1. volume and cube

Given a fixed surface area for a cube and a sphere.

A= 1m^2

Determine the volume of the cube and the sphere. Which one has greater volume?

2. Originally Posted by caligyrl4lyfe
Given a fixed surface area for a cube and a sphere.

A= 1m^2

Determine the volume of the cube and the sphere. Which one has greater volume?

Well...the surface area of a square is $\displaystyle SA=6s^2$, where s is the side of the cube, and $\displaystyle SA=4\pi r$, where r is the radius of the sphere.

First find s and r, given that SA=1, then find the volume, noting that $\displaystyle V_{\text{cube}}=s^3$, and $\displaystyle V_{\text{sphere}}=\tfrac{4}{3}\pi r^3$.

Do these processes make sense?

--Chris

3. I'm still a little confused.

So, would I solve SA= 6s^2 for s.

And would I find r by solving $\displaystyle SA=4\pi r$

4. Originally Posted by caligyrl4lyfe
I'm still a little confused.

So, would I solve SA= 6s^2 for s.

And would I find r by solving $\displaystyle SA=4\pi r^{\color{red}2}$
Yes, where SA=1.

So you would be solving $\displaystyle 1=6s^2$ for s, and $\displaystyle 1=4\pi r^{\color{red}2}$ for r.

Then you would plug these values into the corresponding volume equations I gave you. Then see which one is larger.

--Chris

EDIT: I caught a tiny mistake on my part. The surface area of a sphere is given by $\displaystyle SA=4\pi r^{\color{red}2}$, not $\displaystyle SA=4\pi r$.

My mistake!

5. So, is

s= 2.24

and

r=3.40

6. Originally Posted by Chris L T521
Yes, where SA=1.

So you would be solving $\displaystyle 1=6s^2$ for s, and $\displaystyle 1=4\pi r^{\color{red}2}$ for r.

Then you would plug these values into the corresponding volume equations I gave you. Then see which one is larger.

--Chris

EDIT: I caught a tiny mistake on my part. The surface area of a sphere is given by $\displaystyle SA=4\pi r^{\color{red}2}$, not $\displaystyle SA=4\pi r$.

My mistake!
Originally Posted by caligyrl4lyfe
So, is

s= 2.24

and

r=3.40
How did you get those values?

$\displaystyle 1=6s^2\implies \tfrac{1}{6}=s^2\implies s=\sqrt{\tfrac{1}{6}}\approx \color{red}0.41$

$\displaystyle 1=4\pi r^2\implies r^2=\frac{1}{4\pi}\implies r=\sqrt{\frac{1}{4\pi}}\approx \color{red}0.28$.

Now that we know what s and r are, find $\displaystyle V_{\text{cube}}$ and $\displaystyle V_{\text{sphere}}$.

--Chris

7. WHEW!

I finally got it!

The sphere has a larger volume...

Thanks Chris~