which has nothing to do with the areas of circles.

Last edited by JaneFairfax (2008-02-04 13:19:00)

LuisRodg

Real Member

Registered: 2007-10-23

Posts: 322

Re: Surface area of a solid of rotation.

Ok but I dont seem to understand why I cant use the circumference method. Im just trying to learn.

Lets say you have a region R bounded by the functions y = x^2, x=0, x=2, and y=0. You rotate R around the x-axis to obtain a solid. You need to find the surface area of this solid:

You take a cross-section off the solid which is a circle and the formula for the circumference is

Now take the integral of this:

I put all this so you can see my thought process and tell me what Im doing wrong etc.

Last edited by LuisRodg (2008-02-04 13:26:45)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Surface area of a solid of rotation.

Let’s say you have a region R bounded by y = √(1−x[sup]2[/sup]) (semicircle), x = −1, x = 1, and y = 0. Rotate this around the x-axis. Then the solid generated is obviously a sphere and so its surface area is 4π(1)[sup]2[/sup] = 4π.

However, using your method,

Last edited by JaneFairfax (2008-02-05 00:41:36)

gnits

Member

Registered: 2006-02-27

Posts: 3

Re: Surface area of a solid of rotation.

Hi LuisRodg,

I sympathise with you on this one. The reason is subtle. Here is why you can't use the circumference:

Section 1

If f(x)=k, where k is constant, then the area under the curve y = f(x) from x = a to x = b is rectangular and has the value k(b-a). The idea of integration extends this simple concept by defining a value for the area when f is any continuous function. The situation is typical of many where:

(a) Corresponding to an interval a <= x <= b, we wish to define a value

(say) of some quantity Q (in the usual case Q refers to area).

(b)

depends on the values of a continuous function q, and

if q(x) = k where k is constant.

(c) For intuitive reasons we require that Q be additive, i.e. if the interval a <= x <= b is subdivided by points x1,x2,x3,......,xn,xn+1 where

a = x1 < x2 < x3 < ..... < xn < xn+1 = b

then

or writing

,

(d) For intuitive reasons we require that if

for values of x in the interval

then

where

The particular case where

is the area under the curve y = q(x) from x = a to x = b is the mose familiar one. In all such cases we define:

Section 2

In Section 1 we have described the situation in which the value

of a quantity Q is given by the integral:

In practice the required integrand q(x) is usually found by seeking an approximate value of the form:

for

For example, in the case of the area under the curve y = f(x) the most obvious approximation for the area

of the rth strip is:

(Call this "Equation 1")

and from this approximation f(x) emerges as the likely integrand.

Now if:

then writing

we have

(Call this "Equation 2")

Hence if we write

that is

then letting

we see that this agrees with equation 2 if, and only if (THIS IS THE VITAL POINT TO UNDERSTAND IN RELTAION TO WHY THE CIRCUMFERENCE METHOD IS NOT GOOD ENOUGH)

This shows how good the approximation to

must be; IT IS NOT ENOUGH that the error

shoud tend to zero as

and instead WE REQUIRE THAT

Section 3 - Surface Area

Denote by

the surface area swept over by the curve y = f(x) from x = a to x = b when it revolves through 360 degrees about the x-axis. Slicing up this surface area by planes perpendicular to the x-axis, we divide it into a number of bands. Let

be the length of the arc of the curve (say from points P to Q) so that:

The typical band of surface area is the area swept over by the arc PQ, if this area is denoted by

we have:

It follows that:

This is the formula we are all familiar with.

Now, WHY WILL THE CIRCUMFERENCE METHOD NOT WORK in the above derivation?

Because the approximation:

for

DOES NOT satisfy the criterion of Section 2.

You can check it quite simply for yourself. This is the key point.

With the circumference method will will NOT have

Which we MUST have for the integration to be valid.

Last edited by gnits (2008-02-06 04:29:52)

Hunnie78520

Guest

Re: Surface area of a solid of rotation.

Can anyone tell me how I can find the surface area of a sphere when I substitue pi for 562.3. The sphere's circumfrence is 215.489. This was a question on my test. I was completely clueless. I also didn't have a clculater. HELP ME!

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Surface area of a solid of rotation.

Substitue

(where C is the circumference) into the formula

That should give