Center of Mass and Surface Area (multivariable)

In summary: Not exactly. You did the z=sqrt(25-r^2) part. If you do the z=(-sqrt(25-r^2)) part (partial derivatives etc.) you'll get the same integral form.
  • #1
clairez93
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Homework Statement



1. Verify the given moments of inertia and find the center of mass. Assume each lamina has a density of p=1. The problem gives a circle with a radius a.

2. Find the area of the surface of the portion of the sphere x^2 + y^2 + z^2 = 25 inside the cylinder x^2 + y^2 = 9.

Homework Equations





The Attempt at a Solution



1. I already verified the moment, it's the center of mass that's a problem. It gives a picture of a circle with its center at the origin. I would say the center of mass is (0, 0) since its center is at the origin and has a uniform density. Also, I did the math and all and it works out to 0 for x and y as well. The book however says the answer is (a/2, a/2)

2. I wasn't sure about this one. I tried this double integral.

[tex]\int^{2\pi}_{0}\int^{3}_{0}\frac{5}{\sqrt{25 - r^{2}}}r*dr*d\theta[/tex]

(with all the partial derivative and plugging formula and converting to polar steps taken out)
which comes out to 10pi, but the book says the answer is 20pi.
 
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  • #2
I really can't argue with your logic on the first one. But on the second one, I think you are only getting the area for the positive z part. There's also a negative z part.
 
  • #3
Dick said:
I really can't argue with your logic on the first one. But on the second one, I think you are only getting the area for the positive z part. There's also a negative z part.

Does this mean I have to add another double integral with the equation of -5/sqrt 25-r^2 to get the right answer?
 
  • #4
clairez93 said:
Does this mean I have to add another double integral with the equation of -5/sqrt 25-r^2 to get the right answer?

Not exactly. You did the z=sqrt(25-r^2) part. If you do the z=(-sqrt(25-r^2)) part (partial derivatives etc.) you'll get the same integral form.
 

1. What is the center of mass?

The center of mass is a point in an object or system where the mass is evenly distributed in all directions. It is also known as the center of gravity.

2. How is the center of mass calculated?

The center of mass can be calculated by finding the average position of all the mass in an object or system. This is typically done using the formula xcm = (m1x1 + m2x2 + ... + mnxn) / (m1 + m2 + ... + mn), where x represents the position and m represents the mass of each component.

3. What is multivariable calculus?

Multivariable calculus is a branch of calculus that deals with functions of more than one variable. It involves the study of multivariate functions, partial derivatives, multiple integrals, and vector calculus.

4. How is surface area calculated?

Surface area is the measure of the total area that the surface of an object occupies. It can be calculated by finding the sum of the areas of all the individual faces of an object. This can be done using various formulas depending on the shape of the object.

5. How are center of mass and surface area related?

The center of mass and surface area are related in that the center of mass is a point that represents the average position of the mass of an object, while the surface area is a measure of the total area of an object. In some cases, the center of mass can be used to simplify calculations of surface area, such as in the case of a uniform object. However, they are two distinct concepts and one does not directly affect the other.

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