Geometrically demonstrating that the integral of πr^2 matches the volume of a cone with radius and height of r

Does anyone have a geometric demonstration that the integral of πr2 matches the volume of a cone with radius and height of r?

integration geometry

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

I hope the diagram is self-explanatory.

Consider a cone of radius $R$ , height $H$ and semi-vertical angle $\theta$ . Now from the apex point, consider an arbitrary disk at a distance $h$ which is centered around the axis of the cone. Let this disk have a radius of $r$ and a thickness of $dr$ . The volume of this disk is:

$dV = \pi r^2 dh$

Now, note that, $r = h\tan{\theta}$

So from here,

$dV = \pi r^2 dh = \pi (h\tan{\theta})^2 dh$

Integrating with respect to $h$ from $0$ to $H$ , we get:

$V = \frac{1}{3}\pi H^3 \tan^2{\theta}$

Again, for the cone, $R = H\tan{\theta}$ . Using this and the above formula:

$V = \frac{1}{3}\pi H (H^2\tan^2{\theta})$

$\boxed{V = \frac{1}{3}\pi R^2 H}$

So for a cone of radius and height being equal, you can obtain the required expression from here. It just so happens that the integral of the area of a circle gives the volume of a cone in this special case. This is because, when the height is equal to the radius, then $\tan{\theta}=1$ . Then $h = r \implies dh = dr$ and $dV$ becomes:

$dV = \pi (h\tan{\theta})^2 dh=\pi r^2 dr$

Karan Chatrath - 1 year, 9 months ago

Very good! That's right! Can you draw the diagram, please?

Go!Game RJ - 1 year, 9 months ago

I have added a diagram.

Karan Chatrath - 1 year, 9 months ago

Go!Game RJ - 1 year, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$