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Determine the moment of inertia of the area about  the  $x$ axis.

$d I_{x}=d I_{\overline{x}}+d A y^{2}$

$d I_{x}=\frac{1}{12} d x y^{3}+\left(y d x\right)\left(\frac{y}{2}\right)^{2}$

$\frac{y^{3} }{4} d x$

$=\frac{1}{12} d x y^{3}+\frac{3 y^{3}}{12} d x =\frac{4}{12} d x y^{3}=\frac{1}{3} d x y^{3}$

$I_{x}= \int_{0}^{2} \frac{1}{3}(0.25)^{3}\left(x^{9}\right) d x$

$=\frac{1}{192}\left[\frac{x^{10}}{10}\right]_{0}^{2} =0.533 \mathrm{m}^{4}$

Determine the moment of inertia of the area about  the  $y$  axis.

$d I{y}=d I \overline{y}+d A x^{2}$

$=\frac{1}{12} y(d x)^{3}+(y d x) x^{2}$

$=y x^{2} d x=\left(0.25 x^{5}\right) d x$

$I{y}=\int_{0}^{2} 0.25 x^{5} d x=0.25\left[\frac{x^{6}}{6}\right]_{0}^{2}=2.67 m^{4}$

Determine the moment of inertia of the area about the $x$  axis. Solve the problem in two ways, using rectangular  differential elements: (a) having a thickness of  $d x,$  and  (b) having a thickness of  $d y .$

(a) $d I_{x}=\left[\frac{1}{12} d x y^{3}+(y d x)\left(\frac{y}{2}\right)^{2}\right] * 2$

$=\left[\frac{1}{3} y^{3} d x\right] * 2=\frac{2}{3} y^{3} d x$

$I_{x}=\frac{2}{3} \int_{0}^{1}\left(4-4 x^{2}\right)^{3} d x=19.5 m^{4}$

(b) $d I_{x}=\left[\frac{1}{12} x \left(d y^{3}\right)+x d y\left(y^{2}\right)\right]{* 2}=2 x y^{2} d y$

$\begin{array}{l}{4 x^{2}=4-y} \\ {x^{2}=1-0.25 y} \\ {x=\sqrt{(1-0.25 y)}}\end{array}$

$I_{x}=2 \int_{0}^{4} y^{2} \sqrt{1-0.25 y} \ d y=19.5 m^{4}$