Firstly, the result for y¯ = 2r/π 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 y¯ =∫yds∫ds=∫r⋅rsinθdθπr = 2rπ
Secondly, when a thin ring is considered, differential area of a semicircular ring is its arc length multiplied by thickness.
Area of circle = πr2 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=2πrdr comes at first for thin curved rectangle/ ring and then only comes A=πr2 after performing integration for the full arc.
the geometric 3/4of the straight line from initial to final.
objects moving in fluids faces the fluid friction opposes their motion. so to reduce the fluid friction objects moving in fluids must have special shapes called streamlined shapes.