I'm doing exercise 5/5 from STATICS Meriam 6th to find center of mass Y
that's the resolution:
1º - I couldn't understand well dA. Arc lenght (choosen centroid) is ** $ \pi * dr $ ** but why times R (R $ \pi $ dR)?
2º - From table of commons centroids I know $ Yc = 2 r/\pi $ but how I get into it?
best regards
Firstly, the result for $ \bar y$ = $ 2 r /\pi $ might have been given earlier to the Exercise 5/5 not shown here,for a semi-circular arc, like a wire and not the full area.
It is calculated as $ \bar y $ =$ \dfrac{\int y ds }{ \int ds } = \dfrac{\int r \cdot r \sin \theta d \theta }{ \pi r } $ = $\dfrac {2 r}{\pi} $
Secondly, when a thin ring is considered, differential area of a semicircular ring is its arc length multiplied by thickness.
Area of circle = $ \pi r^2 $ is already an advanced result from integration for full circle.
In methods of differential calculus we have a clear meaning for differential quantities.When area dA and thin radial slice dr are differentials we are allowed to treat area of ring as that of a thin "curved rectangle". When a thin annular semi circle is considered, differentials only are multiplied.
$ dA = 2 \pi r dr $ comes at first for thin curved rectangle/ ring and then only comes $ A = \pi r^2 $ after performing integration for the full arc.
Now
$$ dA = \pi r dr $$
$$ \int y_c dA = \int_{R/2}^R \frac {2 r}{\pi} \cdot \pi r dr $$
etc., hope all else is clear.