check the picture below.
so, if we grab the whole volume of the watermelon, and then subtract the volume of the inner sphere of 9/10, what we're left with is 1/10 of the volume, which is the volume of the rind, recall the rind has a radius of 1/10 of the radius, so we're simply taking off 1/10 from the radius and using the 9/10 and subtracting the volume with that radius.
now, after the 5th week, the radius of the watermelon has then grown to 2*5, or 10 cm, and the rind is 1/10 of that or just 1 cm.
[tex]\bf \cfrac{4\pi \left( \frac{9r}{10} \right)^3}{3}\implies \cfrac{4\pi \cdot \frac{729r^3}{1000}}{3}\implies \cfrac{\frac{2916\pi r^3}{1000}}{3}\implies \cfrac{2916\pi r^3}{3\cdot 1000}\implies \cfrac{243\pi r^3}{250}\\\\
-------------------------------\\\\
V_{rind}=\cfrac{4\pi r^3}{3}~~-~~\cfrac{243\pi r^3}{250}\implies V_{rind}=\pi r^3\left(\cfrac{4}{3}-\cfrac{243}{250} \right)[/tex]
[tex]\bf V_{rind}=\pi r^3\left( \cfrac{271}{250}\right)\implies \boxed{V_{rind}=\cfrac{271\pi }{250}r^3}
\\\\\\
\cfrac{dV_{rind}}{dt}=\cfrac{271\pi }{250}\cdot \stackrel{chain~rule}{3r^2\cdot \cfrac{dr}{dt}}\implies \cfrac{dV_{rind}}{dt}=\cfrac{813\pi r^2\cdot \frac{dr}{dt}}{250}
\\\\\\
\begin{cases}
\frac{dr}{dt}=2\\\\
r=10
\end{cases}\implies \cfrac{dV_{rind}}{dt}=\cfrac{813\pi 10^2\cdot 2}{250}\implies \cfrac{dV_{rind}}{dt}\approx 3043.29[/tex]
