Step 1

We are asked to calculate the moment of inertia of shaded area about y-axis.

 
Step 2

The three sections of shaded area are shown as:

We have the height of first section which is ${h_1} = 3\;{\rm{in}}$.

We have the base of first section which is ${b_1} = 3\;{\rm{in}}$.

We have the height of second section which is ${h_2} = \left( {3 + 3} \right)\;{\rm{in}} = {\rm{6}}\;{\rm{in}}$.

We have the base of second section which is ${b_2} = 6\;{\rm{in}}$.

We have the height of third section which is ${h_3} = 3\;{\rm{in}}$.

We have the base of third section which is ${b_3} = 3\;{\rm{in}}$.

 
Step 3

To calculate the area of first section we use the formula:

\[{A_1} = \frac{1}{2}{h_1}{b_1}\]
 
Step 4

Substitute the known values in the formula:

\begin{array}{c} {A_1} = \frac{1}{2}\left( {3\;{\rm{in}}} \right)\left( {3\;{\rm{in}}} \right)\\ = 4.5\;{\rm{i}}{{\rm{n}}^2} \end{array}
 
Step 5

To calculate the area of second section we use the formula:

\[{A_2} = \frac{1}{2}{h_2}{b_2}\]
 
Step 6

Substitute the known values in the formula:

\begin{array}{c} {A_2} = \frac{1}{2}\left( {6\;{\rm{in}}} \right)\left( {6\;{\rm{in}}} \right)\\ = 18\;{\rm{i}}{{\rm{n}}^2} \end{array}
 
Step 7

To calculate the area of third section we use the formula:

\[{A_3} = {h_3}{b_3}\]
 
Step 8

Substitute the known values in the formula:

\begin{array}{c} {A_3} = \left( {3\;{\rm{in}}} \right)\left( {3\;{\rm{in}}} \right)\\ = 9\;{\rm{i}}{{\rm{n}}^2} \end{array} Step 9

To calculate the moment of inertia of first section about an axis parallel to x-axis we use the formula:

\[{\bar I_1} = \frac{1}{{36}}{b_1}h_1^3\]
 
Step 10

Substitute the known values in the formula:

\begin{array}{c} {{\bar I}_1} = \frac{1}{{36}}\left( {3\;{\rm{in}}} \right){\left( {3\;{\rm{in}}} \right)^3}\\ = 2.25\;{\rm{i}}{{\rm{n}}^4} \end{array}
 
Step 11

To calculate the moment of inertia of second section about an axis parallel to x-axis we use the formula:

\[{\bar I_2} = \frac{1}{{36}}{b_2}h_2^3\]
 
Step 12

Substitute the known values in the formula:

\begin{array}{c} {{\bar I}_2} = \frac{1}{{36}}\left( {6\;{\rm{in}}} \right){\left( {6\;{\rm{in}}} \right)^3}\\ = 36\;{\rm{i}}{{\rm{n}}^4} \end{array}
 
Step 13

To calculate the moment of inertia of third section about an axis parallel to x-axis we use the formula:

\[{\bar I_3} = \frac{1}{{12}}{b_3}h_3^3\]
 
Step 14

Substitute the known values in the formula:

\begin{array}{c} {{\bar I}_3} = \frac{1}{{12}}\left( {3\;{\rm{in}}} \right){\left( {3\;{\rm{in}}} \right)^3}\\ = 6.75\;{\rm{i}}{{\rm{n}}^4} \end{array}
 
Step 15

To calculate the distance of centroid of the first segment from y-axis we use the formula:

\[{d_1} = \frac{2}{3}{b_1}\]
 
Step 16

Substitute the known values in the formula:

\begin{array}{c} {d_1} = \frac{2}{3}\left( {{\rm{3}}\;{\rm{in}}} \right)\\ = 2\;{\rm{in}} \end{array}
 
Step 17

To calculate the distance of centroid of the second segment from y-axis we use the formula:

\[{d_2} = \left( {{b_1} + \frac{{{b_2}}}{3}} \right)\]
 
Step 18

Substitute the known values in the formula:

\begin{array}{c} {d_2} = \left( {3\;{\rm{in}} + \frac{{6\;{\rm{in}}}}{3}} \right)\\ = 3\;{\rm{in}} + {\rm{2}}\;{\rm{in}}\\ = 5\;{\rm{in}} \end{array}
 
Step 19

To calculate the distance of centroid of the third segment from y-axis we use the formula:

\[{d_3} = \frac{{{b_3}}}{2}\]
 
Step 20

Substitute the known values in the formula:

\begin{array}{c} {d_3} = \frac{{3\;{\rm{in}}}}{2}\\ = 1.5\;{\rm{in}} \end{array}
 
Step 21

To calculate the total moment of inertia of the shaded area we use the formula:

\[{I_y} = \left( {{{\bar I}_1} + {{\bar I}_2} + {{\bar I}_3}} \right) + \left( {{A_1}d_1^2 + {A_2}d_2^2 + {A_3}d_3^2} \right)\]
 
Step 22

Substitute the known values in the formula:

\[\begin{array}{c} {I_y} = \left\{ \begin{array}{l} \left( {2.25\;{\rm{i}}{{\rm{n}}^4} + 36\;{\rm{i}}{{\rm{n}}^4} + 6.75\;{\rm{i}}{{\rm{n}}^4}} \right) + \\ \left( {\left( {4.5\;{\rm{i}}{{\rm{n}}^2}} \right){{\left( {2\;{\rm{in}}} \right)}^2} + \left( {18\;{\rm{i}}{{\rm{n}}^2}} \right){{\left( {4\;{\rm{in}}} \right)}^2} + \left( {9\;{\rm{i}}{{\rm{n}}^2}} \right){{\left( {1.5\;{\rm{in}}} \right)}^2}} \right) \end{array} \right\}\\ = 45\;{\rm{i}}{{\rm{n}}^4} + 18\;{\rm{i}}{{\rm{n}}^4} + 450\;{\rm{i}}{{\rm{n}}^4} + 20.25\;{\rm{i}}{{\rm{n}}^{\rm{4}}}\\ = 533.25\;{\rm{i}}{{\rm{n}}^{\rm{4}}} \end{array}\]