The Radius of sphere A is twice that of Sphere B. Find, (i) the ratio among their surface areas. (ii) the ratio among their volumes

I think that this is all that Heureka did.  It is all that I would normally do.  

if the ratio if a 1 dimensional measurement is   $$a:b$$     (the radius is a 1 dimensional measurement)

then the ration of the 2 dimensional measurements (surface area) is just     $$a^2:b^2$$

and the ratio of the 3 dimensional measurement (volume) is just   $$a^3:b^3$$

so if ratio of radius is 1:2

then ratio of SA is  1:4

and ratio of Volume is 1:8

(i) ratio among their surface areas is 4:1.

(ii) the ratio among their volumes is 8:1.

but how plz explain with steps.

(i) ratio among their surface areas is 4:1.

$$r_a=2r_b$$

$$\begin{array}{rcl} Surface_A &=& 4\pi(2r_B)^2 \\ Surface_B &=& 4\pi(r_B)^2 \\ \hline \\ \frac{Surface_A}{Surface_B} &=& \frac{4\pi(2r_B)^2}{4\pi(r_B)^2 } \\\\ \frac{Surface_A}{Surface_B} &=& \frac{4\pi (2)^2 (r_B)^2}{4\pi(r_B)^2 }=2^2 = 4= \frac{4}{1} \\\\ \end{array}$$

(ii) the ratio among their volumes is 8:1.

$$\begin{array}{rcl} V_A &=& \frac{4}{3}\pi(2r_B)^3 \\\\ V_B &=& \frac{4}{3}\pi(r_B)^3 \\\\ \hline \\ \frac{V_A}{V_B} &=& \frac{\frac{4}{3}\pi(2r_B)^3}{\frac{4}{3}\pi(r_B)^3 } \\\\ \frac{V_A}{V_B} &=& \frac{\frac{4}{3}\pi(2)^3(r_B)^3}{\frac{4}{3}\pi(r_B)^3 } = 2^3 = 8 = \frac{8}{1} \\\\ \end{array}$$