Question

Find (x̅, y̅): R = {(x, y): 0 ≤ y ≤ √x^2 + 1 , 0 ≤ x ≤ 1} 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.

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

$$=\dfrac{[\dfrac{x^4}{4}+\dfrac{x^2}{2}]\begin{matrix} 1 \\ 0 \end{matrix}}{[\dfrac{x^3}{3}+x]\begin{matrix} 1 \\ 0 \end{matrix}}=\dfrac{9}{16}$$

$(\dfrac{9}{16}, 0)$