# 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?

• $r_{\text{base}}=\frac{2\pi-\theta}{2\pi}r,\,h=\sqrt{r^2-r_{\text{base}}^2}$ Jun 10, 2015 at 19:18
• For the poor non native English speakers... Can you precise what you mean by "after removing the sector form what's left of the circle into a cone"? Jun 10, 2015 at 19:21
• Dictate clearly as your 2nd sentence is not clear that need be clarified by re-editing. Jun 10, 2015 at 19:30

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

$$V=\frac 13\pi r^3\left(\frac{\Theta}{2\pi}\right)^2 \sqrt{1-\left(\frac{\Theta}{2\pi}\right)^2}$$

• This is exactly what I needed! Thank you so much. Jun 10, 2015 at 20:29

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 $$=\frac{1}{3}\times (\pi R^2)(H)$$ $$=\frac{1}{3}\times \left(\pi \left(\frac{(2\pi-\theta)r}{2\pi}\right)^2\right)\left(\sqrt{(r)^2-\left(\frac{(2\pi-\theta)r}{2\pi} \right)^2}\right)$$ $$=\color{blue}{\frac{1}{3}\pi r^3 \left(1-\frac{\theta}{2\pi }\right)^2\sqrt{1-\left(1-\frac{\theta}{2\pi }\right)^2}}$$

• Blue solution? . Jun 14, 2015 at 22:55
• Yes, you are right. Jun 14, 2015 at 22:56
• In case you have any doubt regarding my answer or even question elsewhere, I will clarify it as much as I can. Jun 14, 2015 at 22:59

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 $$\pi$$.