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I was looking at a water bottle and trying to find the volume for fun, but couldnt remember the formula off the top of my head. So what came to me was i'd just think of the height of the cone and radius of the base as the sides of a right triangle, and then give said triangle a spin around the z-axis/center of the cone overlapping with its height. How do i go about doing that? Possibly integrate using the circumference of the circle on a third axis and then assuming the resulting shape will be of equal volume?

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    $\begingroup$ Hey, welcome to site, give Pappus's centroid theorem a try. $\endgroup$ Jan 19 '20 at 17:32
  • $\begingroup$ Just to clarify: you are assuming that the water bottle is a right circular cone with base in the $xy$-plane and centered on the $z$-axis? $\endgroup$ Jan 19 '20 at 17:33
  • $\begingroup$ That is correct @Sam. Thanks, I'll look it up @ K.K. $\endgroup$ Jan 19 '20 at 17:37
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Let $r$ be the base radius of the cone and $h$ be the height of the cone.

On an $xy$-plane, find the straight-line equation of your orthogonal triangle:

$$\begin{align*} \frac yh + \frac xr &= 1\\ y &= -\frac{h}{r}x+ h \end{align*}$$

Find the volume of revolution around the $y$-axis for the line from $x=0$ to $x=r$,

$$\begin{align*} V &= \int_0^r 2\pi x \left(-\frac{h}{r}x+ h\right) dx\\ &= 2\pi h\int_0^r \left(-\frac{1}{r}x^2+ x\right) dx\\ &= 2\pi h\left[-\frac{x^3}{3r}+ \frac{x^2}2\right]_0^r\\ &= 2\pi h\left(-\frac{r^3}{3r}+ \frac{r^2}2\right)\\ &= 2\pi h\left(\frac{r^2}6\right)\\ &= \frac{1}3\pi r^2 h\\ \end{align*}$$

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This approach uses triple integrals. Typically, one obtains volumes of certain geometric objects by evaluating integrals such as: \begin{eqnarray} \int_{V} dxdydz \tag{1} \label{1} \end{eqnarray} where $V$ is an appropriate domain of integration in $\mathbb{R}^{3}$. Now, let us consider an 'upside-down' circular cone with vetex at the origin, base radius $R=a$ and height $h$. Then, an appropriate domain of integration in (\ref{1}) would be: \begin{eqnarray} \frac{h}{a}\sqrt{x^{2}+y^{2}} \le z \le h \quad \mbox{and} \quad 0 \le x^{2}+y^{2} \le a^{2} \tag{2}\label{2} \end{eqnarray} This reduces the integral (\ref{1}) to: \begin{eqnarray} \int_{x^{2}+y^{2}\le a^{2}}(h- \frac{h}{a}\sqrt{x^{2}+y^{2}})dxdy \tag{3}\label{3} \end{eqnarray}

Using polar coordinates, $x=rcos\theta$, $y=r\sin\theta$ with $0 \le r \le a$, $0 \le \theta \le 2\pi$ and $dxdy \to rdrd\theta$, we get: \begin{eqnarray} \int_{x^{2}+y^{2}\le a^{2}}\frac{h}{a}\sqrt{x^{2}+y^{2}}dxdy = \frac{h}{a}\int_{0}^{a}\int_{0}^{2\pi}r^{2}drd\theta = 2\pi \frac{h}{a} \frac{a^{3}}{3} = \frac{1}{3}2\pi a^{2}h \end{eqnarray} and also: \begin{eqnarray} \int_{x^{2}+y^{2}\le a^{2}} h dxdy = h \frac{a^{2}}{2}2\pi \end{eqnarray} so that (\ref{3}) is now given by: \begin{eqnarray} \pi a^{2} h - \frac{2}{3}\pi a^{2} h = \frac{1}{3}\pi a^{2}h \end{eqnarray}

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I find it easier to integrate the area along the axis of revolution.

Since the cross section here is circular, the area is $$A(r) = \pi r^2 \tag{1}\label{EQ1}$$ Let's say we have a right circular cone of height $h$. The radius is $0$ at $z=0$, and $R$ at $z=h$: $$r(z) = R \frac{z}{h} \tag{2}\label{EQ2}$$ The volume is the integral of the cross sectional area over the total height, $$V = \int_{0}^{h} A(r(z)) d z \tag{3}\label{EQ3}$$ Expanding $\eqref{EQ3}$ we get $$V = \int_{0}^h \pi \left( R \frac{z}{h} \right)^2 d z = \frac{\pi R^2}{h^2} \int_{0}^{h} z^2 d z = \frac{1}{3} \pi h R^2 \tag{4}\label{EQ4}$$

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