I wrote down all the information I could for this problem but I'm having a very difficult time starting it.
Let $D =$ {$(x,y): x^2 + y^2 \le a^2, y\ge 0$} be a semi-circular lamina of radius $a$ above the $x$-axis whose density at any point is $K$ times the distance from the origin. Find the mass and center of mass of $D$.
I was pretty confident in myself until I saw how density was defined. Thanks for any help.
$\endgroup$ 21 Answer
$\begingroup$In Cartesian coordinates the mass and center of mass ($m, \vec s_0$) are found by the integrals: $$\begin{align}m={}&\iint_D \rho(\vec s)~\mathrm d \vec s \\[1ex]={}&\int_{-a}^a\int_0^{\sqrt{a^2-x^2}} K~\sqrt{x^2+y^2}\,\mathrm d x\,\mathrm d y\\[4ex]\vec s_0 ={}& \dfrac 1 m \iint_D \vec s \rho(\vec s)~\mathrm d \vec s\\[1ex]\begin{bmatrix}x_0\\y_0\end{bmatrix} ={}& \dfrac 1 m \int_{-a}^a\int_0^{\sqrt{a^2-x^2}} \begin{bmatrix}x\\y\end{bmatrix} K~\sqrt{x^2+y^2}~\mathrm d x~\mathrm d y\end{align}$$
However, it seems that converting to polar coordinates would have utility. $$D_{\text{polar}}=\{(r,\theta): 0\leq r\leq a, 0\leq \theta\leq \pi\} \\ \rho(r,\theta) = K r \\ \mathrm d x~\mathrm d y = r~\mathrm d r~\mathrm d\theta$$
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