Volume and Surface Area of a Sphere

**benny92000** · October 22nd 2011, 07:59 AM

If the volume of the sphere doubles, what is the ratio of the surface area of the new, larger sphere to the old, smaller sphere? The answer is 2^(2/3). A good method for tackling these types of problems would be great. Should I just plug in numbers and work that way, or is there a more methodical way of doing it?

**TKHunny** · October 22nd 2011, 08:16 AM

Write it down! The notation will help you.

$\frac{4}{3}\pi r_{1}^{3} = V_{1}$

$\frac{4}{3}\pi r_{2}^{3} = V_{2}$

$\frac{\frac{4}{3}\pi r_{2}^{3}}{\frac{4}{3}\pi r_{1}^{3}} = \frac{V_{2}}{V_{1}}$

$\frac{r_{2}^{3}}{r_{1}^{3}} = \frac{V_{2}}{V_{1}}$

$V_{2} = 2\cdot V_{1}$

$\frac{r_{2}^{3}}{r_{1}^{3}} = \frac{2\cdot V_{2}}{V_{1}}$

$\frac{r_{2}^{3}}{r_{1}^{3}} = 2$

Are we getting anywhere?

The question is: $\frac{4 \pi r_{2}^{2}}{4 \pi r_{1}^{2}} = What?$

**benny92000** · October 22nd 2011, 10:24 AM

I'm still a little confused. I get steps 1-3, but I'm not sure exactly what you are doing in 4,5,6, and 7. Thanks for the help!

**bjhopper** · October 22nd 2011, 11:38 AM

Hi benny92000,
SA of sphere = pi d^2
V of sphere = pi/6 * d^3
let d1 =1 SA= pi V=pi/6
V2 = 2* V1 = pi/3
let X = the new diameter
pi/6 *X^3=pi/3
X^3 =2
X= 2^1/3
SA2 = pi*X^2=pi X^2 = pi * 2^2/3

**earboth** · October 22nd 2011, 11:46 AM

Originally Posted by TKHunny

Write it down!
...
$V_{2} = 2\cdot V_{1}$

$\frac{r_{2}^{3}}{r_{1}^{3}} = \frac{2\cdot V_{2}}{V_{1}}$ <-- small typo: Instead of $V_2$ it should be $V_1$

...
Are we getting anywhere?

The question is: $\frac{4 \pi r_{2}^{2}}{4 \pi r_{1}^{2}} = What?$

Originally Posted by benny92000

I'm still a little confused. I get steps 1-3, but I'm not sure exactly what you are doing in 4,5,6, and 7. Thanks for the help!

1. TKHunny showed you how to express the radius of the larger sphere ( $r_2$ ) by the radius of the smaller sphere:

$\frac{r_2^3}{r_1^3}=2~\implies~r_2 = \sqrt[3]{2} \cdot r_1$

2. Now calculate the ratio of the surfaces:

$\frac{4 \pi r_2^2}{4 \pi r_1^2}$

Sub in the term for $r_2$ :

$\frac{4 \pi (\sqrt[3]{2} \cdot r_1)^2}{4 \pi r_1^2} = \frac{4 \pi \sqrt[3]{2^2} \cdot r_1^2}{4 \pi r_1^2}$

3. Cancel equal factors and re-write the cube-root term as a power.

**TKHunny** · October 22nd 2011, 07:29 PM

Originally Posted by earboth

small typo

Gaa!!! You know how long I stared at that?! Anohter pair of eyes never hurts.

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