Answer :
Solution
For this case we can do the following:
[tex]SA=4\pi r^{2}[/tex]Replacing we got:
[tex]SA_1=4\pi(4\operatorname{mm})^2=201.06\operatorname{mm}^{2}[/tex][tex]SA_2=4\pi(5\operatorname{mm})^2=314.16\operatorname{mm}^{2}[/tex]Then the surface area for the first one is 201.06mm² and the second one 314.16mm²
And the volume is:
[tex]V_1=\frac{4}{3}\pi(4\operatorname{mm})^3=268.083\operatorname{mm}^{3}[/tex][tex]V_2=\frac{4}{3}\pi(5\operatorname{mm})^3=523.599\operatorname{mm}^{3}[/tex]