Consider a cone with height [latex]h[/latex] and base radius [latex]r[/latex]. The cone can be viewed as being composed of an infinite number of thin disks stacked along its height.

We slice the cone horizontally at a height [latex]y[/latex] from the base. At this height, the radius of the disk [latex]r(y)[/latex] varies linearly from [latex]0[/latex] at the top to [latex]r[/latex] at the base.

The relationship between [latex]r(y)[/latex] and [latex]y[/latex] can be expressed using similar triangles.

Since the height of the cone is [latex]h[/latex] and the radius at the base is [latex]r[/latex], we have:

[latex]\frac{r(y)}{y} = \frac{r}{h} \implies r(y) = \frac{r}{h} y[/latex]

Each thin disk has a thickness [latex]dy[/latex] and a radius [latex]r(y)[/latex].

The volume [latex]dV[/latex] of a thin disk at height [latex]y[/latex] is the area of the disk times its thickness:

[latex]dV = \pi [r(y)]^2 dy[/latex]

Substitute [latex]r(y) = \frac{r}{h} y[/latex] into the volume element expression:

[latex]dV = \pi \left( \frac{r}{h} y \right)^2 dy = \pi \frac{r^2}{h^2} y^2 dy[/latex]

To find the total volume of the cone, integrate [latex]dV[/latex] from the base ([latex]y = 0[/latex]) to the top ([latex]y = h[/latex]):

[latex]V = \int_0^h \pi \frac{r^2}{h^2} y^2 dy[/latex]

Evaluate the integral:

[latex]V = \pi \frac{r^2}{h^2} \int_0^h y^2 dy[/latex]

The integral of [latex]y^2[/latex] with respect to [latex]y[/latex] is:

[latex]\int y^2 dy = \frac{y^3}{3}[/latex] So, we have: [latex]V = \pi \frac{r^2}{h^2} \left[ \frac{y^3}{3} \right]_0^h = \pi \frac{r^2}{h^2} \left( \frac{h^3}{3} - 0 \right) = \pi \frac{r^2}{h^2} \cdot \frac{h^3}{3} = \frac{1}{3} \pi r^2 h[/latex]

Thus, we have derived the formula for the volume of a circular cone:

[latex]V = \frac{1}{3} \pi r^2 h[/latex]

Visualize the cone with its base on the [latex]xy[/latex]-plane and apex at [latex](0, 0, h)[/latex]. Cross-sections perpendicular to the [latex]z[/latex]-axis are circles

Find the radius of a circular cross-section at height [latex]z[/latex]:

[latex]\frac{r}{h} = \frac{r - R}{h - z}[/latex], where [latex]R[/latex] is the radius at height [latex]z[/latex]

Solving for R:

[latex]R = r - \frac{rz}{h}[/latex]

Area of the circular cross-section:

[latex]A(z) = \pi R^2 = \pi (r - \frac{rz}{h})^2[/latex]

Set up the integral:

[latex]V = \int_0^h A(z) dz = \int_0^h \pi (r - \frac{rz}{h})^2 dz[/latex]

Evaluate the integral:

[latex]\begin{array}{rcl}
V &=& \pi \int_0^h (r^2 - \frac{2r^2z}{h} + \frac{r^2z^2}{h^2}) dz \\
&=& \pi [r^2z - \frac{r^2z^2}{h} + \frac{r^2z^3}{3h^2}]_0^h \\
&=& \pi (r^2h - \frac{r^2h}{1} + \frac{r^2h}{3}) \\
&=& \frac{1}{3}\pi r^2h
\end{array}[/latex]

This result matches the known formula for the volume of a cone, validating our use of the slicing method.

Cross-sectional area:

[latex]A(x) = \pi [f(x)]^2 = \pi (4 - x^2)^2[/latex]

Set up the integral:

[latex]V = \int_0^2 \pi (4 - x^2)^2 dx[/latex]

Evaluate:

[latex]\begin{array}{rcl}
V &=& \pi \int_0^2 (16 - 8x^2 + x^4) dx \\
&=& \pi [16x - \dfrac{8x^3}{3} + \dfrac{x^5}{5}]_0^2 \\
&=& \pi (32 - \dfrac{64}{3} + \dfrac{32}{5}) \\
&=& \frac{256\pi}{15}
\end{array}[/latex]

Cross-sectional area:

[latex]A(x) = \pi [f(x)]^2 = \pi (x + 1)^2[/latex]

Set up the integral:

[latex]V = \int_1^3 \pi (x + 1)^2 dx[/latex]

Evaluate:

[latex]\begin{array}{rcl}
V &=& \pi \int_1^3 (x^2 + 2x + 1) dx \\
&=& \pi [\frac{x^3}{3} + x^2 + x]_1^3 \\
&=& \pi (9 + 9 + 3 - \frac{1}{3} - 1 - 1) \\
&=& \frac{56\pi}{3}
\end{array}[/latex]

Cross-sectional area:

[latex]A(x) = \pi [f(x)]^2 = \pi (\frac{1}{x})^2[/latex]

Set up the integral: [latex]V = \int_1^\infty \pi (\frac{1}{x})^2 dx[/latex]

Evaluate:

[latex]\begin{array}{rcl}
V &=& \pi \int_1^\infty \frac{1}{x^2} dx \\
&=& \pi [-\frac{1}{x}]_1^\infty \\
&=& \pi (0 - (-1)) \\
&=& \pi
\end{array}[/latex]

This example, known as Gabriel’s Horn, shows a solid with finite volume but infinite surface area.

[latex]8\pi[/latex] units³

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end. You can view the transcript for this segmented clip of “2.2 Determining Volumes by Slicing” here (opens in new window).