Question

Find the mass and center of gravity of the lamina lamina with density δ(x, y) = x + y is bounded by the x-axis, the

line x = 1, and the curve y =

√

x..

Answer 1

The mass of a lamina occupying the region $D$ and having density function $\delta(x,y)$ is

$$m=\int\int_D \delta(x, y) dA$$

$$m=\displaystyle\int_{0}^{1}\displaystyle\int_{0}^{\sqrt{x}}(x+y)dydx$$

$$=\displaystyle\int_{0}^{1}[xy+\dfrac{y^2}{2}]\begin{matrix} \sqrt{x} \\ 0 \end{matrix}dx$$

$$=\displaystyle\int_{0}^{1}(x \sqrt{x}+\dfrac{x}{2})dx$$

$$=[\dfrac{2x^{5/2}}{5}+\dfrac{x^2}{4}]\begin{matrix} 1 \\ 0 \end{matrix}$$

$$=\dfrac{2}{5}+\dfrac{1}{4}=\dfrac{13}{20} \ (units \ of\ mass)$$

$$M_y=\displaystyle\int_{0}^{1}\displaystyle\int_{0}^{\sqrt{x}}x(x+y)dydx$$

$$=\displaystyle\int_{0}^{1}x[xy+\dfrac{y^2}{2}]\begin{matrix} \sqrt{x} \\ 0 \end{matrix}dx$$

$$=\displaystyle\int_{0}^{1}(x^2 \sqrt{x}+\dfrac{x^2}{2})dx$$

$$=[\dfrac{2x^{7/2}}{7}+\dfrac{x^3}{6}]\begin{matrix} 1 \\ 0 \end{matrix}$$

$$=\dfrac{2}{7}+\dfrac{1}{6}=\dfrac{19}{42}$$

$$M_x=\displaystyle\int_{0}^{1}\displaystyle\int_{0}^{\sqrt{x}}y(x+y)dydx$$

$$=\displaystyle\int_{0}^{1}[\dfrac{xy^2}{2}+\dfrac{y^3}{3}]\begin{matrix} \sqrt{x} \\ 0 \end{matrix}dx$$

$$=\displaystyle\int_{0}^{1}(\dfrac{x^2}{2}+\dfrac{x\sqrt{x}}{3})dx$$

$$=[\dfrac{x^3}{6}+\dfrac{2x^{5/2}}{15}]\begin{matrix} 1 \\ 0 \end{matrix}$$

$$=\dfrac{1}{6}+\dfrac{2}{15}=\dfrac{3}{10}$$

$$\bar{x}=\dfrac{M_y}{m}=\dfrac{19/42}{13/20}=\dfrac{190}{273}$$

$$\bar{y}=\dfrac{M_x}{m}=\dfrac{3/10}{13/20}=\dfrac{6}{13}$$