# Volume and surface area of the sphere using integration ?

volume and surface area of the sphere using integration ??

i was trying to find the surface area of the sphere using integration, ( by revolving circle on the x axis )

the thing is it doesn't work as the volume problem. i mean in volume problem to get the volume of the sphere, you would start with circle and start slice it into little pieces, then you would multiply the area of that slice with the thickness which is dx

but in the surface problem you would multiply the circumference of the slice with the incremental arch Length which is ds

why i cannot in the surface problem multiply the circumference by the thickness dx, and i have to multiply it by the arc length ds

actually i uploaded a pdf file to clarify things

https://www.physicsforums.com/attachment.php?attachmentid=28266&d=1284531453

## Answers and Replies

jgens
Gold Member

You have to remember how the integral is defined. If you use the thinkness (∆x) instead of the arc length (∆s) when you approximate the integral, both of your upper and lower sums are going to be too small, and so they'll converge to a value that is actually less than the surface area of the sphere.

thank you jgens, but i have another question

why does it work for the volume ( i mean taking the thickness dx ) instead of ds

and for the surface area it doesn't work and i have to use ds

and by the way do you mean by saying " upper and lower sums " limits of integration

jgens
Gold Member

This won't make a lot of sense if you're not familiar with upper and lower sums*, but ...

It works for the volume because if you set up your integral using the thickness, the upper sum will always be greater than the value of the volume and the lower sum will always be less than the volume. This means that they'll converge to the exact volume enclosed by the sphere.

If you set up your upper and lower sums using the thickness when you're trying to calculate the surface area, they're both going to be less than the surface area, so they converge to a value which is less than it.

*http://en.wikipedia.org/wiki/Darboux_integral

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i'm sorry jgens, but my knowledge in calculus is only a Calculus BC course

can you elaborate more on the upper and lower sum
i didn't understand this sentence " the upper sum will always be greater than the value of the volume and the lower sum will always be less than the volume "

i'm sorry for bothering

jgens
Gold Member

Don't worry about asking, it's not a problem, especially when you don't understand. Do you know what a supremum or least upper bound is? If you don't, explaining upper and lower sums is going to be tricky.

i'm sorry i have no idea about the supremum or least upper bound

jgens
Gold Member

I'm not the greatest at explaining this stuff, so just work through it slowly and don't expect to understand everything the instant you read it. Think about it first for a little bit ...

Consider the set [0,1]. Clearly 2 is an upper bound for this set because it's greater than or equal to every member of [0,1]. Similarly, 1.5 is also an upper bound for [0,1] as are the numbers 1.25 and 1. However, 1 is considered the least upper bound (supremum) because it is the smallest number which is greater than or equal to every member of [0,1]. Thus, sup{[0,1]} = 1. This notion can be generalized to other sets too though ...

Suppose that the set $A$ is bounded and non-empty. Then, there is clearly some number $M$ such that $a \leq M$ where $a$ is any member of $A$. However, just like the case above, it's possible to find the least such upper bound. This number is denoted by $\sup{A}$.

Does this make any sense?

yeah sure , it does make sense

jgens
Gold Member

Okay, so I'll assume that you can figure out what a greatest lower bound (infimum) of a set is too. So you know, the greatest lower bound is denoted by inf{A}.

Moving on ...

Let $P=\{t_0,\dots,t_n\}$ be a partition of the interval $[a,b]$ such that $a = t_0 < t_1 < \dots < t_{n-1} < t_n = b$. We define the terms $M_i$ and $m_i$ such that

$$M_i = \sup\{f(x): t_{i-1}<x<t_i\}$$

$$m_i = \inf\{f(x): t_{i-1}<x<t_i\}$$

The upper and lower sums of $f$ for the partition $P$ are then given by

$$U(f,P) = \sum_{i=1}^nM_i(t_i-t_{i-1})$$

$$L(f,P) = \sum_{i=1}^nm_i(t_i-t_{i-1})$$

Does this make sense to you?

yeah i think i got them

so what we are trying to do here is to find the sum of each upper and lower bound for a given interval. is that right ?

jgens
Gold Member

Sort of, yeah. We find the least upper bound of $f$ on the interval $[t_{i-1},t_i]$ and then we multiply it by $t_i - t_{i-1}$ (the distance between the points) to get an area. We then take the sum of all of these areas to get an approximation for the area bounded by the curve. In this case, since we're dealing with the least upper bounds on these intervals, the approximate area given by the upper sum is greater than the area bounded by the curve. However, note that as we refine our partition (add more points), the approximate area get closer and closer to the one bounded by the curve.

You can use similar reasoning with the lower sums to see that they give a lower estimate for the area bounded by the curve.

By the way, it's useful to draw pictures if you have a hard time understanding this.

jgens
Gold Member

I need to get some sleep, so I'll finish my explanation now and hope that you understand everything.

So, at this point, we've defined the upper and lower sums of $f$ for $P$. All that we need to do now is figure out how to define the integral in terms of these sums. As it turns out, a function is defined to be integrable on [a,b] if and only if

$$\inf\{U(f,P): P\;\text{a partition of [a,b]}\} = \sup\{L(f,P): P\;\text{a partition of [a,b]}\}$$.

Moreover, this common number (the least upper bound of $L(f,P)$ and greatest lower bound of $U(f,P)$) is called the integral of $f$ on $[a,b]$.

I know this last bit is really rushed, but it's getting late here ...

Now, when we want to calculate the volume enclosed by a sphere, we can approximate the volume using cylinders. By choosing how we set up these cylinders it's possible to get both an upper and lower sum (try drawing this out). The upper sum is always greater than the volume enclosed by the sphere and the lower sum is always less. Therefore, they converge to the desired volume and so the integral you get also gives you the proper volume.

What you're trying to do with the surface area (by using the thickness), is use the surface area for the side of a cylinder to create an upper and lower sum. However, even with the upper sums, the side thickness of a given cylinder is going to be less than the arc length of the curve on the same interval. This means that both of the upper and lower sums will be less than the value of the surface area, and so they'll converge to a value which is too small.

Hopefully this makes some sense to you. If you get stuck, hopefully someone else can come to your aid. Good luck!

Edit: My LaTeX stuff doesn't look like it's showing up right :( Hopefully it looks okay to you.

thank you very much jgens, it really helped me a lot in understanding the situation. i didn't get it 100 % of course !! but at least i have a small idea now

i'll try to read your notes again for more understanding, because this is the first time i expose to this kind of real analysis in math.

" This means that both of the upper and lower sums will be less than the value of the surface area, and so they'll converge to a value which is too small"

yeah that's true because when i carry out the integration using the thickness the surface area tends to be ( pi^2 * R^2 ) which is missing a factor of ( 4/pi ) which is greater than one