How Is the Center of Mass Calculated for an Astroid?

[skrat]

Homework Statement

Find the center of mass of Astroid $x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$ for $x,y\geq 0$ and $a>0$.

Homework Equations

$x_T=\frac{\int x\left | \dot{\vec{r}}(t) \right |}{\int \left | \dot{\vec{r}}(t) \right |}$

The Attempt at a Solution

$x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$

$(\frac{x}{a})^{\frac{2}{3}}+(\frac{y}{a})^{\frac{2}{3}}=1$

Now is it ok to say that $\frac{x}{a}=\cos^3t$ and $\frac{y}{a}=\sin^3t$ for $t\in \left [ 0,\frac{\pi }{2} \right ]$?

Now $\left | \dot{\vec{r}}(t) \right |=3a \sint \cost$.

Than $x_T=\frac{\int x\left | \dot{\vec{r}}(t) \right |}{\int \left | \dot{\vec{r}}(t) \right |}$ can be written as

$x_T=\frac{\int_{0}^{\frac{\pi }{2}} 3a^2\cos^4t\sint}{\int_{0}^{\frac{\pi }{2}} 3a\cost\sint}=\frac{2}{5}a$

or... is that parameterization wrong?

 

[Dick]

Homework Statement

Find the center of mass of Astroid $x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$ for $x,y\geq 0$ and $a>0$.

Homework Equations

$x_T=\frac{\int x\left | \dot{\vec{r}}(t) \right |}{\int \left | \dot{\vec{r}}(t) \right |}$

The Attempt at a Solution

$x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$

$(\frac{x}{a})^{\frac{2}{3}}+(\frac{y}{a})^{\frac{2}{3}}=1$

Now is it ok to say that $\frac{x}{a}=\cos^3t$ and $\frac{y}{a}=\sin^3t$ for $t\in \left [ 0,\frac{\pi }{2} \right ]$?

Now $\left | \dot{\vec{r}}(t) \right |=3a \sint \cost$.

Than $x_T=\frac{\int x\left | \dot{\vec{r}}(t) \right |}{\int \left | \dot{\vec{r}}(t) \right |}$ can be written as

$x_T=\frac{\int_{0}^{\frac{\pi }{2}} 3a^2\cos^4t\sint}{\int_{0}^{\frac{\pi }{2}} 3a\cost\sint}=\frac{2}{5}a$

or... is that parameterization wrong?

Looks fine to me.