Volumes with Known Cross Sections

Volume from cross sections

Slice a solid with planes perpendicular to the

x

-axis. Suppose the slice at

x

has area

A(x)

. A slab of thickness

\Delta x

there has volume about

A(x)\,\Delta x

, and adding the slabs gives a Riemann sum.

Most solids here stand on a region in the

xy

-plane. Each cross section stands on a segment of the region, of length

s

, and it's the shape the problem names. The area of that shape in terms of

s

A

Cross section on a side $s$ in the base	Area
square	$s^2$
rectangle of height $h$	$sh$
rectangle of height $k$ times its base	$ks^2$
equilateral triangle	$\frac{\sqrt{3}}{4}s^2$
isosceles right triangle, a leg in the base	$\frac{1}{2}s^2$
isosceles right triangle, the hypotenuse in the base	$\frac{1}{4}s^2$
semicircle, the diameter in the base	$\frac{\pi}{8}s^2$

A square, a semicircle, an equilateral triangle, and an isosceles right triangle, each standing on a side s.

The semicircle's radius is

\frac{s}{2}

, so its area is

\frac{1}{2}\pi\left(\frac{s}{2}\right)^2 = \frac{\pi}{8}s^2.

An equilateral triangle of side

s

has height

\frac{\sqrt{3}}{2}s

, which gives its area.

Squares and rectangles

Example 1

The base of a solid is the region between

y = \cos x

and

y = -\cos x

for

-\frac{\pi}{2} \le x \le \frac{\pi}{2}

. Each cross section perpendicular to the

x

-axis is a square with one side in the base. Find the volume of the solid.

On the interval

\cos x \ge 0

, so

y = \cos x

is on top. At

x

, the square stands on the segment from

y = -\cos x

up to

y = \cos x

, so its side is

s = \cos x - (-\cos x) = 2\cos x

The base, with the segment at x = 0.6 that one square stands on. The square rises out of the page.

\begin{aligned} V &= \int_{-\pi/2}^{\pi/2} (2\cos x)^2\,dx \\[4pt] &= 4\int_{-\pi/2}^{\pi/2} \cos^2 x\,dx \\[4pt] &= 2\int_{-\pi/2}^{\pi/2} (1 + \cos 2x)\,dx \\[4pt] &= 2\left(x + \frac{\sin 2x}{2}\right)\Big|_{-\pi/2}^{\pi/2} = 2\pi. \end{aligned}

The volume is

2\pi \approx 6.283

A rectangle needs its height as well as its base. Suppose each cross section on this base were a rectangle 3 times as tall as its base. Its area would be

3s^2

, and the volume would be

3(2\pi) = 6\pi

. Rectangles can stand on slices perpendicular to the

y

-axis too, with the side measured in

y

Semicircles and triangles

Example 2

The base of a solid is the region bounded by

y = \frac{3}{\sqrt{x}}

, the

x

-axis, and the lines

x = 1

and

x = e^2

. Each cross section perpendicular to the

x

-axis is a semicircle with its diameter in the base. Find the volume of the solid.

The base, with the segment at x = 3. Its semicircle stands on it, out of the page.

The diameter at

x

s = \frac{3}{\sqrt{x}}

, so the area is

\frac{\pi}{8}\cdot\frac{9}{x}

\begin{aligned} V &= \int_{1}^{e^2} \frac{9\pi}{8x}\,dx = \frac{9\pi}{8}\ln x\,\Big|_{1}^{e^2} \\[4pt] &= \frac{9\pi}{8}(2 - 0) = \frac{9\pi}{4} \approx 7.069. \end{aligned}

Example 3

The base of a solid is the region between the

y

-axis and the curve

x = y\sqrt{4 - y^2}

for

0 \le y \le 2

. Each cross section perpendicular to the

y

-axis is an equilateral triangle with one side in the base. Find the volume of the solid.

The curve meets the

y

-axis at

y = 0

and

y = 2

. A slice at height

y

crosses the base from

x = 0

to the curve, so

s = y\sqrt{4 - y^2}

, and the integral is in

y

The base, with the horizontal segment at y = 1 that one triangle stands on.

\begin{aligned} V &= \int_{0}^{2} \frac{\sqrt{3}}{4}\,y^2\left(4 - y^2\right) dy \\[4pt] &= \frac{\sqrt{3}}{4}\left(\frac{4y^3}{3} - \frac{y^5}{5}\right)\Big|_{0}^{2} \\[4pt] &= \frac{\sqrt{3}}{4}\cdot\frac{64}{15} = \frac{16\sqrt{3}}{15} \approx 1.848. \end{aligned}

Cross sections given by a formula

Sometimes the cross sections are described directly, with no base region. The volume is still the integral of the area of a slice.

Example 4

A stone column is 6 meters tall. Its horizontal cross section

h

meters above the ground is a square of side

1 + 0.2\cos\left(\frac{\pi h}{3}\right)

meters. Find the volume of the column.

The area of the slice at height

h

is the side squared:

\begin{aligned} V &= \int_{0}^{6} \left(1 + 0.2\cos\tfrac{\pi h}{3}\right)^2 dh \\[4pt] &= \int_{0}^{6} \Big(1 + 0.4\cos\tfrac{\pi h}{3} \\ &\qquad\qquad + 0.04\cos^2\tfrac{\pi h}{3}\Big)\,dh. \end{aligned}

The cosine has period 6, so its integral over

[0, 6]

is 0. The squared cosine averages

\frac{1}{2}

over a period, since

\cos^2\theta = \frac{1}{2}(1 + \cos 2\theta)

, so its integral over

[0, 6]

is half of 6, which is 3.

V = 6 + 0 + 0.04 \cdot 3 = 6.12.

The column holds 6.12 cubic meters of stone.