Answer to Question #185054 in Differential Geometry | Topology for SARALA DEVI

Question

Find the centre of gravity of a thin sheet with density d(x, y) = y, bounded by the

curves y = 4x² and x = 4.

Answer 1

$$m=\displaystyle\int_{0}^4\displaystyle\int_{0}^{4x^2}ydydx$$

$$=\displaystyle\int_{0}^4[\dfrac{y^2}{2}]\begin{matrix} 4x^2 \\ 0 \end{matrix}dx=\displaystyle\int_{0}^48x^4dx=[\dfrac{8x^5}{5}]\begin{matrix} 4 \\ 0 \end{matrix}$$

$$=\dfrac{8192}{5}$$

$$M_x=\displaystyle\int_{0}^4\displaystyle\int_{0}^{4x^2}y^2dydx$$

$$=\displaystyle\int_{0}^4[\dfrac{y^3}{3}]\begin{matrix} 4x^2 \\ 0 \end{matrix}dx=\displaystyle\int_{0}^4\dfrac{64x^6}{3}dx=[\dfrac{64x^7}{21}]\begin{matrix} 4 \\ 0 \end{matrix}$$

$$=\dfrac{1048576}{21}$$

$$M_y=\displaystyle\int_{0}^4\displaystyle\int_{0}^{4x^2}xydydx$$

$$=\displaystyle\int_{0}^4[\dfrac{xy^2}{2}]\begin{matrix} 4x^2 \\ 0 \end{matrix}dx=\displaystyle\int_{0}^48x^5dx=[\dfrac{8x^6}{6}]\begin{matrix} 4 \\ 0 \end{matrix}$$

$$=\dfrac{16384}{3}$$

$$\bar{x}=\dfrac{M_y}{m}=\dfrac{16384(5)}{3(8192)}=\dfrac{10}{3}$$

$$\bar{y}=\dfrac{M_x}{m}=\dfrac{1048576(5)}{21(8192)}=\dfrac{640}{21}$$

$(\dfrac{10}{3}, \dfrac{640}{21})$