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

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SUMMARY

The discussion focuses on calculating the mass and center of mass of a lamina bounded by the curves defined by the equations y = sqrt{a^2 - x^2} and y = 0, with a constant density ρ = k. Participants confirm that k represents the density function and that 'a' is a constant representing the radius of a semicircle. The conversation emphasizes the advantages of converting to polar coordinates for simplification, particularly when dealing with expressions like a^2 - x^2. The mass can be computed using the integral ∫_{-a}^a∫_0^{sqrt{a^2 - x^2}} k dy dx, leading to the conclusion that the mass is k times the area of the semicircle, specifically k(1/2)πa^2.

PREREQUISITES
  • Understanding of polar coordinates and their application in integration.
  • Familiarity with double integrals and their use in calculating area and mass.
  • Knowledge of the equations of curves, specifically semicircles.
  • Basic calculus concepts, including limits of integration and density functions.
NEXT STEPS
  • Study the conversion of Cartesian coordinates to polar coordinates in integration.
  • Learn about calculating mass using double integrals in various coordinate systems.
  • Explore the properties and equations of semicircles and their applications in physics.
  • Investigate the use of graphing calculators for visualizing functions and their intersections.
USEFUL FOR

Students and educators in calculus, particularly those focusing on applications of integration in physics and engineering, as well as anyone interested in understanding the concepts of mass and center of mass in laminae.

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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