Percent change in the volume of a sphere when the radius is doubled

Question

If the radius of a sphere is doubled, then what is the percent increase in volume? Please explain how to tackle this one.

When I double the radius I get $$V=\dfrac{4}{3}\pi(2r)^3$$ But how exactly would you calculate the percent increase?. Note that I am familiar with the percent increase/decrease formula, but have not had experience using that with volumes.

Answer 1

When a dimension double the volume increases 8 times thus the increment is

$$\frac{\Delta V}{V_0}=\frac{V-V_0}{V_0}=\frac{8V_0-V_0}{V_0}=7=700\%$$

Note that the resul is independent from the particular solid considered under an homotetic transformation.

Answer 2

The original volume is $\dfrac 43\pi r^3$.

Doubling the radius where $r\cdot 2=2r$, the volume becomes $V=\dfrac{4}{3}\pi(2r)^3=\dfrac{32}{3}\pi r^3$

What is the amount of increase between $4$ and $32$?

Another way to look at this:

Note that in $1$ dimension, the increase is $2$ times, in $2$ dimensions, the increase is $4$ times, so what is the increase in $3$ dimensions?

And don't forget to subtract $100\%$ after you perform the calculations.

Answer 3

Note: $$V'=\frac43\pi (2R)^3=8\cdot \frac43 \pi R^3=7\cdot \underbrace{\frac43\pi R^3}_{100\%}+\underbrace{\frac43\pi R^3}_{100\%}.$$

Answer 4

Percentages are calculated for small increases/decreases only. However in this case the change is large, but its definition cannot change.

Due to the cubic function dependence we have the result as 8 times multiplied with respect to the original, and by definition of $\%$ (percentage) the increase it is

$$ \dfrac{V2-V_1}{V_1}=700\% $$.

If a question were asked about the surface area with square function dependence, that would be $300\%$ change.