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

In summary: 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
  • #1
harpazo
208
16
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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  • #2
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 $dx\,dy = r\,dr\,d\theta$.
 
  • #3
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.
 
  • #4
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 $y=\sqrt{a^2-x^2}$.
 
  • #5
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 [tex]y= \sqrt{a^2- x^2}[/tex]. If the density is "k" then the mass is given by [tex]\int_{-a}^a\int_0^{\sqrt{a^2- x^2}} k \, dy \, dx[/tex]. Of course, if k is a constant, the mass is just k times the area of a sem-circle with radius a: [tex]k(1/2)\pi a^2[/tex].
 
  • #6
I will work on this problem and only return to this discussion if needed.
 

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

What is mass?

Mass is a measure of the amount of matter in an object. It is typically measured in kilograms (kg) or grams (g).

What is center of mass?

Center of mass is the point at which an object's mass is evenly distributed in all directions. It is also known as the balance point or center of gravity.

How do you find the mass of an object?

The mass of an object can be found by weighing it using a scale or by calculating it using its density and volume. The formula for calculating mass is mass = density x volume.

How is center of mass calculated?

The center of mass can be calculated by finding the average of the x, y, and z coordinates of all the particles that make up the object. Alternatively, it can be found by balancing the object on a pivot point and marking the point where it balances.

Why is finding mass and center of mass important?

Knowing the mass and center of mass of an object is important in many fields of science, such as physics, engineering, and astronomy. It helps us understand an object's stability, motion, and interactions with other objects. It is also crucial in designing structures and vehicles for safety and efficiency.

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