Surface area of a sphere by integration
By Anonymous on Saturday, June 9, 2001 - 11:23 am :
Here is how I tried to prove the surface area of a sphere is 4pr2:

Divide a sphere into an infinite number of small cylinders.

Let R= the radius of the sphere

r= the radius of a cylinder

h= the distance of a cylinder from the cylinder with the largest cross-sectional area (i.e. the cylinder in the middle)

By Pythagoras' theorem, r2=R2-h2

so the surface area of each cylinder =2p(R2-h2)1/2

To sum up, Area =ò0R[2p(R2-h2)] dh

However, I got Area
= 1
2
 p2 R2


I am sure I have made some stupid mistakes, but I just can't spot them

Could anyone help?

Thanks!

By Dan Goodman (Dfmg2) on Saturday, June 9, 2001 - 12:28 pm :
The mistake is when you divided it up into infinitely small cylinders. What you should have done is to divide it up into a shell made of sections of a cone. It's a bit tricky to explain this without a picture, so lo and behold:



Sphere diagram

You break the sphere into objects which look like the object on the right, i.e. a cylinder in which one end has a larger radius than the other end. The reason for this is that the approximation you are using ( cylinders) is too inaccurate; draw a picture for when x is close to -R or R and you'll see what I mean.

The area of this object is dA=2py ds where y is the height of the circle at x and
ds=   _______
Ödx2+dy2
 

as in the diagram. But
y=   _____
ÖR2-x2
 

so
dA/dx=2p   _____
ÖR2-x2
 
   _________
Ö1+(dy/dx)2
 

(dividing by dx and substituting for y and ds).

So we calculate dy/dx and it is
-x   _____
ÖR2-x2
 

, so
  _________
Ö1+(dy/dx)2
 
 =R/   _____
ÖR2-x2
 

. So dA/dx=2pR. So A=ò-RR 2pR dx=4pR2.

Is that OK?

By Anonymous on Saturday, June 9, 2001 - 05:38 pm :

Thanks, Dan!


But when you say dA=2py ds, do you assume that you are working with a cylinder?

Also, I can use my method to prove the volume of a sphere is (4/3)pR3, so I can't see why it is inaccurate.

By Dan Goodman (Dfmg2) on Saturday, June 9, 2001 - 06:45 pm :
Not quite, the calculation is a bit complicated. The object on the diagram above is a strip of a cone, you can work out its area precisely using the formula for the surface area of a cone (cone of radius r and height h has surface area
pr   _____
Ör2 +h2
 

not including the base). The radius of one of the rings is y and the radius of the other is y+dy, the horizontal distance between the rings is dx. If you extended the strip of a cone to a complete cone, it would have height (dx/dy)(y+dy) (you can show this using similar triangles). So the area of the strip of a cone is the area of the cone whose base is the ring of radius y+dy and whose height is (dx/dy)(y+dy) minus the area of the cone whose base is the ring of radius y and whose height is (dx/dy)(y+dy)-dx=y.dx/dy. So, the exact area of the object is
p(y+dy)   _____________________
Ö(y+dy)2 +(dx/dy)2 (y+dy)2
 
 -py.   ____________
Öy2+y2(dx/dy)2
 

which after a bit of algebra (taking common factors outside the square roots) gives
dA = 2py.dy   _________
Ö1+(dx/dy)2
 
 +p(dy)2   _________
Ö1+(dx/dy)2
 

. However, when we integrate dA the (dy)2 term will be sufficiently small that we can just drop it, so we just assume that
dA=2py.dy   _________
Ö1+(dx/dy)2
 
 =2py.dx   _________
Ö1+(dy/dx)2
 

.

So that's how we get dA=2py.ds. The reason it doesn't work if you use cylinders rather than strips of cones, is a bit obscure, but perhaps I can explain with an analogy. Suppose you were trying to find the length of the curve y=x from x=0 to x=1. You know that this is Ö2. However, suppose you tried to approximate the curve with horizontal line segments of width dx. If you add up the lengths of all of these horizontal line segments, you will get 1, even though it seems like they are getting closer and closer to the line y=x. When you try and use the same method to work out the area of the curve under y=x though, you get the right answer. Intuitively, it's because approximation by cylinders doesn't "look like" the surface of a sphere, but it does "look like" a solid sphere. I'm not sure of the precise condition that you need for a good approximation or a bad one, but I'll look it up and get back to you, unless someone else posts first.

By Brad Rodgers (P1930) on Saturday, June 9, 2001 - 07:24 pm :
The reason that it isn't entirely accurate is the same reason that adding up rectangles in 2-D drawing isn't accurate to find the length. In fact, when you add up the cylinders like that, what you are actually finding is the length of y (can you see why).

Here's a rigorous way of showing that dA=2py.ds. Start off assuming that we are working with a cone, instead of a cylinder. In the following diagram, Conic area

The bold green area represents ds, we are trying find the area formed by ds when we rotate it around. First off, lets try to find what L is. By similarity, we know that

(R-ds)/y=R/L

so

L=Ry/(R-ds)

Because the area of a cone is pr l

Total Surface Area=pR2 y/(R-ds)=T (1)

Surface Area made by section left of y line = W=p(R-ds)y (2)

We also know that T-W=dA, so subtracting (2) from (1) and simplifying,

dA=py ×ds(-2R+ds)/(-R+ds)

letting ds tend to zero

dA=py ×ds(-2R)/(-R)

=2py ×ds
Giving the wanted result.

Hope this helps,

Brad

By Brad Rodgers (P1930) on Saturday, June 9, 2001 - 07:25 pm :

Oops, hadn't yet seen your post, Dan.

By Dan Goodman (Dfmg2) on Saturday, June 9, 2001 - 09:49 pm :

That's OK, the diagram is quite useful :)

By Dan Goodman (Dfmg2) on Saturday, June 9, 2001 - 10:21 pm :
OK, I've had a little think about this, and I can explain why the approximation by cylinders won't work. In fact, I'll explain it for a curve y=f(x) rather than for the sphere, but the same ideas hold for a sphere as well, but are considerably more complicated to explain.

You probably already know that the length of the curve f(x) between a and b is
òab   ______________
Ö1+f ' (x)2
 
 dx

, and we get this equation in the same way as for the sphere example above.

Have you heard of Taylor series or Maclaurin series? Basically, Taylor's theorem says that if f is a differentiable function, then f(x+h)=f(x)+h.f ' (x)+(h2 /2)f ' ' (x)+(h3 /3!)f ' ' ' (x)+¼ for small values of h. So if h=dx then we can ignore any term dx2 or above if we're integrating it once. So f(x+dx)=f(x)+dx.f ' (x). In other words, we can approximate f(y) for y near x by a linear function (i.e. f(y)=a(y-x)+b). However, if we try and approximate f(y) by a constant near x, then the error term will be f ' (x).dx, so when we integrate the length of our approximation, we will get a non-zero error term, because òf ' (x)dx is nonzero (whereas òf ' (x)dx2 = 0). That's slightly badly explained, but I hope you get the idea (write back if you didn't follow me, tell me which bits you did and didn't understand). So, approximating using horizontal lines is like approximating f(x) by constants, whereas approximating using lines with a gradient is like approximating f(x+dx)=f(x)+dx.f ' (x). For surfaces in 3D, a similar idea applies. We can approximate a surface sufficiently accurately if we use planes which are at a tangent to the surface, but not if we use horizontal planes. It's a bit more complicated in the above example of the sphere, because we're not using planes, but with a bit more work the same idea applies.

Here's why it's not a problem when we're calculating areas rather than lengths. If we approximate f(y) by a constant near x (i.e. f(x+dx)=f(x)) then the error term is (as before) f ' (x).dx. But this time, we multiply by dx because we're calculating areas, width×height=dx×f(x+dx)=dx×(f(x)+dx.f ' (x)+¼)=f(x)dx+ f ' (x).dx2 +¼ = f(x)dx. So in this case, the approximation is good enough.

Let me know if that helps.

By Anonymous on Monday, June 11, 2001 - 02:59 pm :

Yes, that makes sense.
Thank you for your help, Dan and Brad!