Center of mass of a curved surface

[brkomir]

Homework Statement

Let $\Sigma$ be surface given as $f(x,y)=\frac{1}{2a}(x^2+y^2)$ above circle $K((0,0),a)$ for $a>0$. Calculate the center of mass.

Homework Equations

If you are not familiar with $K((0,0),a)$ , that is a circle in point $(0,0)$ with radius a.

The Attempt at a Solution

I would attempt to get a solution if I would be able to imagine $\Sigma$. If I understand correctly, $f(x,y)$ is also a circle with radius $\sqrt{2a}$ ? or...?

My question here is...: What does the fact, that the $f(x,y)$ is above a circle tell me?

 

[mfb]

I think the surface is meant as 3-dimensional object, so you can write $$z=f(x,y)=\frac{1}{2a}(x^2+y^2)$$ That's a [strike]cone[/strike] paraboloid. The center of mass is trivial for the x- and y-coordinate, and the z-coordinate is interesting.

Edit: fixed, thanks.

 

[LCKurtz]

I think the surface is meant as 3-dimensional object, so you can write $$z=f(x,y)=\frac{1}{2a}(x^2+y^2)$$ That's a cone. The center of mass is trivial for the x- and y-coordinate, and the z-coordinate is interesting.

Actually, it's a paraboloid.

 

[brkomir]

Ok, if the surface is 3-dimensional than i can write $x=rcos\varphi$ and $y=rsin\varphi$ for $r\in \left [ 0,a \right ]$ and $\varphi \in \left [ 0,2\pi \right ]$

Therefore $f(r,\varphi )=\frac{1}{2r}r^2=\frac{r}{2}$

So $z=\frac{\int_{0}^{2\pi }\int_{0}^{a}\frac{r^2}{4}drd\varphi }{\int_{0}^{2\pi }\int_{0}^{a}\frac{r}{2}drd\varphi }=\frac{a}{3}$

Or is this not so easy? :D

 

[mfb]

Why did you replace a by r in the denominator for f?
And I think there are factors of r missing in the integration (r dr dφ).

 

[brkomir]

Blah, I don't know. I shouldn't have done that!

$x=rcos\varphi$ and $y=rsin\varphi$ for $r\in \left [ 0,a \right ]$ and $\varphi \in \left [ 0,2\pi \right ]$

Therefore $f(r,\varphi )=\frac{1}{2a}r^2$

So $z=\frac{\int_{0}^{2\pi }\int_{0}^{a}\frac{r^4}{4a^2}drd\varphi }{\int_{0}^{2\pi }\int_{0}^{a}\frac{r^2}{2a}drd\varphi }=\frac{3}{10}a$