Chemical Forums
Specialty Chemistry Forums => Other Sciences Question Forum => Topic started by: xiankai on May 05, 2006, 11:02:46 PM

there was a particular question that was asking the volume of water that a sphere with a radius of 4.5 m and filled to a depth of 6.5 m.
while i know that the sphere can be a function of its radius, i have yet to find one for its depth.
i have thought of integrating the surface area of the sphere, but the surface doesnt relate to the depth from what i know.

I don't get what the question is about, try to reword it.

how do i find the rate of change of the volume of a liquid when the depth of the liquid in the sphere increases?

I think I've got it now, but I have no idea how to do it :(
I recall some tricks with integration after changing coordinates to cylindrical or something  but that's about all you will get from me atm.

Consider the function f(x) = sqrt(r^2  x^2)
When you plot y = f(x), r<x<r you get the top half of a circle. When you rotate this curve about the xaxis, you get a sphere.
Now consider an infinitely small slice of the curve with dimensions y, dx. If you rotate this small slice about the xaxis, you get a cylindrical disc with area 2Pi ydx.
Therefore if you integrate 2Pi ydx from r to r you get the volume of a sphere. If you integrate ydx from r to h, where r<h<r, you get the volume of a spherical slice with height h.

Aren’t all points on the surface of a sphere equidistant from the center?

Aren’t all points on the surface of a sphere equidistant from the center?
True. But that's on the surface...and what we want here is the volume...it's what's on the INSIDE that counts ;) hehe
What Yggdrasil means is...(see picture)

EDIT :
V = ???dV = ??? [(pSin^{2}(Phi)] dpd(Phi)d?
For an entire sphere centered at (0,0,0) around the zaxis :
0 ? p ? R
0 ? ? ? 2(Pi)
0 ? Phi ? Pi
I assume you know this already.
For a portion of a sphere, however, Phi & p alone change. You can graph the sphere and see this yourself. Notice : Phi changes from 0 to an angle , call it 'a'.
z = pCos(Phi)
Suppose the sphere is cut at z = h , then h = pCos(Phi). Where h = height of the portion of a sphere with respect to the origin along the zaxis.
p = h / Cos(Phi) , this will make the integration a bit more difficult but will give a precise volume of the sphere.
The Variables become :
0 ? p ? h / Cos(Phi)
0 ? ? ? 2(Pi)
0 ? Phi ? a
V = ??? [(p^{2}Sin(Phi)] dpd(Phi)d? = 1/3 ?? [R^{3}  ( h^{3} / Cos^{3}(Phi) ](Sin(Phi)) d(Phi)d?
0 ? Phi ? a
V =  ? 1/3 [ R^{3}Cos(a) + h^{3} tan^{2}(a)/2  R^{3} ] d?
Cos(a) = h/R and tan^{2}a = ( 1  Cos^{2}a / Cos^{2}a ) = ( R^{2}  h^{2} ) / h^{2}
V =  ? 1/3 [ R^{2}h + h(R^{2}  h^{2})  R^{3} ] d?
Integrate w.r.t ? : 0 ? ? ? 2(Pi)
V = 2(Pi)/3 [ R^{2}(R  h) + h( h^{2}  R^{2} ) / 2 ]
This is the gen eq for the volume of a portion of a sphere cut by a plane perpendicular to z. You can manipulate it the way u want. All u have to do is to plug in the value of "h" and radius "R".

The easiest way is to work with Spherical coordinates, as Borek suggested.
Spherical?

Oh sorry I guess I need some new eye glasses ... yeah I worked with spherical while u recommended cylindrical

there was a particular question that was asking the volume of water that a sphere with a radius of 4.5 m and filled to a depth of 6.5 m
In this case, h = 6.5  R = 2 m
V_{T} = 4/3 (Pi)R^{3} = 243 (Pi) / 2
V_{small} = V = 2(Pi)/3 [ R^{2}(R  h) + h( h^{2}  R^{2} ) / 2 ] = 275 (Pi) / 12 m^{3}
Therefore, V_{req} = V_{T}  V_{small} = ( 1458  275 ) Pi / 12 =1183 (Pi) / 12 m^{3}

???
(121.568.75/3)?

I literally did this problem 2 days ago for homework! (except depth was 8cm and sphere has radius of 10cm!)
I did it 'lemonoman's way' (or the 'right' way!). I am not entirely sure what Vant_Hoff did, for some reason I got a different answer to him probably me making some mistake somewhere in the calculation.
There is no problem with integrating a 'many to many' equation you just integrate with respect to y instead of x. HOWEVER, you could think about it differently because it's a sphere, so in fact you could imagine the empty bit being on the side (as if gravity acted horizontally not vertically!), hence it is possible to integrate with respect to x, as you would do normally.
The rate of change bit is another question. You will need to form a couple of differential equations using the chain rule, but to find the rate of change they need to give you more information about the amount of water being added per unit of time.
I am sure I've made a few mistakes so please correct me as I will be tested on this in my Alevel math next month!

Somehow that confirms my result ;)

1st way :
...Therefore, V_{req} = V_{T}  V_{small} = ( 243  54 ) Pi / 2 =189 (Pi) / 2 m^{3}
2nd way :
...Therefore, V_{req} = V_{half sphere} + V_{rem} = 243 (Pi) / 4 + 135 (Pi) / 4 = 378 (Pi) / 4 = 189 (Pi) / 2 m^{3}
So is the volume 98^{7}/_{12} (Pi) m^{3} (= ^{1183}/_{12} (Pi) m^{3}) or ^{189}/_{2} (Pi) m^{3} (= 94^{1}/_{2} (Pi) m^{3}) ? :\

I did it 'lemonoman's way' (or the 'right' way!).
Hahahaha...way to make friends...and enemies...all at the same time ;D lol

lol ;D, I hope I didn't offend Vant_Hoff I think he has got the right idea of finding the volume of a sphere but when you want to find the volume of part of a sphere, the best way to do it is by integrating, rather than trying to work out a fraction of the sphere. (Also, being Canadian, I am more likely to favour lemonoman over Vant_Hoff ;))

when you want to find the volume of part of a sphere, the best way to do it is by integrating, rather than trying to work out a fraction of the sphere.
The thing is  he integrated, just using different approach. However, looks like there is some error in his calculations and I couldn't find it.

For a portion of a sphere, however, Phi alone changes. You can graph the sphere and see this yourself. Notice : Phi changes from 0 to an angle, call it 'a'.
That's were you have gone astray :) You are calculating not only volume of the portion of the sphere, but also volume of the cone attached to it and going down to the sphere centre.
You owe me ?^{2} snacks ;)

I rechecked my work ... oversimplification only works for symmetrical portions , so I have to change the variables again. Thx To borek for the correction. I'll correct my mistake when I have time.

Go for cylindrical as I told you, you will kill yourself with your own fist in spherical ;)

Go for cylindrical as I told you, you will kill yourself with your own fist in spherical ;)
:D. If his mistake was that he also had the volume of the cone in there couldn't he just minus that volume off the calculated one?

If his mistake was that he also had the volume of the cone in there couldn't he just minus that volume off the calculated one?
I simplified p rom 0 to R, while the real value ( h/cos(Phi) should be taken into account. My simplification works for specific cases, I shall correct my mistake in a while. Hold on a bit plz ...
Go for cylindrical as I told you, you will kill yourself with your own fist in spherical
I reached the same exp in spherical, cylindircal , and rectangular coordinates. there r also lots of others ways, but working with cylindrical when handling a sphere is not practical, and in rect coord, the integral would involve lots of squared roots.

I did it 'lemonoman's way' (or the 'right' way!). I am not entirely sure what Vant_Hoff did, for some reason I got a different answer to him probably me making some mistake somewhere in the calculation.
I used spherical coordinates instead of rectangular. Now sure you you went straight into numerical app and solved it . Try finding a general expression using rect coord, u'll see how much easier it is to apply spherical coordinates in this case. This is something known just as when you work with a cylinder , u use cylindrical coord.
lol , I hope I didn't offend Vant_Hoff I think he has got the right idea of finding the volume of a sphere but when you want to find the volume of part of a sphere, the best way to do it is by integrating, rather than trying to work out a fraction of the sphere. (Also, being Canadian, I am more likely to favour lemonoman over Vant_Hoff
No problem mate, I'm not offended at all :)
I use a proportional const only if I must and only when I'm sure of it. Ofcourse, in a general sense, it does not work.
That's were you have gone astray You are calculating not only volume of the portion of the sphere, but also volume of the cone attached to it and going down to the sphere centre.
I've edited my reply and corrected the variable p. When I thought about this problem, I assumed that the distance between the plane that cuts the sphere, say at z=h, and the center can be neglected, but since you've pointed out yesterday that your ans is different, I recheked it today with some app in the book. It turned out the difference can sometimes be large ( my ans was about 296 instead 309 ) , and so I plugged in the correct variable h/Cos(Phi).
You owe me ?2 snacks
If you say so , but still u didn't come up with any correction man!

You owe me ?2 snacks
If you say so , but still u didn't come up with any correction man!
I have posted correct answer and I have found your error. What other kind of correction do you expect?

Imagine the spherical container filled with water up to depth h
The volume of water will be ?pi.r^{2}dh where r is the radius of each cross section of the water body
let r_{o} = 4.5m
If you cut the sphere into half, you can see the height of the water follows the function (h,r) of a circle of centre (r_{o},0), ie.
(h  r_{o})^{2} + r^{2} = r_{o}^{2}
r^{2} = 2r_{o}h  h^{2}
note: r is the radius of the cross section, h is depth.
volume of water = ?pi.r^{2}dh from 0 to 6.5 = pi ? 2r_{o}h  h^{2} dh from 0 to 6.5 = pi.( f(6.5)f(0) ) where f(h) = r_{o}h^{2}  h^{3}/3

I reached the same exp in spherical, cylindircal , and rectangular coordinates. there r also lots of others ways, but working with cylindrical when handling a sphere is not practical, and in rect coord, the integral would involve lots of squared roots.
The integral is not difficult to evaluate in rectangular coordiates. To integrate sqrt(r^{2}x^{2})dx just use trig substitution. Let x = rsin u. Then dx = rcos u du, and sqrt(r^{2}x^{2}) becomes rcos u. cos^{2} u is easy enough to integrate.

wow this thread has attracted alot of replies, yet im still stuck in my working :'(
f (x) = (r^{2} – x^{2})^{1/2}
y = (r^{2} – x^{2})^{1/2}
y^{2} = r^{2} – x^{2}
x = (r^{2} – y^{2})^{1/2}
f (y) = (r^{2} – y^{2})^{1/2}
2 2
? f(y) dx = [ (r^{2} – y^{2})^{3/2} / (2y) (3/2) ]
4.5 4.5
= [ (4.5^{2} – 2^{2})^{3/2} / (4)2 (3/2) ] – [ (4.5^{2} – (4.5) ^{2})^{3/2} / (4.5)^{2} (3/2) ]
well like lemonoman said, i ran into difficulties integrating from 4.5 to 2, so i tried will17's "y as the subject" method, without results. the second term [ (4.5^{2} – (4.5) ^{2})^{3/2} / (4.5)^{2} (3/2) ] = 0
will17, why did u square the function? i dont know any integration method that involves this... can u tell me?
and once im done with the basics i can look at other methods of integration like cylindrical, rectangular, spherical, ... (voice drones off) ;D

The method I used involved integrating the circle which has been rotated by 2(pi) (360^{o}) about the x (or y in this case) axis. When you do this you multiply the function by pi and square it. This is the reason:
You split the rotated function (which you want to find the volume of between x=a and x=b) into tiny little pieces of width dx (which will be the 'height' of the cylinder) and radius y so the volume of this pancakelike slice is apprx.: dV = (pi)y^{2} dx (volume of a cylinder).
Hence the apprx. total volume is the sum of all the slices:
x=b
V= [sum sign] (pi)y^{2} dx
x=a
The more you slice up the 3D shape the more accurate the value of V becomes. The exact value of V is given by the limit as dx > 0:
x=b
V= lim [sum sign] (pi)y^{2} dx
dx>0 x=a
Hence:
_{b}
V= ? (pi)y^{2} dx
^{a}
and you can take the constant, pi, out of the integral sign.
Whether you integrate it with respect to y or x shouldn't make any difference (because its a circle).
Hope this helps. ;)

volume of water = ?pi.r^{2}dh from 0 to 6.5 = pi ? 2r_{o}h  h^{2} dh from 0 to 6.5 = pi.( f(6.5)f(0) ) where f(h) = r_{o}h^{2}  h^{3}/3
Isn't this simplest to integrate?

I assume everyone has done integration of a quarter of a unit circle to get ¼ pi. Why not do something similar with a sphere?

ok so far i've learnt that the general equation of a circle is x^{2} + y^{2} = r^{2},
and that integration of a sphere involves squaring the function and multiplying it by pi.
didnt know that in the first place :o
(h  r_{o})^{2}
umm... why do u subtract the radius of the sphere from the depth... isnt that akin to finding the upper half of the sphere?
while your method works; i've tried it already, i dont understand the above part :(
PS: snacks for everyone by the way, cheers :D

1. cut the sphere into 2 and u see a circle.
2. put that circle on a graph paper, such that the yaxis is the depth of water, and the xaxis is the radius of the cross section.
3. you end up with a circle of origin (0,r_{o})
4. the general formula of a circle is (x  a)^{2} + (y  b)^{2} = R^{2} where (a,b) is the centre and R is the radius of the circle.
5. Hence, I use the formula (h  r_{o})^{2} + r^{2} = r_{o}^{2}

umm... why do u subtract the radius of the sphere from the depth... isnt that akin to finding the upper half of the sphere?
See the graph above and see for yourself what I am actually doing.
the function of the volume of water as a function of height is thus:
volume of water = pi.f(h) = pi (r_{o}h^{2}  h^{3}/3 )
Cheers for the scooby snacks btw.

ok i understand everything now... thanks guys!

ok i still have some questions;
i realised that if we looked at the area of a circle as a line, we can then treat it as a diameter; finding the area of the new circle with this new diameter should give the volume of the sphere from what i visualise.
however, the main catch is that instead of ?(A/2)^{2}, the method outlined above involves ?(A)^{2}.
can anyone explain this discrepancy? :)