[billnotgatez]

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?

[xiankai]

ok so far i've learnt that the general equation of a circle is x² + y² = r²,

and that integration of a sphere involves squaring the function and multiplying it by pi.

didnt know that in the first place 

(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

[Donaldson Tan]

1. cut the sphere into 2 and u see a circle.

2. put that circle on a graph paper, such that the y-axis is the depth of water, and the x-axis 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 = R² where (a,b) is the centre and R is the radius of the circle.

5. Hence, I use the formula (h - r_o)² + r² = r_o²

[Donaldson Tan]

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_oh² -  h³/3 )

Cheers for the scooby snacks btw.

[xiankai]

ok i understand everything now... thanks guys!

[xiankai]

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?