Cone constructed from circular paper w/ 3-inch radius by....

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iheartthemusic29

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A cone is constructed from a circular piece of paper with a 3-inch radius by cutting out a sector of the circle with arc length x. The two edges of the remaining portion are joined together to form a cone with radius r and height h. Express the volume V of the cone as a function of x.

I got to V=1/3[pi(3-(x/2pi)^2)][sq. rt.(9-(3-(x/2pi)^2)]

I foiled after that (is that the next step?) and got:
V=1/3[pi(9-(6pi/2pi)+(x^2/4pi^2)][sq. rt.(9-9-(6x/2pi)+(x^2/(4pi^2))]

I then distributed the 1/3 and simplified and got:
V=[3pi-x+(x^2)/(12pi)][sq. rt.((-6x/2pi)+(x^2)/(4pi^2))]

I don't know where to go from there. Is everything I've done so far correct or did I make a mistake and, if so, where?
I'd really appreciate some help! :shock:
 
iheartthemusic29 said:
A cone is constructed from a circular piece of paper with a 3-inch radius by cutting out a sector of the circle with arc length x. The two edges of the remaining portion are joined together to form a cone with radius r and height h. Express the volume V of the cone as a function of x.

I got to V=1/3[pi(3-(x/2pi)^2)][sq. rt.(9-(3-(x/2pi)^2)]
Click to expand...
How did you "get to" this point? Since you have expressed "V" as a function of "x", and since this is what the exercise specified, what more do you feel you need to do?

Please be complete. Thank you! :D

Eliz.
 
Well, I was trying to avoid typing all this out...
V=1/3(pir^2)(h)
circumference =pi*diameter=2pir=2pi*3=6pi
circumference-arc length=6pi-x
2pir=6pi-x
r=(6pi-x)/(2pi)=3-(x/2pi)
Area of base=pir^2=pi(3-(x/2pi))^2
In the diagram of the cone, there is a right angle. The radius is now "x" and the hypotenuse of the right angle is 3 inches.
3^2-r^2=h^2
h=sq. rt.(9-(r^2))
I then plugged everything in and that is how I got to:
V=1/3[pi(3-(x/2pi)^2)][sq. rt.(9-(3-(x/2pi)^2)]

As for my final answer, I know that I have expressed V as a function of x, but am wondering if it can be simplified more.
Thanks!
 
\(\displaystyle V = \frac{\pi}{3}\left(3 - \frac{x}{2\pi}\right)^2 \sqrt{9 - \left(3 - \frac{x}{2\pi}\right)^2}\)

leave it as is (sometimes it's better to not oversimplify ... you may end up with something that's not so simple)

... or you write it as

\(\displaystyle V = \frac{\pi}{3}r^2 \sqrt{9 - r^2}\)

where

\(\displaystyle r = 3 - \frac{x}{2\pi}\)

which says it all.
 
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