How to Find Mass and Center of Mass of a Lamina Bounded by Curves?

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Discussion Overview

The discussion revolves around finding the mass and center of mass of a lamina bounded by the curves defined by the equation \(y = \sqrt{a^2 - x^2}\) and \(y = 0\), with a constant density denoted as \(k\). Participants explore various approaches, including integration limits and the potential use of polar coordinates.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • Participants inquire about determining the inner and outer limits of integration for the mass calculation.
  • Some participants confirm that \(k\) represents the density function.
  • There is an assumption that the variable \(a\) in the equation is a constant.
  • Several participants suggest that converting to polar coordinates may simplify the problem, particularly when dealing with expressions like \(a^2 - x^2\).
  • One participant proposes replacing \(x\) with \(r \cos(\theta)\) in the polar coordinate transformation.
  • Another participant suggests using \(x = a \cos(\theta)\) and derives that \(y = a \sin(\theta)\) from the transformation.
  • There is a discussion about the necessity of graphing the function \(y = \sqrt{a^2 - x^2}\) to understand the variable \(a\), with some expressing confusion over its representation in the problem.
  • One participant describes the shape of the lamina as a semi-circle with a center at (0, 0) and radius \(a\), providing an integral expression for the mass based on this understanding.

Areas of Agreement / Disagreement

Participants generally agree on the shape of the lamina and the representation of the density function, but there are varying opinions on the best approach to solve the problem, particularly regarding the use of polar coordinates and the interpretation of the variable \(a\). The discussion remains unresolved with multiple competing views on the methodology.

Contextual Notes

There are limitations in the discussion regarding the assumptions about the variable \(a\) and the need for graphing tools to visualize the function. The integration limits and the specific steps for conversion to polar coordinates are also not fully resolved.

harpazo
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Find the mass and center of mass of the lamina bounded by the graphs of the equations for the indicated density.

y = sqrt{a^2 - x^2}, y = 0

Rho = k

1. How do I find the inner and outer limits of integration?

2. Is k the density function?

3. Is the letter a in the radicand constant?

4. Is it simpler to convert to polar coordinates for this problem?
 
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Harpazo said:
Find the mass and center of mass of the lamina bounded by the graphs of the equations for the indicated density.

y = sqrt{a^2 - x^2}, y = 0

Rho = k

1. How do I find the inner and outer limits of integration?

Here I would skip to your fourth question.

Harpazo said:
2. Is k the density function?

Yep!

Harpazo said:
3. Is the letter a in the radicand constant?

I would assume so.

Harpazo said:
4. Is it simpler to convert to polar coordinates for this problem?

Absolutely! Whenever I see expressions like $a^2-x^2$ showing up, I'm already thinking polar coordinates. The $y=0$ means you'll need to think carefully about your angle limits. Don't forget that $dx\,dy = r\,dr\,d\theta$.
 
Ackbach said:
Here I would skip to your fourth question.
Yep!
I would assume so.
Absolutely! Whenever I see expressions like $a^2-x^2$ showing up, I'm already thinking polar coordinates. The $y=0$ means you'll need to think carefully about your angle limits. Don't forget that $dx\,dy = r\,dr\,d\theta$.

So, for y = sqrt{ a^2 - x^2}, you are saying for me to replace x with rcosθ, right?

This yields y = sqrt{a^2 - (rcos θ)^2}.

Shouldn't the first step be to graph y = sqrt{a^2 - x^2}? Again, I guess a graphing calculator is needed to graph this y value because the variable "a" makes no sense to me. In the textbook, the authors use actual numbers not variables in the practice problems explained.
 
Harpazo said:
So, for y = sqrt{ a^2 - x^2}, you are saying for me to replace x with rcosθ, right?

I would go with $x = a \cos(\theta)$, actually. Then...

Harpazo said:
This yields y = sqrt{a^2 - (rcos θ)^2}.

becomes $y=a \sin(\theta)$.

Harpazo said:
Shouldn't the first step be to graph y = sqrt{a^2 - x^2}? Again, I guess a graphing calculator is needed to graph this y value because the variable "a" makes no sense to me. In the textbook, the authors use actual numbers not variables in the practice problems explained.

$a$ is the radius of the semi-circle described by $y=\sqrt{a^2-x^2}$.
 
Surely, you realized that this was a semi-circle, above the x-axis, with center at (0, 0) and radius a? This entire semi-circle lies between the vertical lines x= -a and y= -b. And for each x, y lies above the x-axis, y= 0, and below the semi-circle y= \sqrt{a^2- x^2}. If the density is "k" then the mass is given by \int_{-a}^a\int_0^{\sqrt{a^2- x^2}} k \, dy \, dx. Of course, if k is a constant, the mass is just k times the area of a sem-circle with radius a: k(1/2)\pi a^2.
 
I will work on this problem and only return to this discussion if needed.
 

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