volume and cube

**caligyrl4lyfe** · September 30th 2008, 08:21 PM

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?

**Chris L T521** · September 30th 2008, 08:51 PM

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 $SA=6s^2$ , where s is the side of the cube, and $SA=4\pi r$ , where r is the radius of the sphere.

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

Do these processes make sense?

--Chris

**caligyrl4lyfe** · September 30th 2008, 09:05 PM

I'm still a little confused.

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

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

**Chris L T521** · September 30th 2008, 09:09 PM

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 $ SA=4\pi r^{\color{red}2} $

Yes, where SA=1.

So you would be solving $1=6s^2$ for s, and $1=4\pi r^{\color{red}2}$ 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 $SA=4\pi r^{\color{red}2}$ , not $SA=4\pi r$ .

My mistake!

**caligyrl4lyfe** · September 30th 2008, 09:25 PM

So, is

s= 2.24

and

r=3.40

**Chris L T521** · September 30th 2008, 09:31 PM

Originally Posted by Chris L T521

Yes, where SA=1.

So you would be solving $1=6s^2$ for s, and $1=4\pi r^{\color{red}2}$ 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 $SA=4\pi r^{\color{red}2}$ , not $SA=4\pi r$ .

My mistake!

Originally Posted by caligyrl4lyfe

So, is

s= 2.24

and

r=3.40

How did you get those values?

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

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

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

--Chris

**caligyrl4lyfe** · September 30th 2008, 09:45 PM

WHEW!

I finally got it!

The sphere has a larger volume...

Thanks Chris~

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