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

4 π r^{2}

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, $r^{2} = R^{2} - h^{2}$

so the surface area of each cylinder $= 2 π (R^{2} - h^{2})^{1 / 2}$

To sum up, Area $= \int_{0}^{R} [2 π (R^{2} - h^{2})] dh$

However, I got Area $= \frac{1}{2} π^{2} R^{2}$
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

R

and you'll see what I mean.

The area of this object is $dA = 2 π y ds$ where $y$ is the height of the circle at $x$ and $ds = \sqrt{{dx}^{2} + {dy}^{2}}$ as in the diagram. But $y = \sqrt{R^{2} - x^{2}}$ so $dA / dx = 2 π \sqrt{R^{2} - x^{2}} \sqrt{1 + (dy / dx)^{2}}$ (dividing by $dx$ and substituting for $y$ and $ds$ ).

So we calculate $dy / dx$ and it is $- x \sqrt{R^{2} - x^{2}}$ , so $\sqrt{1 + (dy / dx)^{2}} = R / \sqrt{R^{2} - x^{2}}$ . So $dA / dx = 2 π R$ . So $A = \int_{- R}^{R} 2 π R dx = 4 π R^{2}$ .

Is that OK?

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

Thanks, Dan!

But when you say

dA = 2 π y 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 $(4 / 3) π R^{3}$ , 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

π r \sqrt{r^{2} + h^{2}}

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

π (y + dy) \sqrt{(y + dy)^{2} + (dx / dy)^{2} (y + dy)^{2}} - π y . \sqrt{y^{2} + y^{2} (dx / dy)^{2}}

which after a bit of algebra (taking common factors outside the square roots) gives

dA = 2 π y . dy \sqrt{1 + (dx / dy)^{2}} + π (dy)^{2} \sqrt{1 + (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 = 2 π y . dy \sqrt{1 + (dx / dy)^{2}} = 2 π y . dx \sqrt{1 + (dy / dx)^{2}}

So that's how we get $dA = 2 π y . ds$ . 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 $\sqrt{2}$ . 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 $dA = 2 π y . ds$ . 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

$(R - ds) / y = R / L$

$L = Ry / (R - ds)$

Because the area of a cone is $π r l$

Total Surface Area $= π R^{2} y / (R - ds) = T$ (1)

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

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

$dA = π y \times ds (- 2 R + ds) / (- R + ds)$

letting $ds$ tend to zero

$dA = π y \times ds (- 2 R) / (- R)$

$= 2 π y \times ds$
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 $\int_{a}^{b} \sqrt{1 + f' (x)^{2}} dx$ , 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 $f (x + h) = f (x) + h . f' (x) + (h^{2} / 2) f'' (x) + (h^{3} / 3!) f''' (x) + \dots$ for small values of $h$ . So if $h = dx$ then we can ignore any term ${dx}^{2}$ 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 $\int f' (x) dx$ is nonzero (whereas $\int f' (x) {dx}^{2} = 0$ ). 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, $width \times height = dx \times f (x + dx) = dx \times (f (x) + dx . f' (x) + \dots) = f (x) dx + f' (x) . {dx}^{2} + \dots = f (x) dx$ . 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!