[lemonoman]

I did it 'lemonoman's way' (or the 'right' way!).

Hahahaha...way to make friends...and enemies...all at the same time  lol

[Will]

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 )

[Borek]

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.

[Borek]

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

[Hunt]

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.

[Borek]

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

[Will]

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

 . 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?

[Hunt]

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.

[Hunt]

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!

[Borek]

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?

[Donaldson Tan]

Imagine the spherical container filled with water up to depth h

The volume of water will be ?pi.r²dh 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 + r² = r_o²
r² = 2r_oh - h²
note: r is the radius of the cross section, h is depth.

volume of water = ?pi.r²dh from 0 to 6.5 = pi ? 2r_oh - h² dh from 0 to 6.5 = pi.( f(6.5)-f(0) ) where f(h) = r_oh² -  h³/3

[Yggdrasil]

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(r²-x²)dx just use trig substitution.  Let x = rsin u.  Then dx = rcos u du, and sqrt(r²-x²) becomes rcos u.  cos² u is easy enough to integrate.

[xiankai]

wow this thread has attracted alot of replies, yet im still stuck in my working 

f (x) = (r² – x²)^1/2

     y = (r² – x²)^1/2

    y² = r² – x²

       x = (r² – y²)^1/2

f (y) = (r² – y²)^1/2

     2                                                                 2
    ?    f(y) dx =  [  (r² – y²)^3/2 / (-2y) (3/2)  ]
-4.5                                                                  -4.5

                     =  [  (4.5² – 2²)^3/2 / (-4)2 (3/2) ] – [  (4.5² – (-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 [  (4.5² – (-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)

[Will]

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 pancake-like slice is apprx.: dV = (pi)y² dx (volume of a cylinder).
Hence the apprx. total volume is the sum of all the slices:
          x=b
V= [sum sign] (pi)y² 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)y² dx
      dx->0     x=a

Hence:
       _b
V= ? (pi)y² 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.

[Donaldson Tan]

volume of water = ?pi.r²dh from 0 to 6.5 = pi ? 2r_oh - h² dh from 0 to 6.5 = pi.( f(6.5)-f(0) ) where f(h) = r_oh² -  h³/3

Isn't this simplest to integrate?