Chapter 1, Problem 1.23P

The surfaces $p=3,p=5,\phi ={130}^o,z=3andz=4.5$ define a closed surface. Find (a) the enclosed volume; (b) the total area of the enclosing surface; (c) the total length of the 12 edges of the surfaces; (a) the Length of the longest straight line that lies entirely within the volume.

To determine

(a)

The enclosed volume.

Answer to Problem 1.23P

The enclosed volume is 6.28

Explanation of Solution

Given information:

For the closed surface,

$\begin{array}{c}\rho =3\\ \rho =5\\ \varphi =100°\\ \varphi =130°\\ z=3\\ z=4.5\end{array}$

Concept used:

Volume= ${\displaystyle {\int }_{\rho }^{\rho }{\displaystyle {\int }_{\varphi }^{\varphi }{\displaystyle {\int }_z^z\rho d\rho d\varphi dz}}}$ ......... (1)

Calculation:

Substituting the values in equation (1), we get

   $\text{Volume}={\displaystyle {\int }_3^{4.5}{\displaystyle {\int }_{{100}^{\circ }}^{{130}^{\circ }}{\displaystyle {\int }_3^5\rho d\rho d\varphi dz}}}$

   $={\displaystyle {\int }_3^{4.5}{\displaystyle {\int }_{1.74}^{2.26}{\displaystyle {\int }_3^5\rho d\rho d\varphi dz}}}$

   $={{\displaystyle {\int }_3^{4.5}{\displaystyle {\int }_{1.74}^{2.26}\left[\frac{{\rho }^2}{2}\right]}}}_3^{5}d\varphi dz$

   $={{\displaystyle {\int }_3^{4.5}{\displaystyle {\int }_{1.74}^{2.26}\left[\frac{{\rho }^2}{2}\right]}}}_3^{5}d\varphi dz$

$\begin{array}{c}={\displaystyle {\int }_3^{4.5}{\displaystyle {\int }_{1.74}^{2.26}\left[\frac{16}{2}\right]}}d\varphi dz\\ ={{\displaystyle {\int }_3^{4.5}\left[8\varphi \right]}}_{1.74}^{2.26}dz\\ ={\displaystyle {\int }_3^{4.5}\left[8\times 2.26-8\times 1.74\right]}dz\\ ={\displaystyle {\int }_3^{4.5}4.16dz}\end{array}$

$\begin{array}{l}={\left[4.16z\right]}_3^{4.5}\\ =4.5\times 4.16-3\times 4.16\\ =6.28\end{array}$

Conclusion:

Therefore, the enclosed volume is 6.28.

To determine

(b)

Total area of the enclosing surface.

Answer to Problem 1.23P

Total area of the enclosing surface is 20.7.

Explanation of Solution

Given information:

For the closed surface,

$\begin{array}{c}\rho =3\\ \rho =5\\ \varphi =100°\\ \varphi =130°\\ z=3\\ z=4.5\end{array}$

Concept used:

Area= $2{\displaystyle {\int }_{\varphi }^{\varphi }{\displaystyle {\int }_{\rho }^{\rho }\rho d\rho d\varphi }}+{\displaystyle {\int }_z^z{\displaystyle {\int }_{\varphi }^{\varphi }3d\varphi dz}}+{\displaystyle {\int }_z^z{\displaystyle {\int }_{\varphi }^{\varphi }5d\varphi dz}}+2{\displaystyle {\int }_z^z{\displaystyle {\int }_{\varphi }^{\varphi }d\varphi dz}}$ ........... (2)

Calculation:

Substituting the values in equation (2), we get

   $\text{Area}=2{\displaystyle {\int }_{{100}^{\circ }}^{{130}^{\circ }}{\displaystyle {\int }_3^5\rho d\rho d\varphi }+{\displaystyle {\int }_3^{4.5}{\displaystyle {\int }_{{100}^{\circ }}^{{130}^{\circ }}3d\varphi dz+}}}{\displaystyle {\int }_3^{4.5}{\displaystyle {\int }_{{100}^{\circ }}^{{130}^{\circ }}5d\varphi dz+}}2{\displaystyle {\int }_3^{4.5}{\displaystyle {\int }_3^{5}d\rho dz}}$

   $=2{\displaystyle {\int }_{1.74}^{2.26}{\displaystyle {\int }_3^5\rho d\rho d\varphi }}+{\displaystyle {\int }_3^{4.5}{\displaystyle {\int }_{1.74}^{2.26}3d\varphi dz}}+{\displaystyle {\int }_3^{4.5}{\displaystyle {\int }_{1.74}^{2.26}5d\varphi dz}}+2{\displaystyle {\int }_3^{4.5}{\displaystyle {\int }_3^{5}d\rho dz}}$

   $\left[\because \text{converting degree to radian for }\varphi \right]$

   $=2{{\displaystyle {\int }_{1.74}^{2.26}\left[\frac{{\rho }^2}{2}\right]}}_3^{5}d\varphi +{{\displaystyle {\int }_3^{4.5}\left[3\varphi \right]}}_{1.74}^{2.26}dz+{{\displaystyle {\int }_3^{4.5}\left[5\varphi \right]}}_{1.74}^{2.26}dz+2{\displaystyle {\int }_3^{4.5}{\left[\rho \right]}_3^{5}dz}$

   $=2{\displaystyle {\int }_{1.74}^{2.26}\left[\frac{25}{2}-\frac{9}{2}\right]}d\varphi +{\displaystyle {\int }_3^{4.5}\left[3\times 2.26-3\times 1.74\right]}dz+{\displaystyle {\int }_3^{4.5}\left[5\times 2.26-5\times 1.74\right]}dz+2{\displaystyle {\int }_3^{4.5}\left[5-3\right]}dz$

Further integrating we have,

   $\begin{array}{c}\text{Area}=2{\displaystyle {\int }_{1.74}^{2.26}8d\varphi +{\displaystyle {\int }_3^{4.5}1.56dz+}}{\displaystyle {\int }_{2.26}^{4.5}2.6dz}+2{\displaystyle {\int }_3^{4.5}2dz}\\ =2{\left[8\varphi \right]}_{1.74}^{2.26}+{\left[1.56\right]}_3^{4.5}+{\left[2.6z\right]}_3^{4.5}+2{\left[2z\right]}_3^{4.5}\\ =2\left[8\times 2.26-8\times 1.74\right]+\left[1.56\times 4.5-1.56\times 3\right]+\left[2.6\times 4.5-2.6\times 3\right]+2\left[2\times 4.5-2\times 3\right]\\ =2\times 4.16+2.34+3.9+6\\ =\text{2}0.\text{7}\end{array}$

Conclusion:

Therefore, the total area of the enclosing surface is 20.7

To determine

(c)

Total length of the twelve edges of the surfaces.

Answer to Problem 1.23P

Total length of the twelve edges of the surfaces is 22.4.

Explanation of Solution

Given information:

For the closed surface,

$\begin{array}{c}\rho =3\\ \rho =5\\ \varphi =100°\\ \varphi =130°\\ z=3\\ z=4.5\end{array}$

Concept used:

$\text{Length}=4\times 1.5+4\times 2+2\left[\frac{{30}^{\circ }}{{360}^{\circ }}2\pi \times 3+\frac{{30}^{\circ }}{{360}^{\circ }}2\pi \times 5\right]$    ..........(3)

Calculation:

From equation (3), we get

$\begin{array}{c}\text{Length}=4\times 1.5+4\times 2+2\left[\frac{{30}^{\circ }}{{360}^{\circ }}2\pi \times 3+\frac{{30}^{\circ }}{{360}^{\circ }}2\pi \times 5\right]\\ =6+8+2\left[{0.083}^{\circ }\times 6\pi +{0.083}^{\circ }\times 10\pi \right]\\ =14+2\times 4.17\\ =22.4\end{array}$

Conclusion:

Therefore, the total length is 22.4.

To determine

(d)

The length of the longest straight line that lies entirely within the volume.

Answer to Problem 1.23P

The length of the longest straight line that lies entirely within the volume is 3.21.

Explanation of Solution

Given information:

For the closed surface,

$\begin{array}{c}\rho =3\\ \rho =5\\ \varphi =100°\\ \varphi =130°\\ z=3\\ z=4.5\end{array}$

Concept used:

$\text{Length}= |B-A|$    ..........(4)

Calculation:

The longest straight line will be between the points $A(\rho =3,\varphi ={100}^{\circ },z=3)$ and

$B(\rho =5,\varphi ={130}^{\circ },z=4.5)$

Now converting A and B to Cartesian co-ordinates, $x=\rho \cos \varphi ,y=\rho \sin \varphi ,z=z$

For A ,

$\begin{array}{c}x=3\cos 100°\\ =-0.519\\ y=3\sin 100°\\ =2.952\\ z=3\end{array}$

Therefore, $A(x=-0.52,y=2.95,z=3)$

For B ,

$\begin{array}{c}x=5\cos {130}^{\circ }\\ =-3.21\\ y=5\sin {130}^{\circ }\\ =3.83\\ z=4.5\end{array}$

Therefore, $B(x=-3.21,y=3.83,z=4.5)$

So, from equation (4),

$\begin{array}{c}\text{Length}=|B-A|\\ =|(-3.21+0.51,3.83-2.95,4.5-3)|\\ =\sqrt{{2.69}^2+{0.88}^2+{1.5}^2}\\ =\sqrt{9.8888}\\ =\text{3}.\text{21}\end{array}$

Conclusion:

Therefore, the total length is 3.21.