How can I find the change in the surface area of this problem?
I tried finding the different areas of the cylinder.
For its sides, I get:
$$A_\mathrm{side}=2\pi rh,$$
for the half sphere:
$$A_\mathrm{sphere}=2\pi r^3,$$
for the circle in the bottom
$$A_\mathrm{circle}=\pi r^2,$$
for total surface area:
$$S=2\pi rh+2\pi r^3+\pi r^2,$$
since
$$\frac{dS}{dt}=\frac{\partial{S}}{\partial{r}}\frac{dr}{dt}+\frac{\partial{S}}{\partial{h}}\frac{dh}{dt}.$$
So:
$$\frac{dS}{dt}=2\pi\left[-\frac{t}{2}(h+3r^2+r)+6rt\right]\label{1}\tag{1}$$
Given the values
$$t=4;\:r(4)=4;\:h(4)=50,$$
when I evaluate \eqref{1} I get:
$$\frac{dS}{dt}=-216\pi$$
while the correct answer is
$$
\frac{dS}{dt}=-56\pi.
$$
Now, what am I doing wrong?