Find Mass & Center of Mass

[harpazo]

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?

 

[Ackbach]

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.

2. Is k the density function?

Yep!

3. Is the letter a in the radicand constant?

I would assume so.

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 $dx\,dy = r\,dr\,d\theta$.

 

[harpazo]

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.

 

[Ackbach]

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...

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

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

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 $y=\sqrt{a^2-x^2}$.

 

[HallsofIvy]

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].

 

[harpazo]

I will work on this problem and only return to this discussion if needed.