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 \(0 \leq t < 2\pi.\)

\[\mathbf{h}(t) = \left(16 \sin^3 t,~ 13 \cos t - 5 \cos 2t - 2 \cos 3t - \cos 4t \right).\]

\[\mathbf{h}(t) = \left(16 \sin^3 t,~ 13 \cos t - 5 \cos 2t - 2 \cos 3t - \cos 4t \right).\]

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,

\[\mathbf{r}_{cm} = \frac{1}{A} \int_{\mathbf{r} \in \mathcal{D}} \mathbf{r} ~dx~dy.\]

(3) The radius of gyration,

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

(4) Comment on the accuracy of your numerical estimates.

(1) The area of \(\mathcal{D}\), given by,

\[A = \int_{\mathbf{r} \in \mathcal{D}} dx~dy.\]

(2) The center of mass,

\[\mathbf{r}_{cm} = \frac{1}{A} \int_{\mathbf{r} \in \mathcal{D}} \mathbf{r} ~dx~dy.\]

(3) The radius of gyration,

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

(4) Comment on the accuracy of your numerical estimates.

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