Surface Area of Sphere as a Riemann Sum

1. Jan 7, 2010

ryoma

1. The problem statement, all variables and given/known data
How do you solve the surface area of a sphere using Riemann Sums?

2. Relevant equations

3. The attempt at a solution
I started out with
2 * (lim n->∞ [ (i=1 to n) ∑ [ 2*pi*(√(r^2 - (i/rn)^2))*(r/n) ] ])
where the summation is the surface area of the cylinders (or discs) inside a hemisphere, and simplified to
2 * (lim n->∞ [ (2*pi/n^2) * (i=1 to n)) ∑ [ √((nr^2)^2 - i^2) ] ])

I'm pretty sure I'm doing this right, but I don't know what to do now. Is it possible to use something like a trig substitution like in integrals except with summations? Thank you for any help. Oh, and this isn't for class or anything.

EDIT:

n is the number of cylinders

Last edited: Jan 7, 2010
2. Jan 8, 2010

tiny-tim

Welcome to PF!

Hi ryoma! Welcome to PF!

(have a pi: π and try using the X2 tag just above the Reply box )
(did you mean (i/n)r ?)

No, this is the sum for the volume of a sphere …

you're adding the surface area times thickness of each cylinder, which is its volume.

3. Jan 8, 2010

ryoma

oh, yeah it would be (i/n)r or (ir/n) xP thanks for pointing that out. that would change the summation to
2 * (lim n->∞ (2πr^2 / n^2) * ( (i=1 to n) ∑ √(n^2 - i^2) ) )
sorry for making such a stupid mistake. and wouldn't the volume of a sphere be:
2 * (lim n->∞ (i=1 to n) ∑ π * (r/n) * (r^2 - (i^2)(r^2)/n^2))
because the formula is for the volume of a cylinder would be π(r^2)h and r/n is the height and r is √[ r^2 - (i^2)(r^2)/n^2 ] so r^2 would be (r^2 - (i^2)(r^2)/n^2) ?

4. Jan 9, 2010

tiny-tim

Hi ryoma!

(just got up :zzz: …)

Sorry, but I'm finding it very difficult to read your posts.

Can you please use the X2 tag (just above the Reply box). ?

5. Jan 9, 2010

ryoma

Like this?:

summation for surface area:
$$2 \times \lim_{n\rightarrow\infty} (\frac{2\pi \times r^2}{n^2} \sum_{i=1}^\infty \sqrt{n^2 - i^2})$$

summation for volume:
$$2 \times \lim_{n\rightarrow\infty} (\sum_{i=1}^\infty (\pi\frac{r}{n} \times (r^2 - \frac{i^2 \times r^2}{n^2})))$$

Or like this:
surface area:
2 * (lim n->∞ (2πr2 / n2) * ( (i=1 to n) ∑ √(n2 - i2) ) )

and

volume:
2 * (lim n->∞ (i=1 to n) ∑ π * (r/n) * (r2 - (i2)(r2)/n2))

Last edited: Jan 9, 2010
6. Jan 9, 2010

tiny-tim

(in LaTeX, go back to using ^, and for big brackets type "\left(" and "\right)" )

√(n2 - i2) is h, but apart from that both the surface and the volume look wrong.

See you in the morning. :zzz:

7. Jan 9, 2010

ryoma

I made mistakes with the latex, so I kept on going back and making changes every minute or so, but this summation to express the volume of a sphere, I know is correct because I have tested it and got (4/3) * pi * r2 :

$$2 \times \lim_{n\rightarrow\infty} (\sum_{i=1}^\infty (\frac{r\pi}{n} \times (r^2 - \frac{(ir)^2}{n^2})))$$

oh, and what I got for the summation for the surface area was after some simplifying; what I originally had was:
$$2 \times \lim_{n\rightarrow\infty} (\sum_{i=1}^\infty (2\pi \times \sqrt(r^2 - \frac{(ir)^2}{n^2}) \times (r/n))$$

these are probably gonna be wrong, so I'll edit them until they are correct.

Last edited: Jan 9, 2010
8. Jan 9, 2010

ryoma

I thought h would be r/n? and if the summation for the surface area is wrong, how would it be set up? I used the lateral surface area of the cylinders, and I thought I did it all right...

9. Jan 10, 2010

tiny-tim

(just got up :zzz: …)
oh i see now … you're using discs, not cylinders!

Yes, the thickness of each disc is r/n, and your formula is correct.

(except the ∞ over the ∑ should be an n )
Almost, but you're adding the surface area of the vertical sides of the discs, so you haven't allowed for the slope of the surface of the sphere.

10. Jan 10, 2010

ryoma

wait a second. What's the difference between a disc and a cylinder? until now, I was thinking they were the same thing. and about the surface area, so the discs have to be sort of like the frustum of a cone? Why does it have to be like that, because in the end, wouldn't the lateral surface area of both a frustum and a disc just go to zero?

oh, and how do you have "n" instead of infinity above the summation symbol? when i tried replacing \infty with \n , it just messed it up xP

11. Jan 10, 2010

tiny-tim

ok I agree technically a disc is a cylinder, but it's usual to call a cylinder a disc if the height is extremely small. Because you called it a cylinder, I assume you were slicing the sphere with cylindrical "cookie-cutters" (which actually is quite a sensible way to solve the problem).
Yes, like a frustrum. Try finding the surface area of a cone that way … yes, they'll both go to zero, but one will always be sinθ times the other.
It's not "\n", it's just "n" … and a "\" without a proper function after it just makes LaTeX go crazy, looking for one.

12. Jan 10, 2010

ryoma

this is what I got so far:
lateral surface area of frustum
r1 = small radius
r2 = big radius
h = vertical height of frustum
lateral surface area = $$\pi \times (r_1 + r_2) \times \sqrt{h^2 + (R - r)^2}$$

and then I set up the summation with
(R = Radius of sphere)
r1 = $$\sqrt{R^2 - \frac{(Ri)^2}{n^2}}}$$
r2 = $$\sqrt{R^2 - \frac{R(i - 1)^2}{n^2}}$$
h = R/n
and just plugged in r1 , r2 , and h into the equation for the surface area of a frustum with the summation from i=1 to n. Am I doing it right?

Last edited: Jan 10, 2010
13. Jan 11, 2010

tiny-tim

hmm …

why are you introducing r1 and r2? just use 2πr

and, for the area, you need h (= R/n) divided by the sine or cosine of the angle.

14. Jan 12, 2010

ryoma

Wouldn't you have to use r1 and r2 because the frustums have different length radii at their bases? and for the height, why do you divide by sine or cosine of the angle, and i'm not really understanding what angle it is.

15. Jan 13, 2010

tiny-tim

No, because r2 = r1 + ∆r, where ∆r is very small.

For the surface area of the sphere, you want the surface area of the frustrum, which is the height of the frustrum times the sine of the angle it makes with the horizontal.

(try calculating the surface area of a cone using this method, but without the sine!)

16. Jan 13, 2010

ryoma

but if the radii of the two bases of the frustum are basically the same, wouldn't that mean that it's like the same as calculating the surface area of a cylinder? and i'm still not understanding the angle, like, if the top base was smaller than the bottom base of a frustum, is it the angle that's created from from making a line perpendicular to the bases and going through the edge of the top base?

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