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Finding the volume of a wine glass when placed on it's side - solids of revolution

Jipsonburger

New member
Aug 5, 2012
2
Apparently when a "snifter" glass is placed on it's side and filled up to the tip, this volume is the optimum amount that should be poured to make a shot.

Hence i have put a glass on an axis and modelled an equation for the top half of the glass...

\(\displaystyle f(x)= -0.00393x^4+0.0843x^3-0.693x^2+2.198x+1.246\)

I know how to find the volume of the whole glass (using solids of revolution) which is 188mL (188cm^3), but i am unable to find a method for working out the volume when its tipped on its side and filled up (i.e) the region between the glass and the line \(\displaystyle g(x)=2.46\).

So in effect my question is:

"How do i work out the volume of the liquid that can be poured into this glass when it is placed on its side, using solids of revolution?

If anyone could shed some light on the issue it would be greatly appreciated =) Thanks in advance =)
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Apparently when a "snifter" glass is placed on it's side and filled up to the tip, this volume is the optimum amount that should be poured to make a shot.
Really, on what basis is this an optimum? .. and how would a bar-person know what volume that was?

Hence i have put a glass on an axis and modelled an equation for the top half of the glass...

\(\displaystyle f(x)= -0.00393x^4+0.0843x^3-0.693x^2+2.198x+1.246\)

I know how to find the volume of the whole glass (using solids of revolution) which is 188mL (188cm^3), but i am unable to find a method for working out the volume when its tipped on its side and filled up (i.e) the region between the glass and the line \(\displaystyle g(x)=2.46\).

So in effect my question is:

"How do i work out the volume of the liquid that can be poured into this glass when it is placed on its side, using solids of revolution?

If anyone could shed some light on the issue it would be greatly appreciated =) Thanks in advance =)
When finding a volume like this you use the contour of the interior of the glass not the exterior.

CB
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Apparently when a "snifter" glass is placed on it's side and filled up to the tip, this volume is the optimum amount that should be poured to make a shot.

Hence i have put a glass on an axis and modelled an equation for the top half of the glass...

\(\displaystyle f(x)= -0.00393x^4+0.0843x^3-0.693x^2+2.198x+1.246\)

I know how to find the volume of the whole glass (using solids of revolution) which is 188mL (188cm^3), but i am unable to find a method for working out the volume when its tipped on its side and filled up (i.e) the region between the glass and the line \(\displaystyle g(x)=2.46\).

So in effect my question is:

"How do i work out the volume of the liquid that can be poured into this glass when it is placed on its side, using solids of revolution?

If anyone could shed some light on the issue it would be greatly appreciated =) Thanks in advance =)
Let the internal contour be \(y=f(x)\), then the volume above \(y=2.46\) is:

\[V= \int_{x_{min}}^{x_{max}} A(x) dx\]

where \(A(x)\) is the area of a segment of a circle of radius \(r(x)=f(x)\), and semi angle \(\theta(x)=\arccos(2.46/y(x))\)

\[A(x)=\frac{1}{2} (r(x))^2 (2\theta(x)-\sin(2\theta(x)) )\]

CB
 
Last edited:

Jipsonburger

New member
Aug 5, 2012
2
Let the internal contour be \(y=f(x)\), then the volume above \(y=2.46\) is:

\[V= \int_{x_{min}}^{x_{max}} A(x) dx\]

where \(A(x)\) is the area of a segment of a circle of radius \(r(x)=f(x)\), and semi angle \(\theta(x)=\arccos(2.46/y(x))\)

\[A(x)=\frac{1}{2} (r(x))^2 (2\theta(x)-\sin(2\theta(x)) )\]

CB
Thanks so much =)

Just 1 more quick question, what is "y(x)" that you refer to?
 

CaptainBlack

Well-known member
Jan 26, 2012
890