In Problems 17 to 30, for the curve $\mathit{y}\mathbf{=}\sqrt{\mathbf{x}}$, between$\mathit{x}\mathbf{=}\mathbf{0}$and$\mathit{x}\mathbf{=}\mathbf{2}$, find:

The moments of inertia about the x axis of a lamina in the shape of the plane area under the curve;

The sum of the products of each element of mass in the body and the square of the element's distance from the axis is a measure of a body's resistance to rotational acceleration about a certain axis.

Examine a tiny section of wire at $f\left(x\right)$from the -axis,

Determine the small element.

$$\begin{gathered} \mathrm{dl}=\mathrm{dM}\mathrm{f}{\left(\mathrm{x}\right)}^2 \\ =\mathrm{xdM} \\ =\mathrm{x\lambda ds}\left(\mathrm{x}\right) \\ =\mathrm{\lambda x}\sqrt{1+\left({\mathrm{f}}^{\cancel{\mathrm{c}}}\left(\mathrm{x}\right)\right)\mathrm{dx}} \\ =\mathrm{\lambda }\sqrt{{\mathrm{x}}^2+\frac{\mathrm{x}}{4}\mathrm{dx}} \\ =\mathrm{\lambda }\sqrt{{\left(\mathrm{x}+\frac{1}{8}\right)}^2-\frac{1}{64}\mathrm{dx}} \end{gathered}$$

Combining all of the bits and integrate in the limits 0 to 2.

$$\begin{gathered} l={\int }_0^{2}dl\left(x\right) \\ =\lambda {\int }_0^2\sqrt{{\left(x+\frac{1}{8}\right)}^2-\frac{1}{64}dx} \end{gathered}$$

It is known that$u=x+\frac{1}{8}$ and role="math" localid="1658834364864" $du=dx$. Substitute it in the above expression.

$l=\lambda {\int }_{1/8}^{17/8}\sqrt{u^2-\frac{1}{64}du}$

It is known that $u=\frac{1}{8}\cosh \theta$and $du=\frac{1}{8}\sin h\theta d\theta$. Substitute it in the above expression.

$$\begin{gathered} l=\frac{\lambda }{64}{\int }_{\cos h-1\left(1\right)}^{\cos h-1\left(17\right)}\sqrt{\cos h^2\theta -1\sin h\theta d\theta } \\ =\frac{\lambda }{64}{\int }_{\cos h-1\left(1\right)}^{\cos h-1\left(17\right)}\sin h^2\theta d\theta \\ =\frac{\lambda }{128}{\int }_{\cos h-1\left(1\right)}^{\cos h-1\left(17\right)}\left(\cos h\left(2\theta \right)-1\right)d\theta \\ =\frac{\lambda }{128}{\left[\frac{1}{2}\sin h\left(2\theta \right)-\theta \right]}_{\cos h-1\left(1\right)}^{\cos h-1\left(17\right)} \\ =2.226\lambda \end{gathered}$$

Write the expression for the mass of the wire.

$M=\lambda L$

So, the value of L is 2.562.

Thus, the moment of inertia in terms of mass is $0.869M$.