Topic 3 of 7

Area, volume, and arc length

Step 16 · Calculus 2 topics

Use integrals to find the area between two curves, the volume of a solid made by spinning a region, and the length of a curve.

Area between curves

The area between y = f(x) above and y = g(x) below, from a to b, is the integral of the height f(x) − g(x). Find where the curves cross to get the limits.

A = \int_{a}^{b} \left(f(x) - g(x)\right)dx

The region between y = 2x above and y = x² below, from x = 0 to x = 2, has area 4/3.

Example: A line and a parabola

Find the area between y = 2x and y = x².

They meet where x² = 2x: at x = 0 and x = 2.
Between them the line is on top.
$\int_{0}^{2} (2x - x^{2})\,dx = 4 - \frac{8}{3}$ .

Answer $\frac{4}{3}$

Volumes of solids of revolution

Spin a region about an axis and slice the solid. Slices across the axis are discs, or washers with a hole, of area π(R² − r²). Or use cylindrical shells around the axis, with volume 2π × radius × height × thickness.

V = \int_{a}^{b} \pi\left(R(x)^{2} - r(x)^{2}\right)dx \qquad V = \int_{a}^{b} 2\pi x\,h(x)\,dx

Example: Discs

Spin the region under y = √x from 0 to 4 about the x-axis. Find the volume.

Each slice is a disc of radius √x, with area πx.
$\int_{0}^{4} \pi x\,dx = 8\pi$ .

Answer8π, about 25.1

Example: Shells

Spin the region under y = x² from 0 to 2 about the y-axis. Find the volume.

A shell at x has radius x, height x² and thickness dx.
$\int_{0}^{2} 2\pi x \cdot x^{2}\,dx = 2\pi \cdot \frac{16}{4}$ .

Answer8π

Example: A cone, as a check

Spin y = x from 0 to h about the x-axis. Find the volume.

Discs of radius x: $\int_{0}^{h} \pi x^{2}\,dx = \frac{\pi h^{3}}{3}$ .
The solid is a cone with radius h and height h.

Answer $\frac{1}{3}\pi h^{3}$ , matching the cone formula

Arc length

A short piece of curve with run dx and rise dy has length √(dx² + dy²), by the Pythagorean theorem. Adding the pieces gives the length of the curve.

L = \int_{a}^{b} \sqrt{1 + f'(x)^{2}}\,dx

Example: A straight line, as a check

Find the length of y = 2x from x = 0 to x = 3.

f′ = 2, so $L = \int_{0}^{3} \sqrt{5}\,dx = 3\sqrt{5}$ .
The distance from (0, 0) to (3, 6) is √45, which is also 3√5.

Answer3√5, about 6.71

Example: A curve

Find the length of $y = \frac{2}{3}x^{3/2}$ from x = 0 to x = 3.

f′(x) = √x, so the integrand is $\sqrt{1 + x}$ .
$\int_{0}^{3} \sqrt{1 + x}\,dx = \frac{2}{3}(8 - 1)$ .

Answer $\frac{14}{3}$

Try it yourself

Work each one on paper first, then open the answer to check.

Find the area between y = x and y = x² from 0 to 1.
Show answer
$\frac{1}{6}$
Find the area between y = 4 − x² and the x-axis.
Show answer
$\frac{32}{3}$
Spin the region under y = x from 0 to 3 about the x-axis. Find the volume.
Show answer
9π
Spin the region under y = 2 from 0 to 5 about the x-axis. What solid is it, and what is its volume?
Show answer
A cylinder, with volume 20π
Spin the region under y = x from 0 to 1 about the y-axis, using shells. Find the volume.
Show answer
$\frac{2\pi}{3}$
Find the length of y = 3x + 1 from x = 0 to x = 4.
Show answer
4√10, about 12.65