# sphere cone problem

• January 31st 2010, 09:45 AM
sphere cone problem
If a cone with perpendicular height of 6h and radius 2h has the same volume as a sphere of radius r, show that r= cuberoot 6h.

I got:

1/3 pi r squared x h = 4/3 pi x r cubed substitute

1/3 pi (6h) squared = 4/3 pi x r cubed divide by 3

pi (2h) squared x 6h = 4 pi r cubed divide by pi

(2h) squared x 6h = 4 r cubed

24h cubed = 4r cubed

...where am i going wrong?
• January 31st 2010, 10:18 AM
1005
Quote:

If a cone with perpendicular height of 6h and radius 2h has the same volume as a sphere of radius r, show that r= cuberoot 6h.

I got:

1/3 pi r squared x h = 4/3 pi x r cubed substitute

1/3 pi (6h) squared = 4/3 pi x r cubed divide by 3

pi (2h) squared x 6h = 4 pi r cubed divide by pi

(2h) squared x 6h = 4 r cubed

24h cubed = 4r cubed

...where am i going wrong?

Things seem to be going wrong once you've substituted. Why are you squaring the height (6*h)?

Let me show you how to do it:
$\frac{1}{3}\pi (R_1)^2 H_1=\frac{4}{3}\pi (R_2)^3$
$R_1=2h$
$H_1=6h$
$R_2=r$
$\frac{1}{3}\pi [2h]^2 [6h]=\frac{4}{3}\pi [r]^3$
Where brackets signify substitution.

First, I'll multiply both sides by $\frac{4}{3}$ and divide both sides by $\pi$:
$\frac{1}{4} (2h)^2 (6h)=r^3$
Next, I'll distribute the squared out:
$\frac{1}{4} 4h^2 (6h)=r^3$
The four from what I just did cancels with the $\frac{1}{4}$. Also, I will combine the $h^2$ and the $h$.
$6h^3=r^3$
Take the third root of both sides, and you're done!
$r=h\sqrt[3]{6}$