Example: Quarter of a Circle
$
\def \half {\frac{1}{2}}
\def \kwart {\frac{1}{4}}
\def \hieruit {\quad \Longrightarrow \quad}
\def \EN {\quad \mbox{and} \quad}
\def \spekhaken {\iint}
$
As an example, consider the area which is delimited by:
$$
x^2 + y^2 \le R^2 \EN x \ge 0 \EN y \ge 0
$$
The area of this quarter of a circle is simply given by:
$$
A = \kwart \pi R^2
$$
The first order moments are calculated as follows:
$$
\overline{x} = \frac{ \spekhaken x \, dx dy }{ \spekhaken dx dy }
= \frac{ \int_0^R \int_0^{\pi/2} r.cos(\phi) \, r.dr.d\phi}{ \pi.R^2/4 }
= \frac{4}{\pi R^2} \int_0^R r^2\,dr \int_0^{\pi/2} cos(\phi) \, d\phi
$$ $$
= \frac{4}{\pi R^2} \left[ \frac{1}{3} r^3 \right]_0^R
\left[ \, sin(\phi) \, \right]_0^{\pi/2}
= \frac{4 R}{3 \pi} \approx 0.42441 \, R
$$
Likewise:
$$
\overline{y} = \frac{ \spekhaken y \, dx dy }{ \spekhaken dx dy }
= \frac{4}{\pi R^2} \left[ \frac{1}{3} r^3 \right]_0^R
\left[ \, - cos(\phi) \, \right]_0^{\pi/2}
= \frac{4 R}{3 \pi}
$$
The second order moments are calculated as follows:
$$
\overline{x^2} = \frac{ \spekhaken x^2 \, dx dy }{ \spekhaken dx dy }
= \frac{ \int_0^R \int_0^{\pi/2} \left[r.cos(\phi)\right]^2
r.dr.d\phi}{ \pi.R^2/4 }
= \frac{4}{\pi R^2} \int_0^R r^3\,dr \int_0^{\pi/2} cos^2(\phi) \, d\phi
$$
Where:
$$
cos(2\phi) = 2.cos^2(\phi) - 1 \hieruit cos^2(\phi) = \frac{cos(2\phi)+1}{2}
\hieruit
$$ $$
\int_0^{\pi/2} cos^2(\phi) \, d\phi =
\kwart \int_0^{\pi/2} cos(2\phi)\,d(2\phi) \, + \, \half \int_0^{\pi/2}d\phi
= \kwart \left[ \, sin(\theta) \, \right]_0^{\pi}
+ \half (\pi/2)
$$ $$
\hieruit \overline{x^2} = \frac{4}{\pi R^2} \left[ \frac{r^4}{4} \right]_0^R
. \, \frac{\pi}{4} = \frac{R^2}{4}
$$
In very much the same way we would find:
$$
\overline{y^2} = \frac{R^2}{4}
$$
This result can also be predicted, however, by considerations of symmetry.
The whole figure is symmetric around the line $y = x$, namely, and so any result
for $x$ can also be used for $y$ . The cross moment of inertia is:
$$
\overline{xy} = \frac{4}{\pi R^2} \int_0^R r^2.r.dr
\int_0^{\pi/2} cos(\phi) sin(\phi) \, d\phi
= \frac{4}{\pi R^2} \left[ \frac{r^4}{4} \right]_0^R
\half \left[ sin^2(\phi) \right]_0^{\pi/2}
$$
We find that it has only part of the magnitude of the two other second order
moments:
$$
\overline{xy} = \frac{R^2}{2 \pi} < \frac{R^2}{4}
$$
However, all second order moments must be calculated with respect to the center
of mass:
$$
\overline{x^2} - \overline{x}^2 =
\overline{y^2} - \overline{y}^2
= \frac{R^2}{4} - \left(\frac{4 R}{3 \pi}\right)^2
\EN
\overline{xy} - \overline{x}\overline{y}
= \frac{R^2}{2 \pi} - \left(\frac{4 R}{3 \pi}\right)^2
$$
Recall the formulas:
$$
\sigma'_{xx} = cos^2(\theta) \sigma_{xx} + sin^2(\theta) \sigma_{yy}
+ 2 sin(\theta) cos(\theta) \sigma_{xy}
$$ $$
\sigma'_{yy} = sin^2(\theta) \sigma_{xx} + cos^2(\theta) \sigma_{yy}
- 2 sin(\theta) cos(\theta) \sigma_{xy}
$$
In our case, we have typically the situation where the angle of rotation must
be $45^o$ ($\theta = \pi/4$), in order to arrive at the main moments of inertia:
$$
\sigma'_{xx} = \half \sigma_{xx} + \half \sigma_{yy} + 2 \half \sigma_{xy}
$$ $$
\sigma'_{yy} = \half \sigma_{xx} + \half \sigma_{yy} - 2 \half \sigma_{xy}
$$
Substitute the above values:
$$
\sigma'_{xx} = \frac{R^2}{4} - \left(\frac{4 R}{3 \pi}\right)^2
+ \left[ \frac{R^2}{2 \pi} - \left(\frac{4 R}{3 \pi}\right)^2 \right]
\approx 0.04890 \, R^2
$$ $$
\sigma'_{yy} = \frac{R^2}{4} - \left(\frac{4 R}{3 \pi}\right)^2
- \left[ \frac{R^2}{2 \pi} - \left(\frac{4 R}{3 \pi}\right)^2 \right]
\approx 0.09085 \, R^2
$$