[xiankai]

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.

[Borek]

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

[xiankai]

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

[Borek]

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.

[Yggdrasil]

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 x-axis, 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 x-axis, 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.

[billnotgatez]

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

[lemonoman]

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)

[Hunt]

EDIT :

V = ???dV = [(pSin²(Phi)] dpd(Phi)d?

For an entire sphere centered at (0,0,0) around the z-axis :

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 z-axis.

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 = [(p²Sin(Phi)] dpd(Phi)d? = 1/3 ?? [R³ - ( h³ / Cos³(Phi) ](Sin(Phi)) d(Phi)d?

0 ? Phi ? a

V = - ? 1/3 [ R³Cos(a) + h³ tan²(a)/2 - R³ ] d?

Cos(a) = h/R and tan²a = ( 1 - Cos²a / Cos²a ) = ( R² - h² ) / h²

V = - ? 1/3 [ R²h + h(R² - h²) - R³ ] d?

Integrate w.r.t ? : 0 ? ? ? 2(Pi)

V = 2(Pi)/3 [ R²(R - h) + h( h² - R² ) / 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".

[Borek]

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

Spherical?

[Hunt]

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

[Hunt]

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)R³ = 243 (Pi) / 2

V_small = V = 2(Pi)/3 [ R²(R - h) + h( h² - R² ) / 2 ] = 275 (Pi) / 12 m³

Therefore, V_req = V_T - V_small = ( 1458 - 275 ) Pi / 12 =1183 (Pi) / 12 m³

[Borek]

(121.5-68.75/3)?

[Will]

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 A-level math next month!

[Borek]

Somehow that confirms my result

[Will]

1st way :
...Therefore, V_req = V_T - V_small = ( 243 - 54 ) Pi / 2 =189 (Pi) / 2 m³

2nd way :
...Therefore, V_req = V_{half sphere} + V_rem = 243 (Pi) / 4 + 135 (Pi) / 4 = 378 (Pi) / 4 = 189 (Pi) / 2 m³

So is the volume 98⁷/₁₂ (Pi) m³ (= ¹¹⁸³/₁₂ (Pi) m³) or ¹⁸⁹/₂ (Pi) m³ (= 94¹/₂ (Pi) m³) ?