Volume of a Sphere if You Know Surface Area Base
Volume and expanse of the sphere using integration ?
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i was trying to find the surface area of the sphere using integration, ( by revolving circumvolve on the x axis )
the thing is it doesn't piece of work as the volume trouble. i hateful in volume problem to get the volume of the sphere, you would showtime with circle and outset slice information technology into little pieces, so you would multiply the surface area of that slice with the thickness which is dx
but in the surface problem yous 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 information technology by the arc length ds
actually i uploaded a pdf file to clarify things
https://world wide web.physicsforums.com/attachment.php?attachmentid=28266&d=1284531453
Answers and Replies
You have to remember how the integral is defined. If you employ the thinkness (∆x) instead of the arc length (∆s) when you lot approximate the integral, both of your upper and lower sums are going to be too small, and then they'll converge to a value that is actually less than the surface surface area of the sphere.
thank you jgens, but i have some other question
why does it piece of work for the book ( i mean taking the thickness dx ) instead of ds
and for the surface area it doesn't work and i take to use ds
and by the way practice you mean by saying " upper and lower sums " limits of integration
This won't make a lot of sense if you're non familiar with upper and lower sums*, merely ...
It works for the book because if you lot gear up up your integral using the thickness, the upper sum will e'er be greater than the value of the volume and the lower sum will always exist less than the volume. This means that they'll converge to the exact volume enclosed past the sphere.
If you lot prepare your upper and lower sums using the thickness when you're trying to summate the surface surface area, they're both going to be less than the surface area, and so they converge to a value which is less than it.
*http://en.wikipedia.org/wiki/Darboux_integral
i'm sad jgens, but my noesis in calculus is just a Calculus BC course
can you elaborate more than on the upper and lower sum
i didn't empathise this judgement " the upper sum will always be greater than the value of the book and the lower sum will always be less than the volume "
i'yard distressing for bothering
Don't worry about asking, information technology's not a problem, especially when you don't sympathise. Practice you lot know what a supremum or least upper jump is? If yous 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
I'm not the greatest at explaining this stuff, so simply work through it slowly and don't expect to empathize everything the instant you read it. Think nigh it first for a footling bit ...
Consider the set up [0,1]. Clearly two is an upper bound for this set because it'southward greater than or equal to every member of [0,i]. Similarly, 1.five is as well an upper bound for [0,1] as are the numbers ane.25 and 1. However, ane is considered the to the lowest degree upper bound (supremum) considering it is the smallest number which is greater than or equal to every member of [0,ane]. Thus, sup{[0,i]} = i. This notion can be generalized to other sets as well though ...
Suppose that the set [itex]A[/itex] is divisional and non-empty. Then, there is clearly some number [itex]M[/itex] such that [itex]a \leq M[/itex] where [itex]a[/itex] is any member of [itex]A[/itex]. However, just like the case above, it'southward possible to find the least such upper jump. This number is denoted by [itex]\sup{A}[/itex].
Does this make any sense?
yes sure , information technology does make sense
Okay, then I'll assume that yous tin figure out what a greatest lower bound (infimum) of a set up is too. So you know, the greatest lower jump is denoted by inf{A}.
Moving on ...
Let [itex]P=\{t_0,\dots,t_n\}[/itex] be a segmentation of the interval [itex][a,b][/itex] such that [itex]a = t_0 < t_1 < \dots < t_{n-i} < t_n = b[/itex]. Nosotros ascertain the terms [itex]M_i[/itex] and [itex]m_i[/itex] such that
[tex]M_i = \sup\{f(10): t_{i-1}<x<t_i\}[/tex]
[tex]m_i = \inf\{f(x): t_{i-ane}<x<t_i\}[/tex]
The upper and lower sums of [itex]f[/itex] for the partitioning [itex]P[/itex] are then given past
[tex]U(f,P) = \sum_{i=ane}^nM_i(t_i-t_{i-one})[/tex]
[tex]L(f,P) = \sum_{i=ane}^nm_i(t_i-t_{i-ane})[/tex]
Does this make sense to you?
yeah i call up i got them
and then what we are trying to do here is to find the sum of each upper and lower spring for a given interval. is that correct ?
Sort of, aye. We notice the least upper bound of [itex]f[/itex] on the interval [itex][t_{i-1},t_i][/itex] and and then we multiply it by [itex]t_i - t_{i-1}[/itex] (the altitude between the points) to get an area. We then have the sum of all of these areas to get an approximation for the surface area divisional by the curve. In this example, since we're dealing with the least upper bounds on these intervals, the estimate area given by the upper sum is greater than the area bounded past the curve. Nonetheless, note that as we refine our partition (add more points), the approximate area become closer and closer to the one bounded by the curve.
You lot can use similar reasoning with the lower sums to see that they give a lower estimate for the area divisional by the curve.
Past the way, it's useful to draw pictures if you lot have a difficult time agreement this.
I need to go some slumber, so I'll stop my explanation at present and hope that you empathize everything.
So, at this point, we've defined the upper and lower sums of [itex]f[/itex] for [itex]P[/itex]. All that we need to do now is figure out how to define the integral in terms of these sums. Equally information technology turns out, a function is defined to be integrable on [a,b] if and only if
[tex]\inf\{U(f,P): P\;\text{a partition of [a,b]}\} = \sup\{L(f,P): P\;\text{a partition of [a,b]}\}[/tex].
Moreover, this mutual number (the least upper bound of [itex]L(f,P)[/itex] and greatest lower jump of [itex]U(f,P)[/itex]) is called the integral of [itex]f[/itex] on [itex][a,b][/itex].
I know this last scrap is really rushed, but it'southward getting late here ...
At present, when we desire to calculate the volume enclosed past a sphere, we can judge the volume using cylinders. By choosing how we fix these cylinders information technology'south possible to become 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 e'er less. Therefore, they converge to the desired volume and then the integral you get also gives you the proper volume.
What you're trying to exercise with the surface area (past using the thickness), is use the surface expanse for the side of a cylinder to create an upper and lower sum. However, fifty-fifty 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 ways that both of the upper and lower sums volition be less than the value of the area, and then they'll converge to a value which is likewise small.
Hopefully this makes some sense to you. If you get stuck, hopefully someone else can come to your aid. Skilful luck!
Edit: My LaTeX stuff doesn't expect like it'southward showing up correct :( Hopefully it looks okay to y'all.
thank you very much jgens, information technology really helped me a lot in agreement the situation. i didn't get it 100 % of course !! but at least i take a small idea at present
i'll try to read your notes once again for more understanding, because this is the get-go 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 expanse, and so they'll converge to a value which is too minor"
yeah that's truthful because when i conduct out the integration using the thickness the surface surface area tends to exist ( pi^2 * R^2 ) which is missing a factor of ( 4/pi ) which is greater than one
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