Question

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

The moment of inertia about y the axis of the solid of revolution if the density is $\mathbf{\left|}\mathit{x}\mathit{y}\mathit{z}\right|$.

Short answer

The moment of inertia about the -axis of the solid is,$I=\frac{32}{5}$ .

Step-by-step solution

Definition of Density Function

For obtaining the probabilities associated with the random variable which is continuous in nature, the integral of the density function is determined.

Representation of the area bounded by the curve

Draw the bounded region for the curve.

Calculation of the moment of inertia

As the density function is symmetric in x, y and z integrate over the first octant and then multiply the result by 4.

Determine the moment of inertia about the y-axis of the solid.

$$\begin{gathered} l=4{\int }_0^{2}xdx{\int }_0^{\sqrt{x}}ydy{\int }_0^{\sqrt{x-y^2}}zdz\left(x^2+z^2\right) \\ =4{\int }_0^{2}xdx{\int }_0^{\sqrt{x}}ydy{\overline{)\left(\frac{1}{2}{x}^2{z}^2+\frac{1}{4}{z}^4\right)}}_0^{\sqrt{x-y^2}} \\ =4{\int }_0^{2}xdx{\int }_0^{\sqrt{x}}ydy\left(\frac{1}{2}{x}^2\left(x-y^2\right)+\frac{1}{4}{\left(x-y^2\right)}^2\right) \\ =4{\int }_0^{2}xdx{\int }_0^{\sqrt{x}}ydy\left(\frac{1}{2}{x}^3-\frac{1}{2}{x}^2{y}^2+\frac{1}{4}{x}^2-\frac{1}{4}{y}^4\right) \end{gathered}$$

Integrate over .

$$\begin{gathered} l=4{\int }_0^{2}xdx{\overline{)\left(\frac{1}{4}{x}^3{y}^2-\frac{1}{8}{x}^2{y}^4+\frac{1}{8}{x}^2{y}^2-\frac{1}{8}x{y}^4+\frac{1}{24}{y}^6\right)}}_0^{\sqrt{x}} \\ =4{\int }_0^{2}xdx\left(\frac{1}{8}{x}^4+\frac{1}{24}{x}^3\right) \\ =\left[\frac{1}{12}{x}^6+\frac{1}{30}{x}^3\right] \\ =\frac{32}{5} \end{gathered}$$

Thus, the moment of inertia about the -axis of the solid is, $l=\frac{32}{5}$.