Finding the Mass and the Center of Mass of a Lamina

Find the mass and center of mass of a lamina in the shape of an isosceles right triangle. The mass density function \(\rho =\rho (x,y) \) is directly proportional to the square of the distance from the vertex at the right angle.

Solution We position the lamina in the \(xy\)-plane so that the vertex at the right angle is at the origin and the two equal sides lie along the positive coordinate axes. See Figure 33. Suppose the equal sides of the triangle each measure \(a.\)

Since the distance of a point \((x,y) \) from the origin is \(\sqrt{ x^{2}+y^{2}},\) the mass density of the lamina is \[ \rho (x,y)=k(x^{2}+y^{2}) \]

where \(k\) is the constant of proportionality. Since \(\rho =\rho (x,y)\) is continuous on \(R,\) the mass \(M\) of this lamina is given by a double integral. \[ \begin{eqnarray*} M& =&\displaystyle\iint\limits_{\kern-3ptR}\rho (x,y)\,{\it dA}=\displaystyle\iint\limits_{\kern-3ptR}k(x^{2}+y^{2})\,d\!A=k\int_{0}^{a} \int_{0}^{a-x}(x^{2}+y^{2})\,{\it dy}\,{\it dx}\\[4pt] &=&k\int_{0}^{a}\left[ \left( x^{2}y+\dfrac{ y^{3}}{3}\right) \right] _{0}^{a-x}{\it dx} = k\int^a_0 \left[x^2a -x^3+\dfrac{(a-x)^3}{3}\right] {\it dx}\\[4pt] & =& k\int_{0}^{a}\dfrac{1}{3}(a^{3}-3a^{2}x+6ax^{2}-4x^{3})\,{\it dx} =\dfrac{k}{3} \left[a^3x-3a^2 \dfrac{x^2}{2} +2ax^3 -x^4\right]^a_0\\[4pt] &=&\dfrac{k}{3}\left( a^{4}-\dfrac{3a^{4}}{2}+2a^{4}-a^{4}\right) =\dfrac{ka^{4}}{6} \end{eqnarray*} \]

Due to the symmetry of both the region \(R\) and the mass density, the center of mass \((\bar{x},\bar{y})\) lies on the line \(y=x\). So, once we find \(\bar{ x}\), we also know \(\bar{y}\). The moment of mass with respect to the \(y\)-axis, \(M_{y}\), is \[ \begin{eqnarray*} M_{y}& =& \displaystyle\iint\limits_{\kern-3ptR}x\rho (x,y)\,{\it dA}=\displaystyle\iint\limits_{\kern-3ptR}xk(x^{2}+y^{2})\,d\!A=k\int_{0}^{a} \int_{0}^{a-x}(x^{3}+xy^{2})\,{\it dy}\,{\it dx} \\[4pt] & =&\dfrac{k}{3}\int_{0}^{a}(a^{3}x-3a^{2}x^{2}+6ax^{3}-4x^{4})\,{\it dx}=\dfrac{ka^{5}}{15} \end{eqnarray*} \]

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The center of mass \((\bar{x},\bar{y})\) is: \[ \bar{x}=\dfrac{M_{y}}{M}=\dfrac{\dfrac{ka^{5}}{15}}{\dfrac{ka^{4}}{6}}=\dfrac{2}{5}a=\overline{y} \]

The center of mass of the lamina is the point \(\left( \dfrac{2}{5}a,\dfrac{2}{5}a\right) \).