Question

Find the center of mass of the solid generated by the area bounded by x = 1, x = 3, y = 0

and y = x^2 by revolving about the x-axis.

Answer 1

When a curve $y=f(x)$ is rotated about the $x$ - axis, the centre of mass of the solid generated will lie on the $x$ – axis because of the symmetry of the curve.

$$\bar{x}=\dfrac{\displaystyle\int_{1}^{3}xy^2dx}{\displaystyle\int_{1}^{3}y^2dx}=\dfrac{\displaystyle\int_{1}^{3}x^5dx}{\displaystyle\int_{1}^{3}x^4dx}$$

$$=\dfrac{[\dfrac{x^6}{6}]\begin{matrix} 3 \\ 1 \end{matrix}}{[\dfrac{x^5}{5}]\begin{matrix} 3 \\ 1 \end{matrix}}=\dfrac{910}{363}$$

$(\dfrac{910}{363}, 0)$