# sphere cone problem

• March 8th 2009, 09:19 AM
sphere cone problem
The radius of a sphere is 3 cm. The radius of the base of a cone is also 3 cm.
The volume of the sphere is 3 times the volume of the cone.

Work out the curved surface area of the cone, giving your answer as a multiple of pi.

Thanks.
• March 8th 2009, 09:51 AM
Volume of sphere and come
Quote:

The radius of a sphere is 3 cm. The radius of the base of a cone is also 3 cm.
The volume of the sphere is 3 times the volume of the cone.

Work out the curved surface area of the cone, giving your answer as a multiple of pi.

Thanks.

The volume of a sphere is $\frac{4}{3}\pi r^3$, and the volume of a cone is $\frac{1}{3}\pi r^2h$. So:

$\frac{4}{3}\pi r^3 = 3\times \frac{1}{3}\pi r^2h = \pi r^2h$

$\Rightarrow h = \frac{4}{3}r = 4$ cm.

Can you finish it now?

• March 8th 2009, 11:13 AM
earboth
Quote:

The radius of a sphere is 3 cm. The radius of the base of a cone is also 3 cm.
The volume of the sphere is 3 times the volume of the cone.

Work out the curved surface area of the cone, giving your answer as a multiple of pi.

Thanks.

Let s denote the slanted height, r the radius of the base circle and h the height of the cone. Then the curved surface of a cone is a sector of a circle with radius s. The area of the curved surface, the area of the complete circle, the circumference of the base circle = arc of the sector and the circumference of the complete circle are directly proportional:

$\dfrac{a_{cs}}{\pi s^2} = \dfrac{2\pi r}{2\pi s}~\implies~a_{cs} = \dfrac{2\pi r}{2\pi s} \cdot \pi s^2 = \boxed{\pi r s}$
• March 8th 2009, 11:17 AM
Quote:

Hello GAdamsThe volume of a sphere is $\frac{4}{3}\pi r^3$, and the volume of a cone is $\frac{1}{3}\pi r^2h$. So:

$\frac{4}{3}\pi r^3 = 3\times \frac{1}{3}\pi r^2h = \pi r^2h$

$\Rightarrow h = \frac{4}{3}r = 4$ cm.

Can you finish it now?

Hmm...not sure.
• March 8th 2009, 11:25 AM
Surface area of cone

earboth has given you the formula for the curved surface area of a cone. You now need to work out the value of $s$, which you can do using Pythagoras' Theorem. (It's the easiest right-angled triangle of all!)

Then plug the values of $r$ and $s$ into this formula, and you're there.

• March 9th 2009, 12:37 PM
Quote:

earboth has given you the formula for the curved surface area of a cone. You now need to work out the value of $s$, which you can do using Pythagoras' Theorem. (It's the easiest right-angled triangle of all!)

Then plug the values of $r$ and $s$ into this formula, and you're there.

s = root 7

Is teh answer 3 pi root7 ?
• March 9th 2009, 12:44 PM
earboth
Quote:

... The radius of the base of a cone is also 3 cm.
...

Quote:

[SIZE=3]...

$\Rightarrow h = \frac{4}{3}r = 4$ cm.

...

Quote:

Hmm...not sure.

Since r = 3 cm and h = 4 cm the slanted height s is:

$s^2 = r^2 + h^2~\implies~ s = 5\ cm$

Therefore the curved surface area is:

$a_{cs} = \pi\cdot 3\ cm \cdot 5\ cm = 15 \pi\ cm^2$
• March 9th 2009, 01:09 PM
Quote:

Originally Posted by earboth
Since r = 3 cm and h = 4 cm the slanted height s is:

$s^2 = r^2 + h^2~\implies~ s = 5\ cm$

Therefore the curved surface area is:

$a_{cs} = \pi\cdot 3\ cm \cdot 5\ cm = 15 \pi\ cm^2$

Ah yes, I made a basic error. I have it now. Thanks!
• March 14th 2009, 04:48 AM
osmosis786
Dont understand
Quote:

Hello GAdamsThe volume of a sphere is $\frac{4}{3}\pi r^3$, and the volume of a cone is $\frac{1}{3}\pi r^2h$. So:

$\frac{4}{3}\pi r^3 = 3\times \frac{1}{3}\pi r^2h = \pi r^2h$

$\Rightarrow h = \frac{4}{3}r = 4$ cm.

Can you finish it now?

Hi i dont understand why you hve multiplied the volume of the cone by 3

• March 14th 2009, 05:58 AM
Sphere and cone
Hello osmosis786
Quote:

Originally Posted by osmosis786
Hi i dont understand why you hve multiplied the volume of the cone by 3

I think you're going to kick yourself!

Quote:

The radius of a sphere is 3 cm. The radius of the base of a cone is also 3 cm.

Quote:

The volume of the sphere is 3 times the volume of the cone.

Work out the curved surface area of the cone, giving your answer as a multiple of pi.

Thanks.