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This is something I remember as a standard problem given to college math and physics students ... I've been hunting for a model answer online but no luck: everyone is happy to do the cylinder on it's side or a truncated cone or the intersection of two objects with a lot of symmetry in common but not this one for some reason.
I can brute-force it but was wondering if there is an elegant method.
Basic Description
Imagine a glass whose inner surface can be modeled as a truncated cone open at the wide end (eg. a martini glass or a conical beer glass). Fill it with water. Tilt the glass so some water spills ... keep going, until the level of the water reaches the top limb of the bottom surface of the glass.
Q. what is the volume of water remaining?
My reasoning
- this is a truncated cone intersected by an inclined plane:
cone: [itex]x^2+y^2=mz: b<z<t[/itex]
so b is the bottom of the glass and t is the top, and m is the radial gradient of the sides.
If I tilt the glass towards +y about x, but rotate the axis with the tilt, then gravity rotates the other way.
The plane would be: [itex](x,y,ay+c): a=m(t-b)/(t+b), c=t(1-a/m)[/itex];
strategy
A. Take a volume integral between the cone and the plane.
temptation is to slice the volume along z, thicknesses dz.
The cross-section at each z will be a circle radius R(z) cut by a chord at y=r(z):|r|≤R.
This area can be evaluated with a trig substitution so the volume integral becomes:
[tex]\int_b^t \pi R^2 dz - R^2 \cos^{-1} \left ( \frac{r}{R} \right ) -r \sqrt{R^2-r^2}dz[/tex]
Where [itex]R^2=mz[/itex] and [itex]r=\frac{b}{m}-\frac{z-c}{a}[/itex]
(The first term is just the volume of the cone.)
Which looks ... cosy.
So you see why I feel I may have missed a simplification?... hmmm...
B: take the volume under the cone and subtract it from the volume under the plane inside the ellipse (the intersection of the plane and the cone is an ellipse - so it's projection onto a horizontal plane at, say, z=b, would also be something eggy... this doesn't look any easier).
the real challenge
I've been asked for a model answer...
I need to be able to show this to someone who is not very comfy with calculus.
Probably I really want the solution in terms of the depth of the glass and the radii or the top and bottom ... I figured I could plug that in later.
I can brute-force it but was wondering if there is an elegant method.
Basic Description
Imagine a glass whose inner surface can be modeled as a truncated cone open at the wide end (eg. a martini glass or a conical beer glass). Fill it with water. Tilt the glass so some water spills ... keep going, until the level of the water reaches the top limb of the bottom surface of the glass.
Q. what is the volume of water remaining?
My reasoning
- this is a truncated cone intersected by an inclined plane:
cone: [itex]x^2+y^2=mz: b<z<t[/itex]
so b is the bottom of the glass and t is the top, and m is the radial gradient of the sides.
If I tilt the glass towards +y about x, but rotate the axis with the tilt, then gravity rotates the other way.
The plane would be: [itex](x,y,ay+c): a=m(t-b)/(t+b), c=t(1-a/m)[/itex];
strategy
A. Take a volume integral between the cone and the plane.
temptation is to slice the volume along z, thicknesses dz.
The cross-section at each z will be a circle radius R(z) cut by a chord at y=r(z):|r|≤R.
This area can be evaluated with a trig substitution so the volume integral becomes:
[tex]\int_b^t \pi R^2 dz - R^2 \cos^{-1} \left ( \frac{r}{R} \right ) -r \sqrt{R^2-r^2}dz[/tex]
Where [itex]R^2=mz[/itex] and [itex]r=\frac{b}{m}-\frac{z-c}{a}[/itex]
(The first term is just the volume of the cone.)
Which looks ... cosy.
So you see why I feel I may have missed a simplification?... hmmm...
B: take the volume under the cone and subtract it from the volume under the plane inside the ellipse (the intersection of the plane and the cone is an ellipse - so it's projection onto a horizontal plane at, say, z=b, would also be something eggy... this doesn't look any easier).
the real challenge
I've been asked for a model answer...
I need to be able to show this to someone who is not very comfy with calculus.
Probably I really want the solution in terms of the depth of the glass and the radii or the top and bottom ... I figured I could plug that in later.
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Cosy... but the integral is quite doable.