Connect with knowledgeable experts and enthusiasts on IDNLearn.com. Join our community to receive timely and reliable responses to your questions from knowledgeable professionals.
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]