Find Volume of Wine Glass on Side with Solids of Revolution

In summary: Thank you for the clarification, CB. In summary, the conversation discusses the concept of an optimum volume for a "snifter" glass when placed on its side and filled up to the tip. The speaker mentions their attempt to model an equation for the top half of the glass and their knowledge of finding the volume of the whole glass using solids of revolution. They then ask for help in finding the volume between the glass and a given line when the glass is placed on its side. Another person suggests using the internal contour of the glass and provides an equation for finding the volume above a certain height. The speaker asks for clarification on a variable used in the equation.
  • #1
Jipsonburger
2
0
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 modeled 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 =)
 
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  • #2
Jipsonburger said:
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?

Jipsonburger said:
Hence i have put a glass on an axis and modeled 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
 
  • #3
Jipsonburger said:
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 modeled 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
 
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  • #4
CaptainBlack said:
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?
 
  • #5
Jipsonburger said:
Thanks so much =)

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

The value of y corresponding to x, so y(x)=f(x)

CB
 

FAQ: Find Volume of Wine Glass on Side with Solids of Revolution

What is the purpose of finding the volume of a wine glass on its side using solids of revolution?

Finding the volume of a wine glass on its side using solids of revolution allows us to calculate the exact volume of the glass, which is useful for determining the amount of liquid it can hold. This information can be used in the design and production of wine glasses, as well as in the serving and measuring of wine.

What is meant by "solids of revolution" in this context?

Solids of revolution refer to three-dimensional shapes that are created by rotating a two-dimensional shape around an axis. In the case of a wine glass, the two-dimensional shape is the cross-section of the glass when viewed from the side, and the axis is the stem of the glass.

What are the steps involved in finding the volume of a wine glass on its side using solids of revolution?

The first step is to determine the equation of the two-dimensional shape that represents the cross-section of the glass. Then, using calculus, we can find the volume of the solid by integrating the cross-sectional area with respect to the axis of rotation. This involves using the formula for the volume of a solid of revolution, which is π times the integral of the cross-sectional area squared. Finally, we can plug in the values and solve for the volume of the wine glass.

How accurate is this method of finding the volume of a wine glass on its side?

This method is quite accurate, as it takes into account the curvature of the wine glass and provides a more precise measurement compared to simpler methods like using the formula for the volume of a cylinder. However, it is important to note that this method assumes the wine glass has a perfectly symmetrical shape and does not take into account any variations in thickness or irregularities in the glass.

Can this method be applied to other objects besides wine glasses?

Yes, this method can be applied to any object that can be represented by a two-dimensional shape and rotated around an axis. It is commonly used in engineering and manufacturing for objects with complex shapes, such as bottles, vases, and even car parts.

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