# Volume of a Cone

#### bilalsaeedkhan

Can someone help me with calculating the volume of a Cone by taking a vertical cross-sectional area of the cone and integrating it? I know how to solve it for the horizontal cross-section, which gives an area of circle. But if we take a vertical cross-section then we get a Hyperbola, am I correct?

#### earboth

MHF Hall of Honor
Can someone help me with calculating the volume of a Cone by taking a vertical cross-sectional area of the cone and integrating it? I know how to solve it for the horizontal cross-section, which gives an area of circle. But if we take a vertical cross-section then we get a Hyperbola, am I correct? <=== No

View attachment 18295
If the cut is parallel to the axis of the cone then you'll get parabolas as cross-sections.

Have a look here: Conic section - Wikipedia, the free encyclopedia

#### bilalsaeedkhan

If the cut is parallel to the axis of the cone then you'll get parabolas as cross-sections.

Have a look here: Conic section - Wikipedia, the free encyclopedia
My case is the third figure in your link, which is a hyperbola. Is the cross-section that I have taken in my original attachment a parabola or a hyperbola?

#### fobos3

It is a lot easier to do it the other way but here it goes.

The equation of a cone is $$\displaystyle x^2+y^2=R^2z^2$$.

Suppose the radius of the base at $$\displaystyle z=h$$ is $$\displaystyle r$$. Then we have $$\displaystyle Rh=r$$ or $$\displaystyle R=\dfrac{r}{h}$$

Now we let $$\displaystyle z\in\left[0,h\right]$$

Next we have the equation of a general plane $$\displaystyle Ax+By+Cz=D$$. We want to construct a plane $$\displaystyle \mathcal{P}||(\hat{\textbf{j}},\hat{\textbf{k}})$$

This would mean that $$\displaystyle x=const$$ or $$\displaystyle \mathcal{P}\rightarrow x=x_0$$.

Now lets intersect this plane with our cone. This gives us the curve

$$\displaystyle R^2z^2-y^2=x_0^2$$ or $$\displaystyle y=\pm\sqrt{R^2z^2-x_0^2}$$

Now at $$\displaystyle z=h$$ $$\displaystyle y=\pm\sqrt{r^2-x_0^2}$$ which will be our limit of integration. Keep in mind that the vertex of the cone is at $$\displaystyle z=0$$. We know that $$\displaystyle z>0$$ therefore $$\displaystyle z=\dfrac{\sqrt{y^2+x_0^2}}{R}$$

Next we find the area $$\displaystyle A(x_0)=2hr-\int_{-\sqrt{r^2-x_0^2}}^{\sqrt{r^2-x_0^2}}\dfrac{\sqrt{y^2+x_0^2}}{R}\,dy=$$ and I'm going to stop here as it is getting quite messy. I guess that helps to a point but I don't see any point in continuing.

bilalsaeedkhan

#### bilalsaeedkhan

Thank you! It is complicated beyond my scope but thanks nonetheless.

#### earboth

MHF Hall of Honor
My case is the third figure in your link, which is a hyperbola. Is the cross-section that I have taken in my original attachment a parabola or a hyperbola?
Of course you are right and I apologize for the confusion.

Obviously a hyperbolic cross-section is hard to handle as fobos3 has demonstrated. So I suggest to use parabolic cross-sections.

According to Archimedes a segment of a parabola has an area

$$\displaystyle A = \frac43 \cdot \underbrace{\frac12 \cdot 2 b \cdot h}_{area\ of\ triangle} = \frac43 b h$$

Use the proportion $$\displaystyle \dfrac hs = \dfrac{2R-x}{2R}~\implies~h=\dfrac{s(2R-x)}{2R}$$

Use Pthagorian theorem to calculate the length of $$\displaystyle b = \sqrt{R^2-(R-x)^2}$$

Caution: This step probably contains a mistake which causes a wrong final result! I wasn't able to spot my error.

Now the colume of the cone should be:

$$\displaystyle V=\displaystyle{\int}_0^{2R}\left(\frac43 \cdot \dfrac{s(2R-x)}{2R} \cdot \sqrt{R^2-(R-x)^2}\right) dx$$

My computer tells me that $$\displaystyle V = \dfrac \pi3 \cdot R^2 \cdot s$$ which is definitely too large.

(... but it looks like a very near miss)

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bilalsaeedkhan

#### bilalsaeedkhan

Thanks, I'll try it again using your setup. hopefully i'll get something.