91Ó°ÊÓ

For the disk.

Figure 1:

Trigonometric property: $\mathbf{cos}\mathbf{θ}\mathbf{=}\frac{\mathbf{sid}\mathbf{e}}{\mathbf{hypotenuse}}$

The moment of inertia about the diameter is calculated as:

$\mathit{I}\mathbf{=}\mathbf{∫}{\mathit{y}}^{\mathbf{2}}\mathit{d}\mathit{m}$ …(1)

The area of the disk using integral is calculated as:role="math" localid="1659193881542" $\mathit{d}\mathit{A}\mathbf{=}{\mathbf{∫}}_{\mathbf{0}}^{\mathbf{a}}{\mathbf{∫}}_{\mathbf{0}}^{\mathbf{2}\mathbf{τ}\mathbf{τ}}\mathit{r}\mathit{d}\mathit{r}\mathit{d}\mathit{θ}$

The centroid of one quadrant of the disk is calculated as,role="math" localid="1659193929004" $\overline{\mathbf{x}}\mathbf{=}\frac{\mathbf{∫}\mathbf{x}\mathbf{d}\mathbf{m}}{\mathbf{∫}\mathbf{d}\mathbf{m}}$

The circumference of the circle is calculated as:$\mathit{C}\mathbf{=}\mathbf{∫}\mathit{d}\mathit{s}\mathbf{=}{\mathbf{∫}}_{\mathbf{0}}^{\mathbf{2}\mathbf{τ}\mathbf{τ}}\mathit{r}\mathit{d}\mathit{θ}$

The centroid of the quarter arc circle is calculated as,

role="math" localid="1659194052588" $\overline{\mathbf{x}}\mathbf{=}\frac{\mathbf{∫}\mathbf{x}\mathbf{d}\mathbf{m}}{\mathbf{∫}\mathbf{d}\mathbf{m}}$ …(2)

And,

$\overline{\mathit{y}}\mathbf{=}\frac{\mathbf{∫}\mathit{y}\mathbf{dm}}{\mathbf{∫}\mathbf{dm}}$ …(3)

The length of the arc is calculated as,$\mathit{S}\mathbf{=}{\mathbf{∫}}_{\mathbf{0}}^{\mathbf{τ}\mathbf{τ}\mathbf{/}\mathbf{2}}\sqrt{\mathbf{4}{\mathbf{a}}^{\mathbf{2}}\mathbf{}\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{2}}\mathbf{}\mathbf{θ}\mathbf{+}\mathbf{4}{\mathbf{a}}^{\mathbf{2}}\mathbf{}\mathbf{c}\mathbf{o}{\mathbf{s}}^{\mathbf{2}}\mathbf{}\mathbf{θ}}\mathit{d}\mathit{θ}$

The center of mass is calculated as,$\mathbf{∫}\mathit{d}\mathit{M}\mathbf{=}{\mathbf{∫}}_{\mathbf{-}\mathbf{Ï„}\mathbf{Ï„}\mathbf{/}\mathbf{2}}^{\mathbf{Ï„}\mathbf{Ï„}\mathbf{/}\mathbf{2}}{\mathbf{∫}}_{\mathbf{0}}^{\mathbf{2}\mathbf{}\mathbf{a}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{θ}}\left(\mathrm{ÒÏ}r\right)\mathit{r}\mathit{d}\mathit{r}\mathit{d}\mathit{θ}\mathit{M}$

(a)

Consider the figure shown below.

Consider$â–¡OPQ$in Figure (1).

$cos\mathrm{θ}=\frac{sid\mathrm{e}}{\mathrm{hypotenuse}}$ …(1)

Substitute for side and for hypotenuse in equation (1).

localid="1659194754727" $\begin{array}{l}\cos \mathrm{θ}=\frac{f}{2a}\\ r=2a\mathrm{³¦´Ç²õθ}\end{array}$

Thus, proved the equation of the given circle is $2a\mathrm{³¦´Ç²õθ}$.

(b)

Consider figure (1).

The region of disk is:$\left\{\left(r,\mathrm{θ}\right)∈-\frac{\mathrm{τ}\mathrm{τ}}{2}≤\mathrm{θ}≤\frac{\mathrm{τ}\mathrm{τ}}{2},0≤r≤2a\cos \mathrm{θ}\right\}$

The area of the disk using integral is calculated as,

$$\begin{gathered} dA={∫}_{-\mathrm{τ}\mathrm{τ}/2}^{\mathrm{τ}\mathrm{τ}/2}{∫}_0^{2a\cos \mathrm{θ}}rdrd\mathrm{θ} \\ ={∫}_{-\mathrm{τ}\mathrm{τ}\mathrm{τ}/2}^{\mathrm{τ}\mathrm{τ}/2}{\left[\frac{r^2}{2}\right]}_0^{2a\cos \mathrm{θ}}d\mathrm{θ} \\ ={∫}_{-\mathrm{τ}\mathrm{τ}/2}^{\mathrm{τ}\mathrm{τ}/2}\frac{2a{\cos }^2\mathrm{θ}}{2}d\mathrm{θ} \\ =\frac{4a^2}{4}{\left[\mathrm{θ}+\frac{\sin 2\mathrm{θ}}{2}\right]}_{-\mathrm{τ}\mathrm{τ}/2}^{\mathrm{τ}\mathrm{τ}/2} \end{gathered}$$

Solve further,

$$\begin{gathered} A=a^2\left(\frac{\mathrm{Ï„}\mathrm{Ï„}}{2}+\frac{\sin \mathrm{Ï„}\mathrm{Ï„}}{2}+\frac{\mathrm{Ï„}\mathrm{Ï„}}{2}+\frac{\sin \mathrm{Ï„}\mathrm{Ï„}}{2}\right) \\ =a^2\left(\frac{2\mathrm{Ï„}\mathrm{Ï„}}{2}\right) \\ =\mathrm{Ï„}\mathrm{Ï„}a \end{gathered}$$

Thus, the area of the disk is $\mathrm{Ï„}\mathrm{Ï„}{a}^2$.

(c)

The coordinate of the centroid is calculated as,

$$\begin{gathered} \overline{\mathrm{y}}=\left(\frac{{\mathrm{Ï„Ï„²¹}}^2}{2}\right)={∫}_0^{\mathrm{Ï„Ï„}/2}{∫}_0^{2\mathrm{a}\mathrm{³¦´Ç²õθ}}{\mathrm{r}}^2\mathrm{³¦´Ç²õθ»å°ù»åθ} \\ ={∫}_0^{\mathrm{Ï„Ï„}/2}\left(\frac{{\left(2\mathrm{a}\mathrm{³¦´Ç²õθ}\right)}^3}{3}-0\right)\cos \mathrm{θ»åθ} \\ =\frac{8{\mathrm{a}}^3}{3}{∫}_0^{\mathrm{Ï„Ï„}/2}{\cos }^4\mathrm{θ»åθ} \\ =\left(\frac{8{\mathrm{a}}^2}{3}\right)\left(\frac{3\mathrm{Ï„Ï„}}{16}\right) \end{gathered}$$

Solve further,

$$\begin{gathered} \overline{\mathrm{x}}\left(\frac{{\mathrm{Ï„Ï„²¹}}^2}{2}\right)=\frac{1}{2}{\mathrm{Ï„Ï„²¹}}^2 \\ \overline{\mathrm{x}}=\frac{1}{2}{\mathrm{Ï„Ï„²¹}}^3\left(\frac{2}{{\mathrm{Ï„Ï„²¹}}^2}\right) \\ =\mathrm{a} \end{gathered}$$

The coordinate of the centroid is calculated as,

localid="1659198143305" $$\begin{gathered} \overline{\mathrm{y}}=\left(\frac{{\mathrm{Ï„Ï„²¹}}^2}{2}\right)={∫}_0^{\mathrm{Ï„Ï„}/2}{∫}_0^{2\mathrm{a}\mathrm{³¦´Ç²õθ}}{\mathrm{r}}^2\sin \mathrm{θ»å°ù»åθ} \\ ={∫}_0^{\mathrm{Ï„Ï„}/2}\left(\frac{{\left(2\mathrm{a}\mathrm{³¦´Ç²õθ}\right)}^3}{3}-0\right)\sin \mathrm{θ»åθ} \\ =\frac{8{\mathrm{a}}^3}{3}{∫}_0^{\mathrm{Ï„Ï„}/2}{\sin }^4\mathrm{θ»åθ} \\ =\left(\frac{8{\mathrm{a}}^2}{3}\right)\left(\frac{1}{4}\right) \end{gathered}$$

Solve further,

localid="1659196005060" $$\begin{gathered} \overline{\mathrm{y}}\left(\frac{{\mathrm{Ï„Ï„²¹}}^2}{2}\right)=\frac{2}{3}{\mathrm{a}}^3 \\ \overline{\mathrm{y}}=\frac{2}{3}{\mathrm{a}}^3\left(\frac{2}{{\mathrm{Ï„Ï„²¹}}^2}\right) \\ =\frac{4}{3}\mathrm{Ï„Ï„²¹} \end{gathered}$$

Thus, the required centroid is localid="1659196019520" $\left(\mathrm{a},\frac{4\mathrm{a}}{\mathrm{Ï„Ï„}}\right)$.

(d)

The moment of inertia about the x axis is calculated as,

$l_x=∫y^2\mathrm{ÒÏ}rdrd\mathrm{θ}$

Rewrite the equation in polar coordinate form.

$$\begin{gathered} {\mathrm{l}}_{\mathrm{x}}={∫}_{-\mathrm{Ï„Ï„}/2}^{\mathrm{Ï„Ï„}/2}{∫}_0^{2\mathrm{a}\mathrm{³¦´Ç²õθ}}{\left(\mathrm{r}\sin \mathrm{θ}\right)}^2\mathrm{ÒÏr»å°ù»åθ} \\ =\mathrm{ÒÏ}{∫}_{-\mathrm{Ï„Ï„}/2}^{\mathrm{Ï„Ï„}/2}{∫}_0^{2\mathrm{a}\mathrm{³¦´Ç²õθ}}{\mathrm{r}}^3{\sin }^2\mathrm{θ»å°ù»åθ} \\ =\mathrm{ÒÏ}{∫}_{-\mathrm{Ï„Ï„}/2}^{\mathrm{Ï„Ï„}/2}\left(\frac{{\mathrm{r}}^4}{4}\right){\mathrm{r}}^3{\sin }^2\mathrm{θ»åθ} \\ =\frac{\mathrm{ÒÏ}}{4}{∫}_{-\mathrm{Ï„Ï„}/2}^{\mathrm{Ï„Ï„}/2}{\left(2\mathrm{a}\cos \mathrm{θ}\right)}^4{\sin }^2\mathrm{θ»åθ} \end{gathered}$$

Solve further,

localid="1659201192573" $$\begin{gathered} l_{\mathit{x}}=\frac{16{\mathrm{a}}^4\mathrm{ÒÏ}}{4}{∫}_{-\mathrm{Ï„Ï„}/2}^{\mathrm{Ï„Ï„}/2}{\cos }^4\mathrm{θ}{\sin }^2\mathrm{θ»åθ} \\ =\frac{16{\mathrm{a}}^4\mathrm{ÒÏ}}{4}\left(\frac{\mathrm{Ï„Ï„}}{16}\right) \\ =\frac{\mathrm{ÒÏ}}{4}{\mathrm{Ï„Ï„²¹}}^4 \end{gathered}$$

Calculate the moment of inertia about the y axis.

$l_y=∫x^2\mathrm{ÒÏ}r\mathrm{d}rd\mathrm{θ}$

Rewrite the equation in polar coordinate form.

$$\begin{gathered} l_y={∫}_{-\mathrm{Ï„Ï„}/2}^{\mathrm{Ï„Ï„}/2}{∫}_0^{2\mathrm{a}\mathrm{³¦´Ç²õθ}}{\left(\mathrm{r}\cos \mathrm{θ}\right)}^2\mathrm{ÒÏ}rdrd\mathrm{θ} \\ =\mathrm{ÒÏ}{∫}_{-\mathrm{Ï„Ï„}/2}^{\mathrm{Ï„Ï„}/2}{∫}_0^{2\mathrm{a}\mathrm{³¦´Ç²õθ}}{\mathrm{r}}^3{\cos }^2\mathrm{θ}drd\mathrm{θ} \\ =\mathrm{ÒÏ}{∫}_{-\mathrm{Ï„Ï„}/2}^{\mathrm{Ï„Ï„}/2}\left(\frac{{\mathrm{r}}^4}{4}\right){\mathrm{r}}^3{\cos }^2\mathrm{θ»åθ} \\ =\frac{\mathrm{ÒÏ}}{4}{∫}_{-\mathrm{Ï„Ï„}/2}^{\mathrm{Ï„Ï„}/2}{\left(2\mathrm{a}\cos \mathrm{θ}\right)}^4{\cos }^2\mathrm{θ}d\mathrm{θ} \end{gathered}$$

Solve further,

$$\begin{gathered} l_y=\frac{16{\mathrm{a}}^4\mathrm{ÒÏ}}{4}{∫}_{-\mathrm{Ï„Ï„}/2}^{\mathrm{Ï„Ï„}/2}{\cos }^6\mathrm{θ} \\ \mathrm{θ}d\mathrm{θ}=\frac{16{\mathrm{a}}^4\mathrm{ÒÏ}}{4}\left(\frac{5\mathrm{Ï„Ï„}}{16}\right) \\ =\frac{5\mathrm{ÒÏ}}{4}{\mathrm{Ï„Ï„²¹}}^4 \end{gathered}$$

The final moment of inertia about the z axis is given by,

$l_z=l_x+l_y$ …..(1)

Substitute$\frac{5p}{4}\mathrm{Ï„}\mathrm{Ï„}{a}^4$for$l_y$and$\frac{\mathrm{ÒÏ}}{4}{\mathrm{Ï„Ï„²¹}}^4$for$l_x$in equation (1).

$$\begin{gathered} l_z=\frac{\mathrm{ÒÏ}}{4}{\mathrm{Ï„Ï„²¹}}^4+\frac{5\mathrm{ÒÏ}}{4}\mathrm{Ï„}\mathrm{Ï„}{a}^6 \\ =\frac{6\mathrm{ÒÏ}}{4}{\mathrm{Ï„Ï„²¹}}^4 \\ =\frac{3\mathrm{ÒÏ}}{2}{\mathrm{Ï„Ï„²¹}}^4 \end{gathered}$$

Calculate the value M as follows,

localid="1659201902008" $$\begin{gathered} ∫dM={∫}_{-\mathrm{Ï„}\mathrm{Ï„}/2}^{\mathrm{Ï„}\mathrm{Ï„}/2}{∫}_0^{2acos\mathrm{θ}}\mathrm{ÒÏ}rdrd\mathrm{θ} \\ M=\mathrm{ÒÏ}{∫}_{-\mathrm{Ï„}\mathrm{Ï„}/2}^{\mathrm{Ï„}\mathrm{Ï„}/2}{\left(\frac{r^2}{4}\right)}_0^{2acos\mathrm{θ}}d\mathrm{θ} \\ M=\frac{4a^2\mathrm{ÒÏ}}{2}{∫}_{-\mathrm{Ï„}\mathrm{Ï„}/2}^{\mathrm{Ï„}\mathrm{Ï„}/2}co{s}^2\mathrm{θ}d\mathrm{θ} \\ =\left(\frac{4a^2\mathrm{ÒÏ}}{2}\right)\left(\frac{\mathrm{Ï„}\mathrm{Ï„}}{2}\right) \end{gathered}$$

Solve further,

$M={\mathrm{ÒÏÏ„Ï„²¹}}^2$

Thus, the moment of inertia about the coordinate axis are$l_x=\frac{1}{4}M{a}^2,l_y=\frac{5}{4}M{a}^2$and $l_z=\frac{3}{2}M{a}^2$.

(e)

The length of the arc is calculated as,

$$\begin{gathered} S={∫}_0^{\mathrm{ττ}/2}\sqrt{4a^2{\sin }^2\mathrm{θ}+4a^2{\cos }^2\mathrm{θ}}d\mathrm{θ} \\ ={∫}_0^{\mathrm{ττ}/2}\sqrt{4a^2\left(1\right)}d\mathrm{θ} \\ =2a{∫}_0^{\mathrm{ττ}/2}1d\mathrm{θ} \\ =2a\left(\frac{\mathrm{ττ}}{2}\right) \end{gathered}$$

Solve further,

$s=\mathrm{Ï„}\mathrm{Ï„}a$

The x coordinate of the centroid is calculated as:$∫\stackrel{¯}{x}dS=∫xds$

$$\begin{gathered} \overline{x}{∫}_0^{\mathrm{τ}\mathrm{τ}/2}\sqrt{{\left(\frac{dr}{d\mathrm{θ}}\right)}^2+r^2}={∫}_0^{\mathrm{τ}\mathrm{τ}/2}x\sqrt{{\left(\frac{dr}{d\mathrm{θ}}\right)}^2+r^{2}d\mathrm{θ}} \\ \overline{x}{∫}_0^{\mathrm{τ}\mathrm{τ}/2}2a={∫}_0^{\mathrm{τ}\mathrm{τ}/2}r\cos \mathrm{θ}2ad\mathrm{θ} \\ \overline{x}\left(2a\right)\left(\frac{\mathrm{ττ}}{2}\right)={∫}_0^{\mathrm{τ}\mathrm{τ}/2}\left(2a\cos \mathrm{θ}\cos \mathrm{θ}\right)2ad\mathrm{θ} \end{gathered}$$

Solve further,

$$\begin{gathered} \overline{x}\left(\mathrm{²¹Ï„Ï„}\right)=4a^2{∫}_0^{\mathrm{Ï„Ï„}/2}\cos \mathrm{θ}d\mathrm{θ} \\ \overline{x}=\frac{4a^2\left({\displaystyle \frac{\mathrm{Ï„Ï„}}{4}}\right)}{a\mathrm{Ï„Ï„}} \\ =a \end{gathered}$$

The y coordinate of the centroid is calculated as,

$$\begin{gathered} ∫\overline{y}ds=∫yds\overline{y}{∫}_0^{\mathrm{ττ}/2}\sqrt{{\left(\frac{dr}{d\mathrm{θ}}\right)}^2+r^2} \\ ={∫}_0^{\mathrm{ττ}/2}y\sqrt{{\left(\frac{dr}{d\mathrm{θ}}\right)}^2+r^{2}d\mathrm{θ}}\overline{y}{∫}_0^{\mathrm{τ}\mathrm{τ}/2}2a \\ ={∫}_0^{\mathrm{ττ}/2}r\sin \mathrm{θ}2ad\mathrm{θ}\overline{y}\left(2a\right)\left(\frac{\mathrm{ττ}}{2}\right) \\ ={∫}_0^{\mathrm{ττ}/2}\left(2a\cos \mathrm{θ}\sin \mathrm{θ}\right)2ad\mathrm{θ} \end{gathered}$$

Solve further,

$$\begin{gathered} \overline{y}\left(\mathrm{²¹Ï„Ï„}\right)=4a^2{∫}_0^{\mathrm{Ï„Ï„}/2}\cos \mathrm{θ}\sin \mathrm{θ}d\mathrm{θ} \\ \overline{y}=\frac{4a^2\left({\displaystyle \frac{1}{2}}\right)}{a\mathrm{Ï„Ï„}} \\ =\frac{2a}{\mathrm{Ï„Ï„}} \end{gathered}$$

Thus, the length of the arc is and the centroid of the arc is $\left(a,\frac{2}{\mathrm{Ï„Ï„}}a\right)$.

localid="1659330007314" $$\begin{gathered} d\overline{y}=\frac{{∫}_0^{\mathrm{ττ}/2}r\sin \mathrm{θ}\mathrm{λ}rd\mathrm{θ}}{{∫}_0^{\mathrm{ττ}/2}\mathrm{λ}rd\mathrm{θ}} \\ =\frac{r^2}{r}\frac{{\left[-\cos \mathrm{θ}\right]}_0^{\mathrm{ττ}/2}}{{\left[\mathrm{θ}\right]}_0^{\mathrm{ττ}/2}} \\ =\frac{r\left(\cos 0\right)}{{\displaystyle \frac{\mathrm{ττ}}{2}}} \\ =\frac{2r}{\mathrm{ττ}} \end{gathered}$$

Substitute a for r in the above expression, $\stackrel{¯}{y}=\frac{2a}{\mathrm{ττ}}$

Thus, the required centroid is $\left(\frac{2\mathrm{a}}{\mathrm{Ï„Ï„}},\frac{2\mathrm{a}}{\mathrm{Ï„Ï„}}\right)$.

(f)

The center of mass is calculated as,

$$\begin{gathered} ∫\mathrm{dM}={∫}_{-\mathrm{Ï„Ï„}/2}^{\mathrm{Ï„Ï„}/2}{∫}_0^{2\mathrm{a}\mathrm{³¦´Ç²õθ}}\left(\mathrm{ÒÏr}\right)\mathrm{»å°ù»åθ} \\ =\mathrm{ÒÏ}{∫}_{-\mathrm{Ï„Ï„}/2}^{\mathrm{Ï„Ï„}/2}{\left(\frac{{\mathrm{r}}^3}{3}\right)}_0^{2\mathrm{a}\mathrm{³¦´Ç²õθ}}\mathrm{»åθ} \\ =\frac{8{\mathrm{a}}^3\mathrm{ÒÏ}}{3}{∫}_{-\mathrm{Ï„Ï„}/2}^{\mathrm{Ï„Ï„}/2}{\cos }^3\mathrm{θ»åθ} \\ =\frac{8{\mathrm{a}}^3\mathrm{ÒÏ}}{3}\left(\frac{4}{3}\right) \end{gathered}$$

Solve further,

$M=\frac{32a^3\mathrm{ÒÏ}}{9}$

Calculate the value of .

$$\begin{gathered} \overline{x}=\frac{1}{M}{∫}_{-\mathrm{Ï„Ï„}/2}^{\mathrm{Ï„Ï„}/2}{∫}_0^{2\mathrm{a}\mathrm{³¦´Ç²õθ}}\left(x\mathrm{ÒÏ}r\right)rdrd\mathrm{θ} \\ =\frac{\mathrm{ÒÏ}}{\left({\displaystyle \frac{3{\mathrm{a}}^2\mathrm{ÒÏ}}{9}}\right)}{∫}_{-\mathrm{Ï„Ï„}/2}^{\mathrm{Ï„Ï„}/2}{∫}_0^{2\mathrm{a}\mathrm{³¦´Ç²õθ}}\left(\mathrm{r}\cos \mathrm{θ}\right)r^{2}drd\mathrm{θ} \\ =\frac{9}{32{\mathrm{a}}^3}{∫}_{-\mathrm{Ï„Ï„}/2}^{\mathrm{Ï„Ï„}/2}{\left(\frac{{\mathrm{r}}^4}{4}\right)}_0^{2\mathrm{a}\mathrm{³¦´Ç²õθ}}\cos \mathrm{θ»åθ} \\ =\frac{9}{4.32a^3}{∫}_{-\mathrm{Ï„Ï„}/2}^{\mathrm{Ï„Ï„}/2}{\left(2\mathrm{a}\cos \mathrm{θ}\right)}^4{\cos }^2\mathrm{θ}d\mathrm{θ} \end{gathered}$$

Solve further,

$$\begin{gathered} \overline{x}=\frac{9\left(16\right)a^4}{4.32a^3}{∫}_{-\mathrm{Ï„Ï„}/2}^{\mathrm{Ï„Ï„}/2}{∫}_0^{2\mathrm{a}\mathrm{³¦´Ç²õθ}}{\mathrm{r}}^2{\cos }^5\mathrm{θ»åθ} \\ =\frac{9\mathrm{a}}{18}\left(\frac{16}{15}\right) \\ =\frac{6a}{5} \end{gathered}$$

The moment of inertia about the axis is calculated as,

$l_x=∫y^2\left(r\right)rdrd\mathrm{θ}$

Rewrite the equation in polar coordinate form.

$$\begin{gathered} l_x={∫}_{-\mathrm{Ï„Ï„}/2}^{\mathrm{Ï„Ï„}/2}{∫}_0^{2\mathrm{a}\mathrm{³¦´Ç²õθ}}{\left(\mathrm{r}\sin \mathrm{θ}\right)}^2{r}^{2}drd\mathrm{θ} \\ ={∫}_{-\mathrm{Ï„Ï„}/2}^{\mathrm{Ï„Ï„}/2}{∫}_0^{2\mathrm{a}\mathrm{³¦´Ç²õθ}}{\mathrm{r}}^4{\sin }^2\mathrm{θ}drd\mathrm{θ} \\ ={∫}_{-\mathrm{Ï„Ï„}/2}^{\mathrm{Ï„Ï„}/2}{\left(\frac{{\mathrm{r}}^5}{5}\right)}_0^{2\mathrm{a}\mathrm{³¦´Ç²õθ}}{\sin }^2\mathrm{θ»åθ} \\ =\frac{1}{5}{∫}_{-\mathrm{Ï„Ï„}/2}^{\mathrm{Ï„Ï„}/2}{\left(2\mathrm{a}\cos \mathrm{θ}\right)}^5{\sin }^2\mathrm{θ}d\mathrm{θ} \end{gathered}$$

Solve further,

$$\begin{gathered} l_x=\frac{32a^5}{5}{∫}_{-\mathrm{τ}\mathrm{τ}/2}^{\mathrm{τ}\mathrm{τ}/2}{\cos }^5\mathrm{θ}{\sin }^2\mathrm{θ}d\mathrm{θ} \\ =\frac{32a^5}{5}\left(\frac{16}{105}\right) \\ =\frac{512}{525}{a}^5 \end{gathered}$$

Calculate the moment of inertia about the $y$ axis.

$l_y=∫x^2{r}^{2}drd\mathrm{θ}$

Rewrite the equation in polar coordinate form.

$$\begin{gathered} l_y={∫}_{-\mathrm{Ï„Ï„}/2}^{\mathrm{Ï„Ï„}/2}{∫}_0^{2\mathrm{a}\mathrm{³¦´Ç²õθ}}{\left(\mathrm{r}\cos \mathrm{θ}\right)}^2{r}^{2}drd\mathrm{θ} \\ ={∫}_{-\mathrm{Ï„Ï„}/2}^{\mathrm{Ï„Ï„}/2}{∫}_0^{2\mathrm{a}\mathrm{³¦´Ç²õθ}}{\mathrm{r}}^4{\cos }^2\mathrm{θ}drd\mathrm{θ} \\ ={∫}_{-\mathrm{Ï„Ï„}/2}^{\mathrm{Ï„Ï„}/2}{\left(\frac{{\mathrm{r}}^5}{5}\right)}_0^{2\mathrm{a}\mathrm{³¦´Ç²õθ}}{\cos }^2\mathrm{θ»åθ} \\ =\frac{1}{5}{∫}_{-\mathrm{Ï„Ï„}/2}^{\mathrm{Ï„Ï„}/2}{\left(2\mathrm{a}\cos \mathrm{θ}\right)}^5{\cos }^2\mathrm{θ}d\mathrm{θ} \end{gathered}$$

Solve further,

$$\begin{gathered} l_y=\frac{32a^5}{5}{∫}_{-\mathrm{τ}\mathrm{τ}/2}^{\mathrm{τ}\mathrm{τ}/2}{\cos }^2\mathrm{θ}d\mathrm{θ} \\ =\frac{32a^5}{5}\left(\frac{32}{35}\right) \\ =\frac{1024}{175}{a}^5 \end{gathered}$$

The final moment of inertia about the axis is given by,

$l_z=l_x+l_y$ …(2)

Substitute$\frac{1024}{175}{a}^5$ for $l_y$and $\frac{512}{525}{a}^5$for$l_x$ in equation (2).

$\begin{array}{l}{I}_z=\frac{512}{525}{a}^5+\frac{1024}{175}{a}^5\\ =\frac{512}{75}{a}^5\end{array}$

The center of mass is calculated as,

$$\begin{gathered} ∫\mathrm{dM}={∫}_{-\mathrm{Ï„Ï„}/2}^{\mathrm{Ï„Ï„}/2}{∫}_0^{2\mathrm{a}\mathrm{³¦´Ç²õθ}}\left(r\right)r\mathrm{»å°ù»åθ}M \\ ={∫}_{-\mathrm{Ï„Ï„}/2}^{\mathrm{Ï„Ï„}/2}{\left(\frac{{\mathrm{r}}^3}{3}\right)}_0^{2\mathrm{a}\mathrm{³¦´Ç²õθ}}\mathrm{»åθ}M \\ =\frac{8{\mathrm{a}}^3\mathrm{ÒÏ}}{3}{∫}_{-\mathrm{Ï„Ï„}/2}^{\mathrm{Ï„Ï„}/2}{\cos }^3\mathrm{θ»åθ} \\ =\frac{8{\mathrm{a}}^3}{3}\left(\frac{4}{3}\right) \end{gathered}$$

Solve further,

$M=\frac{32}{9}{a}^3$

Thus, the moment of inertia about the coordinate axis are$l_x=\frac{48}{175}M{a}^2,l_y=\frac{288}{175}M{a}^2,l_z=\frac{48}{25}M{a}^2$ and the center of mass is $\frac{6a}{5}$.

(g)

Consider the Figure (1)

The two curves intersect$\frac{\mathrm{Ï„Ï„}}{3}$ at and $-\frac{\mathrm{Ï„Ï„}}{3}$.

The area of the bounded region is calculated as,

$$\begin{gathered} ∫dA=2{∫}_{\frac{\mathrm{Ï„Ï„}}{3}}^{\frac{\mathrm{Ï„Ï„}}{2}}{∫}_0^{2\mathrm{a}\mathrm{³¦´Ç²õθ}}\mathrm{r»å°ù»åθ}+{∫}_{\frac{\mathrm{Ï„Ï„}}{3}}^{\frac{\mathrm{Ï„Ï„}}{3}}{∫}_0^{\mathrm{a}}\mathrm{r»å°ù»åθ} \\ =2{∫}_{\frac{\mathrm{Ï„Ï„}}{3}}^{\frac{\mathrm{Ï„Ï„}}{2}}{\left(\frac{{\mathrm{r}}^2}{2}\right)}_0^{2\mathrm{a}\mathrm{³¦´Ç²õθ}}\mathrm{»åθ}+{∫}_{\frac{\mathrm{Ï„Ï„}}{3}}^{\frac{\mathrm{Ï„Ï„}}{3}}{\left(\frac{{\mathrm{r}}^2}{2}\right)}_0^{\mathrm{a}}\mathrm{»åθ} \\ =\frac{2}{2}{∫}_{\frac{\mathrm{Ï„Ï„}}{3}}^{\frac{\mathrm{Ï„Ï„}}{2}}4{\mathrm{a}}^2{\cos }^3\mathrm{θ»åθ}+\frac{{\mathrm{a}}^2}{2}{∫}_{\frac{\mathrm{Ï„Ï„}}{3}}^{\frac{\mathrm{Ï„Ï„}}{3}}1\mathrm{»åθ} \\ =4{\mathrm{a}}^2\left(\frac{2\mathrm{Ï„Ï„}-3\sqrt{3}}{24}\right)\left(\frac{2\mathrm{Ï„Ï„}}{3}\right) \end{gathered}$$

Solve further,

$$\begin{gathered} ∫dA=-\frac{\sqrt{3}}{2}{a}^2+\frac{2}{3}{a}^2\mathrm{ττ} \\ ={\mathrm{a}}^2\left(\frac{2}{3}\mathrm{ττ}-\frac{\sqrt{3}}{2}\right) \end{gathered}$$

Thus, the area of the region common to both the curves is ${\mathrm{a}}^2\left(\frac{2}{3}\mathrm{Ï„Ï„}-\frac{\sqrt{3}}{2}\right)$.