Originally posted by: BigPoppa
It is a perfect hemisphere no matter how much water has been removed from the bowl. But, it is not a perfect hemisphere in relation to the original bowl, which is where I believe the problem of my interpretation is.
Wrong. If the height of the water, h, is not equal to the radius of the sphere, r, containing the hemisphere, then the water does not make a hemisphere. You can think of this many ways, but here are a few simple examples:
When h = 2*r, then the water makes an entire sphere
When h = r/2, the it could be described as half a pointy tear drop
To solve this problem, the most simple approach is the method of washers.:
Step 1: Find the radius of the circle describing the top of the water (rd) at an arbitrary height (h)
Draw a circle and the vertical diameter.
Draw a horizontal line lower than the center of the circle, and define it as height h from the bottom of the circle
Draw a a radius from the center of the shpere to where the horizontal line intersects the circle.
This makes a right triangle, by the sides rd, r-h, and the hypotenuse r.
Therefore rd^2 = r^2 - (r-h)^2 = 2*r*h - h^2
Step 2: Find the area of the circle describing the top of the water (A)
A = pi * rd^2 = 2*r*h*pi - h^2*pi
Step 3: Find the volume of the water (V)
V = integral ( A*dh ) from 0 to h
V = pi*r*h^2-1/3*h^3*pi
Step 4: Verify some known solutions:
V(h) = pi*r*h^2-1/3*h^3*pi
V(2r) = pi*r*(2r)^2-1/3*(2r)^3*pi = 4/3*pi*r^3
(correct)
V(r) = pi*r*(r)^2-1/3*(r)^3*pi = 2/3*pi*r^3
(correct)
V(0) = pi*r*(0)^2-1/3*(0)^3*pi = 0
(correct)
Crypticburn
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