How can I find the volume of a cone in terms of theta? I have been given these instructions:


*

*Cut out sector from a circle having central angle $\theta$ and radius r

*Form a cone from what's left of the circle (I thought of it as taking a circular piece of paper, cutting a pizza shape out of it, then push the rest of the paper together to make a cone)

*Then find volume of the cone in terms of $\theta$


These are the things I've done so far:


*

*Found the circumference of the cone by subtracting r $\theta$ from 2$\pi$r 

*Subtracted the area of the cut sector from the area of a whole circle


Now I am lost, I have a feeling that finding the radius in terms of $\theta$ is the next step I should take.
How can I do this?
 A: The angle of the sector not taken is $2\pi-\theta$. The length of the circular arc with central angle $2\pi-\theta$ radians and radius $r$ is $r(2\pi-\theta)=2\pi r-\theta r$ (by the definition of radian measure). When you form the cone, this will be the circumference of the circle at the base of the cone. The radius of that circle is $\frac{2\pi r-\theta r}{2\pi}=r\left(1-\frac{\theta}{2\pi}\right)$ (call it $R$).
The "slant height" of the cone is the radius of the original circle, $r$, and is the hypotenuse of a right triangle with base $R=r\left(1-\frac{\theta}{2\pi}\right)$ and height $h$ (let's say). Therefore
$$h=\sqrt{r^2-\left[ r\left(1-\frac{\theta}{2\pi}\right) \right]^2}$$
$$=r\sqrt{\frac{\theta}{\pi}-\frac{\theta^2}{4\pi^2}}$$
Therefore the volume of the cone will be
$$V=\frac 13\pi R^2h=\frac 13\pi\left[r\left(1-\frac{\theta}{2\pi}\right)\right]^2\left[r\sqrt{\frac{\theta}{\pi}-\frac{\theta^2}{4\pi^2}}\right]$$
$$=\frac 13\pi r^3\left(1-\frac{\theta}{2\pi}\right)^2\left(\sqrt{\frac{\theta}{\pi}-\frac{\theta^2}{4\pi^2}}\right)$$
You could simplify that further, as you like.
Note that the answer would have been significantly easier if you defined $\theta$ to be the central angle of the sector left after cutting rather than of the sector that was cut. If we define $\Theta$ as that central angle that was left, we end up with 
$$V=\frac 13\pi r^3\left(\frac{\Theta}{2\pi}\right)^2
\sqrt{1-\left(\frac{\Theta}{2\pi}\right)^2}$$
A: The radius $R$ of the cone is given as
$$2\pi R=(2\pi-\theta)r$$$$ \implies R=\frac{(2\pi-\theta)r}{2\pi}\tag 1$$ & the slant height of the cone will be $r$ hence its vertical height $H$ is given as $$H=\sqrt{(\text{slant height})^2-(\text{radius})^2}$$ $$=\sqrt{(r)^2-\left(\frac{(2\pi-\theta)r}{2\pi} \right)^2}$$ Hence, the volume of the cone $$=\frac{1}{3}\times (\pi R^2)(H)$$ $$=\frac{1}{3}\times \left(\pi \left(\frac{(2\pi-\theta)r}{2\pi}\right)^2\right)\left(\sqrt{(r)^2-\left(\frac{(2\pi-\theta)r}{2\pi} \right)^2}\right)$$ $$=\color{blue}{\frac{1}{3}\pi r^3 \left(1-\frac{\theta}{2\pi }\right)^2\sqrt{1-\left(1-\frac{\theta}{2\pi }\right)^2}}$$
A: in your sector, we have $\theta=\frac{l}{r}$ where $l$ is sector's arc length. so $l=r\theta$.
$l$ will become the cone's base perimeter. so if we call the base's radius as $R$, we have $r\theta=2\pi R$. now we can find $R$ to be $\frac{r\theta}{2\pi}$.
there is a right triangle in our cone which its smallest side is $\frac{r\theta}{2\pi}$, its vertical side is the cone's height and its hypotenuse (cone's lateral side) is $r$ (sector's radius).
use the Pythagorean rule to find the height. It will be $\frac{r}{2\pi} \sqrt{4\pi^2-\theta^2}$.
now you can find the volume: $V=\frac{1}{3} \pi R^2 h$ which will be $\frac{r^3\theta^2}{24\pi^2}\sqrt{4\pi^2-\theta^2}$.
note that $\theta$ is in radians here. so if you want to put it in degrees, you should write $180$ instead of $\pi$.
