Answer
$\dfrac{\pi rh_0 dh}{\sqrt{r^2+h_0^2}}$
Work Step by Step
Here, $\dfrac{ds}{dh}=(\pi r)\dfrac{1}{2\sqrt{r^2+h^2}}(2h)$
and
$\dfrac{ds}{dh}=\dfrac{\pi rh}{\sqrt{r^2+h^2}}$
or, $ds=\dfrac{\pi rh dh}{\sqrt{r^2+h^2}}$
when $h$ changes from $h_0$ to $h_0+dh$, then the change in the lateral surface area becomes: $\dfrac{\pi rh_0 dh}{\sqrt{r^2+h_0^2}}$
