Center of Mass of a Cardioid

[forestmine]

Homework Statement

Find the center of mass of a solid of density [itex]\delta[/itex] = 1 enclosed by the spherical coordinate surface [itex]\rho[/itex] = 1-cos[itex]\phi[/itex].

Homework Equations

The Attempt at a Solution

I'm a bit confused about how to start here, mainly because the surface is defined by spherical coordinates. But here's what I've done so far..

I figured I should start by finding the mass, so, setting up a triple integral over the density, and in this case, it would be a triple integral in spherical coordinates.

$$\int_{0}^{2\pi}\int_{\pi/4}^{\pi}\int_{0}^{1-cos\phi} \rho^2sin\phi d\rho d\phi dθ$$

I'm not sure about the phi limits of integration...the reason I'm thinking beginning at \pi/4 is really just an assumption made by looking at the image. I'm not sure how to mathematically reach that conclusion, however.

I understand how to calculate the center of mass in cartesian coordinates, but the whole spherical surface part is throwing me off. Can I just convert my above integral into a cartesian coordinate system and then just go about finding x-bar, y-bar, and z-bar?

Ah, thank you!

 

[LCKurtz]

Homework Statement

Find the center of mass of a solid of density [itex]\delta[/itex] = 1 enclosed by the spherical coordinate surface [itex]\rho[/itex] = 1-cos[itex]\phi[/itex].

What are the ranges of the variables ##\theta,\phi\, ##?

I'm a bit confused about how to start here, mainly because the surface is defined by spherical coordinates. But here's what I've done so far..

I figured I should start by finding the mass, so, setting up a triple integral over the density, and in this case, it would be a triple integral in spherical coordinates.

$$\int_{0}^{2\pi}\int_{\pi/4}^{\pi}\int_{0}^{1-cos\phi} \rho^2sin\phi d\rho d\phi dθ$$

I'm not sure about the phi limits of integration...the reason I'm thinking beginning at \pi/4 is really just an assumption made by looking at the image. I'm not sure how to mathematically reach that conclusion, however.

Those limits don't define a closed surface for any kind of volume integral.

I understand how to calculate the center of mass in cartesian coordinates, but the whole spherical surface part is throwing me off. Can I just convert my above integral into a cartesian coordinate system and then just go about finding x-bar, y-bar, and z-bar?

No, that is the last thing you would want to do. Once you figure out the correct limits of the solid you will know the limits for the triple integral. Then if you want the ##x## moment, you just put the ##x## moment arm in the integrand like you would do in rectangular. It's just that you do it in spherical coordinates ##x=\rho\sin\phi\cos\theta##. Same idea for the other moments.