Find center of mass

Question

Find center of mass for $$z=x^2+y^2$$ under $z=4$ with density ρ(x,y) = 2+x I have no idea how to find the limits for this. I am trying to do this problem by using $\iint rdrdθ$. Any advice, please help.

Answer 1

I'll give a full answer here since my hint in the comment seems to be incomplete still. I'll show how to calculate the moment for the $y$ coordinate of the centroid, and the rest should be similar.

We have that, $$M_x = M_{y=0} = \iiint_v y \rho(x,y,z) dV$$

To integrate over this volume, we identify the base of the surface, find a way to describe the height of the surface (this will be the $z$ direciton) in terms of the parameters that we use to describe the base, and then integrate over the base itself (which I will call $A$).

The height of the surface can be calculated by starting at your surface, which is given by $z = x^2 + y^2$ or $z = r^2$ in cylindrical coordinates, and moving up to $z =4$. Thus we have, $$M_x = \iint_A \int_{r^2}^{4} y\rho(x,y,z) dz dA$$

Now note that in cylindrical coordinates , $y = r\sin(\theta)$. $$\Rightarrow M_x = \iint_A \int_{r^2}^{4} r\sin(\theta) \rho(x,y,z) dz dA$$

Notice that since this is a paraboloid that is restricted by $z \le 4$, the equation of the base (which I will call $A$) is given by just substituting the equation you have for $z$ into $0 \le z \le 4$: $$0 \le x^2 + y^2 \le 4 \Rightarrow 0 \le r^2 \le 4 \Rightarrow 0 \le r \le 2 \text{ (since r is positive)}$$

We've already identified that the base is a circle, so $0 \le \theta \le 2\pi$. Altogether, $$M_x = \int_{0}^{2\pi} \int_{0}^{2} \int_{r^2}^{4} r\sin(\theta) \rho(x,y,z) r dz dr d\theta$$

Answer 2

Your region is a praboliod. Does that help?

You want to do this in cylindrical? Then $x = r \cos \theta, y = r \sin\theta, z = z $ Now plug these fromula in for $x$ and $y$ in to the equation of the paraboliod, and find an equation for $z$ in terms of $r$ and $\theta$ that is one of your limits. $z=4$ is the other limit for $z$. Plug in $4$ for $z$ into your equation for $z$ and find a limit for $r$.

Take a guess what the limit of $\theta$ is going to be.