Surface Area (Surface of Revolution) - Discrepancy

DoubleMike

So this question has been bothering me for a very long time... but only recently have I mustered up courage to register to ask it.

Before though, let me draw a parallel analogy, so you can see where I'm coming from. When you take the volume of a region enclosed by one or more functions revolved about a line, either the method of diks or shells is used.

Let us examine the volume of a sphere. The simplest way to calculate the volume of a sphere of radius R is to use the disk method with limits of integration -R to R, in respect to either variable (due to the symmetry).

So in essence the Riemann sum is the addition of the volume of infinitely many cylinders. These have parallel sides to the axis of the sphere/integration. Though strictly speaking we are adding very small volumes, I'd rather think of it as adding an infinite number of areas. This goes well with my understanding of infinities (as I've entreated in the General Math Forum). If this is flawed however, please feel free to elucidate.

Anyway.... When asked to take the surface area of a sphere, my book goes about it a very strange method indeed. It breaks the sphere up into an infinite number of partial cones, and uses the formula of their lateral surface area. I understand why this works... it makes sense. But why won't this work:

Breaking the sphere up into an infinite number of cylinders, and using the surface area formula for those (translational surface of a circle).

Here I have supplied a picture of what I mean:

http://www.geocities.com/boeclan/sphere.jpg
*(if the link doesn't work, copy and paste into address bar)

That picture goes for both the surface area and volume integrals. In the case of volume, that is the correct cross-section, whereas for surface area it is not.

Why is it that the function's slope is ignored when calculating volume, but for surface area, one must use the circuitous method described by my book?

*edit = typo

HallsofIvy

So what you are saying is that you are dividing the sphere into very thin disks, of height dh and radius "r" where, since the sphere is given by x²+ y²+ z²= R², r= [itex]\sqrt{R^2- y^2}[/itex]. The lateral area of such a cylinder is given by [itex]2\pir dy= 2\pi\sqrt{R^2-y^2}dy.

Integrate [itex]2\pi\int_{-R}^R\sqrt{R^2-y^2}dy[/itex] and see what you get.
(If you are right, it will be 4πr².)

DoubleMike

Is the integral [itex]2\pi\int_{-R}^R\sqrt{R^2-x^2}dx[/itex]?

How would go integrate that without resorting to trigonometric substitution?
Well anyway, I checked the above integral for a random radius and it differs from the actual surface area!

dextercioby

Write it (leave the 2pi for the moment):
[tex] \int_{-R}^{+R} \frac{R^{2}-x^{2}}{\sqrt{R^{2}-x^{2}}} dx [/tex]

and then use the fact that
[tex] \int \frac{du}{\sqrt{1-u^{2}}}=\arcsin u +C [/tex]

and partial integration to get the result...

Daniel.

P.S.Use the fact that the integrand is even and change the lower integration limit correspondingly.

mathwonk

a smarter way to do the surface area of a sphere is by analogy with the method you used for the volume, i.e. by finding the derivative of the volume formula.

i.e. if one grows the volume of a sphere outwards it follows that the derivative of the volume function is the area of a cylindrical slice. Thus the volume itself is the integral of the area formula for this cylindrical slice, the so called "cylindrical shells" method.

On the other hand if one grows the volume of a sphere radially, it follows that the derivative of the volume function, wrt the radius is the area function of the sphere. hence the derivative of volume wrt radius is area for a sphere.

now this does not help compute the volume since we are not given the area, but since we already know the volume it does let us compute area, i.e. the area formula for a sphere, is the derivative of (4/3)pi r^3, wrt r, namely 4pi r^2.


you might like to read up on pappus theorem as well, for another analog of computing area via slices.

DoubleMike

I'm not entirely sure I know what you mean by even integrand. Also, I think trigonometric substitution is simpler in this case, but I keep messing it up:

[itex]2\pi\int_{-R}^R\sqrt{R^2-x^2}dx[/itex]
I constructed a triangle with hypothenuse R and legs x and [itex]\sqrt{R^2-x^2}[/itex]

so [itex]\sin\theta = \frac{\sqrt{R^2-x^2}}{R}[/itex] and [itex]\cos\theta = \frac{x}{R}[/itex]

hence [itex]-R\sin\theta d\theta = dx[/itex] and [itex]R\sin\theta = \sqrt{R^2-x^2}[/itex]

after the appropriate substitutions I get

[itex]2\pi R^2 \int_{0}^\pi \sin^2 \theta d\theta[/itex]
using a power reducing formula
[itex]\pi R^2 \int_{0}^\pi 1+\cos 2\theta[/itex]
[itex]\pi R^2 [\theta + \frac{1}{2}\sin 2\theta]_{0}^\pi[/itex]
[itex]\pi R^2 [\pi + 0][/itex]
[itex]\pi^2 R^2[/itex]

I'm not entirely sure of my math there, I didn't get the correct answer before, and working it out on itex was even worse... I might very well be wrong.

Regardless, integrating on my calculator, the method of disks differs largely from the actual formula derived using the slope and lateral surface of cones. What gives?

Justin Lazear

You're integrating just fine. Your cylinders model is incorrect, though.

Your model doesn't account for the fact that the length element ds of the cylinder must be longer, since it's at an angle. Compare the arclength traced out by the circle in element dx at x = 0 and x = near R. The first is just about straight, but the second is at a large angle, so it also goes down a good bit in the length dx, and therefore it should be longer.

--J

DoubleMike

Yeah I realize that, but why do you have to account for the angle when finding the surface area, whereas for volume you can pretty much ignore it?

I don't understand that, besides... As the differential of x approaches 0, the significance of slope should be negligible.

Justin Lazear

You do have to account for it. That's why when you change to cylindrical coordinates you get an [itex]r d\theta[/itex] term or an [itex]r^2 \sin{\theta} dr d\theta[/itex] term in spherical coordinates.

--J

mathwonk

i am trying to actually answer the question of how to view the calculation of surface area as analogous to that of volume, but i do not seem to be making any impression on the questioner. oh well. cest la vie. people would rather calculate than think.

Justin Lazear

Thinking is hard. Calculating is easy. Or at least that's what's taught in high school.

--J

DoubleMike

I'm not sure whether to be insulted or not... Perhaps a better explanation is in order...

As it stands, I have to yet see a reason why I have to account for the curvature (besides the fact that the numbers don't match up). Yeah I understand that the curvature becomes more and more severe... But this never seemed to be a problem for calculating volume... And even if it is, why is it that [itex]\pi \int_{-R}^R R^2-X^2 dx[/itex] is a valid integral for volume? I don't see any reference to the slope of line tangent to the sphere's surface...

mathwonk

Basic suggestion: when you ask a question, and someone answers it for you, it may be wise to ask for more details when the answer goes over your head.

DoubleMike

Ok, slow down.. Let's make up. I'm sorry I didn't read your post as carefully as I should have.

I understand what you're saying though, I had thought about it before (I fancied it suspicious that the derivative of volume happened to be the surface area.) I haven't done anything with spherical coordinates yet, but I follow the reasoning.

But still, that doesn't explain why breaking the sphere into small cylindrical segments and taking their lateral surface area doesn't work.

I know that the curvature is important, but there is nothing in the volume's Riemann sum that refers to it... So why should surface area be any different?

Justin Lazear

The slope of the lines in the Cartesian coordinate system is always 1! dxdydz is always just a big box! As I said before, the case is not the same for coordinate systems where your dV element is not always just a big (edit: little) box, and in those coordinate systems, you do have to account for curvature.

--J

mathwonk

what i am saying is there is no reason why what you suggested should work, since it is not really analogous to the method of calculating volume. in my post i give two methods which ARE analogous. i.e. if one works so should the other.

in your post, you give two rather different methods and ask why one works and the other does not. i ask you to think about why the area computation you suggest should work. i.e. what does it have in common with the shells method of computing volume?

and i apologize for getting my nose in a snit. i tried to edit but too slowly.

ill be back later but right now duke / wake is on.

DoubleMike

Ok, so perhaps a better question to ask it why the volume of sphere doesn't require that argument.

as it stands the Riemann sum is
[itex]\pi\sum (R^2 - x^2) \delta x[/itex] (not sure how to make the triangle on itex)

why not a sum of some function that takes the arguments radius and slope to calculate the volume of infinitely small cone sections?

[itex]\pi\sum f(x, \frac{dy}{dx}) \delta x[/itex]

though it came to my mind that the function for the volume of a cone was derived using disks, so it too would be compromised.

Anyway, the point being that for a sphere, as you sum up the disks, shouldn't the fact that they have parallel sides skew the integral? (Because they don't follow the sphere's curvature.)

I don't know how to express my question!