Amount of Air Method?

We have

V = (4/3)*pi*r^3

71 = (4/3) *pi * r^3    Solving for r , we have

71(3/4)/pi  = r^3         Taking the cube root of both sides, we have

r = 2.5687583235794156      And if the design is now 10cm wide, the dimensions of the balloon must have doubled....so doubling the radius, we have  2 * 2.5687583235794156 = 5.1375166471588312

So

V = (4/3)*pi * (5.1375166471588312)*3   = 567.99999cm^3 or  568cm^3

Obviously, Melody's path to the answer is more elegant than mine !!!

I think if a length dimension is 2*original then the volume is 2cubed=8 times original.

$$71*8=568cm^3$$

I'm another mathematician will look at my answer.

I learned that shortcut of Alan.  Nifty isn't it.    

Thanks Alan   ♬       

Yeah...I knew that "trick," too.....I just wanted to try a different approach..but yours is definitely preferable !!!   Just like the balloon......the amount of "hot air" on the forum tends to expand in a "cubic" fashion.....sometimes......

Thanks For Help Melody and CPhill!

$$V_1=\dfrac{4}{3}\pi r_1^3$$

$$V_2=\dfrac{4}{3}\pi r_2^3 \quad | \quad r_2=2r_1$$

$$V_2=\dfrac{4}{3}\pi (2r_1)^3= \underbrace{\dfrac{4}{3}\pi r_1^3}_{V_1}*\underbrace{2^3}_8$$

$$V_2=8V_1=8*71\;cm^3=568\;cm^3$$