Chapter 16.7, Problem 41E

To determine

(a)

To write:

An integral expression for moment of inertia $I_z$ about $z-axis$ of a thin sheet in the shape of a surface $S$ if the density function is $\rho .$

Solution:

$I_z={\iint }_S\left(x^2+y^2\right)\rho \left(x,y,z\right)dS$

Explanation:

1) Concept:

The moment of inertia of a particle of mass m about an axis is defined to be $m{r}^2$ where r is the distance from the particle to the axis.

2) Calculations:

The given body about which the moment of inertia needs to be found is a thin sheet in the shape of a surface $S$.

And density function is $\rho$.

Therefore, the mass of each element is $\rho dS$

And we have to find the moment of inertia about $z-axis.$

The distance of each element from the $z-axis$ is $r=\sqrt{\left(x^2+y^2\right)}$

Therefore,$r^2=x^2+y^2$

Moment of inertia of each element about $z-axis$ is given by

$m{r}^2=\rho dS\left(x^2+y^2\right)$

The moment of inertia of total sheet about $z-axis$ is given by

$I_z={\iint }_S\left(x^2+y^2\right)\rho \left(x,y,z\right)dS$

Conclusion:

An integral expression for moment of inertia $I_z$ about $z-axis$ of a thin sheet in the shape of a surface $S$ if the density function is $\rho$ is given by

$I_z={\iint }_S\left(x^2+y^2\right)\rho \left(x,y,z\right)dS$

(b)

To find:

The moment of inertia about $z-axis$ of a thin funnel in the shape of a cone

$z=\sqrt{x^2+y^2}, 1\le z\le 4$

Solution:

$I_z=\frac{4329}{5}\sqrt{2}\pi$

Explanation:

1) Concept:

Use the formula:

$I_z={\iint }_S\left(x^2+y^2\right)\rho \left(x,y,z\right)dS$

Where $dS=\left|r_x\times r_y\right|dA$

2) Given:

$z=\sqrt{x^2+y^2}, 1\le z\le 4$

Density $=\rho \left(x,y,z\right)=10-z$

3) Calculations:

The given equation of cone is

$z=\sqrt{x^2+y^2}, 1\le z\le 4$

Therefore,

The vector equation of surface $S$ is given by

$r\left(x,y\right)=x i+y j+\sqrt{x^2+y^2}k$

Differentiating with respect to $x \& y,$

$r_x=i+\frac{x}{\sqrt{x^2+y^2}}k$

$r_y=j+\frac{y}{\sqrt{x^2+y^2}}k$

Then,

$\left(r_x\times r_y\right)=\left|\begin{array}{ccc}i& j& k\\ 1& 0& \frac{x}{\sqrt{x^2+y^2}}\\ 0& 1& \frac{y}{\sqrt{x^2+y^2}}\end{array}\right|$

$=i\left(-\frac{x}{\sqrt{x^2+y^2}}\right)-j\left(\frac{y}{\sqrt{x^2+y^2}}\right)+k\left(1\right)$

$=-\frac{x}{\sqrt{x^2+y^2}}i-\frac{y}{\sqrt{x^2+y^2}}j+k$

Therefore,

$\left|r_x\times r_y\right|=\sqrt{{\left(\frac{x}{\sqrt{x^2+y^2}}\right)}^2+{\left(\frac{y}{\sqrt{x^2+y^2}}\right)}^2+1}$

$=\sqrt{\frac{x^2+y^2}{x^2+y^2} +1}=\sqrt{2}$

$\left|r_x\times r_y\right|=\sqrt{2}$

From concept,

The moment of inertia about $z-axis$ of a given funnel is given by

$I_z={\iint }_S\left(x^2+y^2\right)\rho \left(x,y,z\right)dS$

But $\rho =10-z=10-\sqrt{x^2+y^2}$ and

Therefore, above integral becomes,

$I_z={\iint }_S\left(x^2+y^2\right)\left(10-\sqrt{x^2+y^2}\right)dS$

Since, $dS=\left|r_x\times r_y\right|dA$

$I_z={\iint }_{1\le x^2+y^2\le 16}\left(x^2+y^2\right)\left(10-\sqrt{x^2+y^2}\right)\sqrt{2}dA$

Using polar coordinates,

$x=r\cos \theta \& y=r\sin \theta , 0\le \theta \le 2\pi$

$x^2+y^2=r^2{\cos }^2\theta +r^2{\sin }^2\theta =r^2$

The above integral becomes,

$I_z={\int }_0^{2\pi }{\int }_1^4\sqrt{2}{r}^2\left(10-r\right)rdr d\theta$

By integrating with respect to $r,$

$=\sqrt{2}{\int }_0^{2\pi }\left[\frac{10r^4}{4}-\frac{r^5}{5}\right]\genfrac{}{}{0pt}{}{4}{1} d\theta$

By applying the limits of integration,

$=\sqrt{2}{\int }_0^{2\pi }\left[\left(\frac{5}{2}\left(4^4\right)-\frac{4^5}{5}\right)-\left(\frac{5}{2}\left(1^4\right)-\frac{1^5}{5}\right)\right] d\theta$

By simplifying,

$=\sqrt{2}{\int }_0^{2\pi }\left[\frac{5}{2}\left(4^4-1\right)+\frac{1}{5}\left(1-4^5\right)\right] d\theta$

$=\sqrt{2}{\int }_0^{2\pi }\left[\frac{4329}{10}\right] d\theta$

Integrating with respect to $\theta ,$

$=\frac{4329}{10}\sqrt{2}\left[\theta \right]\genfrac{}{}{0pt}{}{2\pi }{0}$

By applying the limits of integration,

$=\frac{4329}{10}\sqrt{2}\left[2\pi -0\right]$

Thus,

$I_z=\frac{4329}{5}\sqrt{2}\pi$

Conclusion:

The moment of inertia about $z-axis$ of a thin funnel in the shape of a cone

$z=\sqrt{x^2+y^2}, 1\le z\le 4$ is

$\frac{4329}{5}\sqrt{2}\pi$