## Calculus 8th Edition

$$128\pi*cm^2/min$$
We can write the surface area as, $$A = 4*\pi*r^2$$ Taking the derivative of both sides with respect to t, $$dA/dt = 8\pi r*dr/dt$$ We know $dr/dr$ is 2 cm/min, and $r=8 cm$. Plugging those in, $$dA/dt = 8*\pi*8*2 cm^2/min = 128\pi*cm^2/min$$