Center of Mass and Surface Area (multivariable)

[clairez93]

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.

 

[Dick]

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.

 

[clairez93]

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?

 

[Dick]

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.