1d:T6d4,

Treat the shaded region as an outer trapezoid minus an inner rectangle.

Outer trapezoid:

Bottom base $$=8$$

Top base $$=4$$

Height $$=3$$

Area of trapezoid:

$$A_t=\frac{(8+4)}{2}(3)=18$$

Centroid of the trapezoid from the bottom:

$$\bar{y}_t=\frac{h(b+2a)}{3(a+b)}$$

With $$a=4$$, $$b=8$$, $$h=3$$:

$$\bar{y}_t=\frac{3(8+2\cdot4)}{3(4+8)}=\frac{16}{12}=\frac{4}{3}$$

Inner rectangle:

Width $$=4$$

Height $$=1$$

Area of rectangle:

$$A_r=4(1)=4$$

Centroid of the rectangle from the bottom:

$$\bar{y}_r=1.5$$

Composite centroid:

$$\bar{y}=\frac{A_t\bar{y}_t-A_r\bar{y}_r}{A_t-A_r}$$

$$\bar{y}=\frac{18\left(\frac{4}{3}\right)-4(1.5)}{18-4}=\frac{24-6}{14}=\frac{9}{7}$$

Now compute moment of inertia about the bottom axis.

For the outer trapezoid, use a rectangle minus two triangles.

Rectangle $$8\times3$$:

$$I_{x,rect}=\frac{bh^3}{3}=\frac{8(3^3)}{3}=72$$

Each side triangle has base $$=2$$ and height $$=3$$:

$$I_{x,tri}=\frac{bh^3}{12}=\frac{2(3^3)}{12}=4.5$$

So, for the trapezoid:

$$I_{x,t}=72-2(4.5)=63$$

For the inner rectangle about the bottom axis:

$$I_{x,r}=\frac{bh^3}{12}+Ad^2$$

$$I_{x,r}=\frac{4(1^3)}{12}+4(1.5)^2=\frac{1}{3}+9=\frac{28}{3}$$

Composite moment of inertia about the bottom axis:

$$I_{x,bottom}=63-\frac{28}{3}=\frac{161}{3}$$

Shift to the centroid $$x$$-axis:

$$I_x=I_{x,bottom}-A(\bar{y})^2$$

Composite area:

$$A=18-4=14$$

Thus:

$$I_x=\frac{161}{3}-14\left(\frac{9}{7}\right)^2$$

$$I_x=\frac{161}{3}-\frac{162}{7}=\frac{641}{21}\approx30.52$$

Final answer: $$30.52$$

1e:T26cbb,

Morning Section