## Friday, February 13, 2015

For this year's prelim exam, I set the following numerical integration question, which seems very appropriate to repost on Valentine's Day.

The radius of gyration $R_G$ of an object is the average distance of points from its center-of-mass, $\mathbf{r}_{cm}$. Consider a heart-shaped domain $\mathcal{D} \in \mathbf{R}^{2}$ enclosed by the curve for $0 \leq t < 2\pi.$
$\mathbf{h}(t) = \left(16 \sin^3 t,~ 13 \cos t - 5 \cos 2t - 2 \cos 3t - \cos 4t \right).$

Assuming $\mathbf{r} = (x, y)$ is a point in $\mathbf{R}^2$, numerically estimate the following:

(1) The area of $\mathcal{D}$, given by,
$A = \int_{\mathbf{r} \in \mathcal{D}} dx~dy.$
(2) The center of mass,
$\mathbf{r}_{cm} = \frac{1}{A} \int_{\mathbf{r} \in \mathcal{D}} \mathbf{r} ~dx~dy.$
$R_G^2 = \frac{1}{A} \int_{\mathbf{r} \in \mathcal{D}} (\mathbf{r} - \mathbf{r}_{cm})^2 ~dx~dy,$ where $(\mathbf{r} - \mathbf{r}_{cm})^2$ is the square of the Euclidean distance between $\mathbf{r}$ and the center of mass.