a=2R

$\frac{b}{2}=\sqrt {R^2-y^2}$

$b^2=4(R^2-y^2)$ (i)

$A \bar y= \int day$

Now, $I=\frac{\pi R^4}{4}$

$A \bar y= \int_{y}^{R} bdy *y$

$b^2=4(R^2-y^2)$

$2b.db=4*(-2y.dy)$

$b.db=-4y.dy$

$y.dy=\frac{-b.db}{4}$

$A \bar y= \int_{b}^{0} b(-\frac{b.db}{4})$

$A \bar y= \frac{b^3}{12}$

$q_{11}=\frac{F}{Ib}*A \bar y$

$=\frac{F}{Ib}*\frac{b^3}{12}= \frac{Fb^2}{12I}$

$\frac{F}{12I}*4*(R^2-y^2)$ from (i)

q will be minimum (zero) when y=R

q will be …