Day 23: Area & Volume with Integrals

Area Between Curves

When two curves enclose a region, its area is the integral of (top curve) minus (bottom curve). The first job is always to find where the curves meet. If the region is described by functions of $y$ , integrate in $y$ instead — right curve minus left curve.

Core Theorem

If

f(x) \ge g(x)

on

[a, b]

, the area between them is

A = \int_a^b \big[\,f(x) - g(x)\,\big]\,dx.

If

p(y) \ge q(y)

on

[c, d]

, then

A = \int_c^d \big[\,p(y) - q(y)\,\big]\,dy \quad (\text{right} - \text{left}).

If the curves cross inside the interval, split at each crossing and add the areas.

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Easy

Find the area of the region bounded by

y = x + 2

and

y = x^2

.

How to think about it

The limits of integration are the

x

-values where the curves meet, so solve

f = g

first. Then integrate (top curve

-

bottom curve) across that interval.

Key idea: Set $x + 2 = x^2$ to find the intersection points. Between them the line is on top.

Full solution

1. Find the intersection points

x^2 = x + 2 \implies x^2 - x - 2 = 0 \implies (x - 2)(x + 1) = 0

, so

x = -1

and

x = 2

.

2. Decide which curve is on top

Test

x = 0

: the line gives

2

, the parabola gives

0

. So

y = x + 2

is the top curve on

[-1, 2]

.

3. Set up and evaluate

A = \int_{-1}^{2} \big[(x + 2) - x^2\big]\,dx = \left[\frac{x^2}{2} + 2x - \frac{x^3}{3}\right]_{-1}^{2} = \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{9}{2}

.

Worked Example 02Easy

Find the area of the region bounded by

y = e^x

,

y = x

,

x = 0

, and

x = 1

.

How to think about it

Key idea: On $[0, 1]$ , $e^x$ is always above $x$ . The vertical lines give the limits.

Full solution

1. Identify top and bottom

For every

x

in

[0, 1]

,

e^x > x

, so top

= e^x

, bottom

= x

.

2. Set up and evaluate

A = \int_0^1 (e^x - x)\,dx = \left[e^x - \frac{x^2}{2}\right]_0^1 = \left(e - \frac{1}{2}\right) - 1 = e - \frac{3}{2} \approx 1.218

.

Step-by-Step SOP

1
Sketch the region
Graph both curves and shade the enclosed region so you can see which function is on top (or on the right).
2
Find the limits of integration
Solve $f = g$ for the intersection points, or read off the given vertical/horizontal boundaries.
3
Write top minus bottom (or right minus left)
Set up the integral. A quick test point confirms the order.
4
Split wherever the curves cross
If the difference changes sign inside the interval, integrate each piece separately and add the (positive) areas.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium

Find the area of the region bounded by

x = y^2

and

x = y + 2

.

Need a hint?

These are functions of

y

, so integrate in

y

. Solve

y^2 = y + 2

for the

y

-limits.

Show solution

1. Find the y-limits

y^2 = y + 2 \implies (y - 2)(y + 1) = 0

, so

y = -1

and

y = 2

.

2. Identify right and left curves

At

y = 0

:

x = y + 2

gives

2

,

x = y^2

gives

0

. So

x = y + 2

is the right curve.

3. Set up and evaluate

A = \int_{-1}^{2} \big[(y + 2) - y^2\big]\,dy = \left[\frac{y^2}{2} + 2y - \frac{y^3}{3}\right]_{-1}^{2} = \frac{9}{2}

.

Practice 02Hard

Find the total area between

y = \sin x

and

y = \cos x

from

x = 0

to

x = \pi/2

.

Need a hint?

The curves cross where

\tan x = 1

, i.e.

x = \pi/4

. On

[0, \pi/4]

cosine is on top; after that, sine is.

Show solution

1. Find the crossing point

\sin x = \cos x \implies \tan x = 1 \implies x = \frac{\pi}{4}

.

2. Split and order each piece

On

[0, \pi/4]

:

\cos x \ge \sin x

. On

[\pi/4, \pi/2]

:

\sin x \ge \cos x

.

3. Two integrals

A = \int_0^{\pi/4} (\cos x - \sin x)\,dx + \int_{\pi/4}^{\pi/2} (\sin x - \cos x)\,dx = (\sqrt{2} - 1) + (\sqrt{2} - 1) = 2\sqrt{2} - 2 \approx 0.828

.

Common Pitfalls

⚠
Not finding the intersection points firstThe limits of integration are usually the values where $f = g$ . Guessing them is the most common source of a wrong answer.
⚠
Forgetting to split when curves crossA single integral of $f - g$ over a sign-changing interval gives net signed area, not total area. Split at every crossing.

Volumes of Revolution: Disk & Washer Method

Revolve a region around an axis and slice the solid perpendicular to that axis. If the region meets the axis, each slice is a solid disk. If there is a gap, each slice is a washer — a big disk with a smaller one removed.

Core Theorem

Disk: each slice is a full disk of radius

R

, so

V = \pi \int_a^b \big[R\big]^2\,dx.

Washer: each slice has outer radius

R

and inner radius

r

, so

V = \pi \int_a^b \Big(\big[R\big]^2 - \big[r\big]^2\Big)\,dx.

Square each radius separately, then subtract — never

(R - r)^2

. Revolving about a horizontal axis → integrate in

x

; about a vertical axis → integrate in

y

.

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Easy

The region under

y = \sqrt{x}

from

x = 0

to

x = 4

is revolved about the

x

-axis. Find the volume.

How to think about it

Slice perpendicular to the axis of revolution. If the region touches the axis, each slice is a disk (

V = \pi\int R^2

); if there's a gap, it's a washer (

V = \pi\int (R^2 - r^2)

, squaring each radius first).

Key idea: Each slice perpendicular to the $x$ -axis is a disk of radius $R(x) = \sqrt{x}$ .

Full solution

1. Identify the radius

The axis is the

x

-axis, so

R(x) = \sqrt{x}

.

2. Set up and evaluate

V = \pi \int_0^4 (\sqrt{x})^2\,dx = \pi \int_0^4 x\,dx = \pi \left[\frac{x^2}{2}\right]_0^4 = 8\pi

.

Worked Example 02Medium

The region bounded by

y = x^3

,

y = 8

, and

x = 0

is revolved about the

y

-axis. Find the volume.

How to think about it

Key idea: Slices perpendicular to the $y$ -axis are disks. Solve $y = x^3$ for $x$ to get the radius as a function of $y$ .

Full solution

1. Write the radius in terms of y

y = x^3 \implies x = y^{1/3}

, so

R(y) = y^{1/3}

, with

y

from

0

to

8

.

2. Set up and evaluate

V = \pi \int_0^8 (y^{1/3})^2\,dy = \pi \int_0^8 y^{2/3}\,dy = \pi \cdot \frac{3}{5}\,(8^{5/3}) = \pi \cdot \frac{3}{5}\cdot 32 = \frac{96\pi}{5}

.

Step-by-Step SOP

1
Sketch the region, the axis, and a representative slice
The slice is perpendicular to the axis of revolution.
2
Check for a gap
Region touches the axis → disk (one radius). Gap between region and axis → washer (two radii).
3
Write each radius as a distance to the axis
For an axis $y = k$ , a radius is $|k - (\text{curve})|$ . Outer = to the farther curve, inner = to the nearer curve.
4
Integrate
$V = \pi \int R^2\,dx$ (disk) or $\pi \int (R^2 - r^2)\,dx$ (washer), along the axis of revolution.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium

The region bounded by

y = x

and

y = x^2

is revolved about the

x

-axis. Find the volume.

Need a hint?

On

[0, 1]

,

y = x

is farther from the

x

-axis (outer) and

y = x^2

is closer (inner).

Show solution

1. Find the limits and radii

x = x^2 \implies x = 0, 1

. Outer

R(x) = x

, inner

r(x) = x^2

.

2. Set up and evaluate

V = \pi \int_0^1 \big(x^2 - x^4\big)\,dx = \pi \left[\frac{x^3}{3} - \frac{x^5}{5}\right]_0^1 = \pi\left(\frac{1}{3} - \frac{1}{5}\right) = \frac{2\pi}{15}

.

Practice 02Hard

The region bounded by

y = x

and

y = x^2

is revolved about the line

y = 2

. Find the volume.

Need a hint?

The axis is above the region. The curve closer to

y = 2

gives the smaller radius. Since

x \ge x^2

on

[0,1]

,

y = x

is closer.

Show solution

1. Distances to the axis y = 2

Outer radius reaches the farther curve

y = x^2

:

R(x) = 2 - x^2

. Inner radius reaches

y = x

:

r(x) = 2 - x

.

2. Expand the integrand

(2 - x^2)^2 - (2 - x)^2 = (4 - 4x^2 + x^4) - (4 - 4x + x^2) = x^4 - 5x^2 + 4x

.

3. Set up and evaluate

V = \pi \int_0^1 (x^4 - 5x^2 + 4x)\,dx = \pi \left[\frac{x^5}{5} - \frac{5x^3}{3} + 2x^2\right]_0^1 = \pi \cdot \frac{8}{15} = \frac{8\pi}{15}

.

Common Pitfalls

⚠
Squaring the difference of the radii $\pi \int (R - r)^2\,dx$ is wrong. The washer area is $\pi R^2 - \pi r^2$ , so the integrand is $R^2 - r^2$ .
⚠
Forgetting the piDisk and washer cross sections are circles with area $\pi r^2$ . The $\pi$ is part of the formula.
⚠
Wrong variable of integrationHorizontal axis → integrate in $x$ with radius a function of $x$ . Vertical axis → integrate in $y$ .

Volume by Known Cross Sections

A solid has a flat base in the plane, and every slice perpendicular to one axis is a known shape — a square, a semicircle, a triangle. If you can write the area of a slice as a function of position, the volume is the integral of that area. No revolution, so usually no $\pi$ .

Core Theorem

If every cross section perpendicular to the

x

-axis has area

A(x)

, then

V = \int_a^b A(x)\,dx.

Build

A

from the base length

s(x) = (\text{upper curve}) - (\text{lower curve})

: square

A = s^2

; semicircle (diameter

s

)

A = \frac{\pi}{8}s^2

; equilateral triangle

A = \frac{\sqrt{3}}{4}s^2

; isosceles right triangle (leg

s

)

A = \frac{1}{2}s^2

.

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Medium

The base of a solid is bounded by

y = x^2

and

y = 4

. Cross sections perpendicular to the

y

-axis are squares. Find the volume.

How to think about it

No revolution here, so usually no

\pi

. Find the base length

s

between the two bounding curves, plug it into the slice's area formula (square, triangle, semicircle), then integrate that area.

Key idea: Slices are perpendicular to the $y$ -axis, so work in $y$ . At height $y$ , the base runs from $x = -\sqrt{y}$ to $x = \sqrt{y}$ .

Full solution

1. Find the side length at height y

x = \pm\sqrt{y}

, so

s(y) = 2\sqrt{y}

.

2. Write the square area

A(y) = s(y)^2 = (2\sqrt{y})^2 = 4y

.

3. Integrate over y from 0 to 4

V = \int_0^4 4y\,dy = \big[2y^2\big]_0^4 = 32

.

Step-by-Step SOP

1
Draw the base and a representative slice
Identify which axis the slices are perpendicular to — that sets the variable of integration.
2
Find the base length s
$s = (\text{upper}) - (\text{lower})$ for slices perpendicular to the $x$ -axis; (right) − (left) for slices perpendicular to the $y$ -axis.
3
Plug s into the slice's area formula
Square $s^2$ ; semicircle $\frac{\pi}{8}s^2$ ; equilateral triangle $\frac{\sqrt{3}}{4}s^2$ ; isosceles right triangle $\frac{1}{2}s^2$ .
4
Integrate A over the interval
$V = \int_a^b A\,dx$ (or $dy$ ). No $\pi$ unless the slice itself is round.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Hard

The base is the region bounded by

y = 1 - \frac{x}{2}

, the

x

-axis, and the

y

-axis. Cross sections perpendicular to the

x

-axis are equilateral triangles. Find the volume.

Need a hint?

At position

x

, the base segment goes from the

x

-axis up to the line, so

s(x) = 1 - \frac{x}{2}

, for

x

from

0

to

2

.

Show solution

1. Write the triangle area

A(x) = \frac{\sqrt{3}}{4}\left(1 - \frac{x}{2}\right)^2

.

2. Integrate

Let

u = 1 - \frac{x}{2}

,

du = -\frac{1}{2}\,dx

:

\int_0^2 \left(1 - \frac{x}{2}\right)^2 dx = 2\int_0^1 u^2\,du = \frac{2}{3}

.

3. Finish

V = \frac{\sqrt{3}}{4}\cdot \frac{2}{3} = \frac{\sqrt{3}}{6}

.

Practice 02Hard

The base of a solid is the disk

x^2 + y^2 \le 1

. Cross sections perpendicular to the

x

-axis are semicircles with diameter across the base. Find the volume.

Need a hint?

At position

x

, the base runs from

y = -\sqrt{1 - x^2}

to

y = \sqrt{1 - x^2}

, so the diameter is

2\sqrt{1 - x^2}

.

Show solution

1. Find the radius of a slice

Diameter

= 2\sqrt{1 - x^2}

, so radius

= \sqrt{1 - x^2}

.

2. Write the semicircle area

A(x) = \frac{1}{2}\pi r^2 = \frac{\pi}{2}(1 - x^2)

.

3. Integrate from x = -1 to x = 1

V = \frac{\pi}{2}\int_{-1}^{1} (1 - x^2)\,dx = \frac{\pi}{2}\left[x - \frac{x^3}{3}\right]_{-1}^{1} = \frac{\pi}{2}\cdot \frac{4}{3} = \frac{2\pi}{3}

.

Common Pitfalls

⚠
Diameter vs. radius for semicircle slicesThe base segment $s$ is the diameter, so the radius is $s/2$ and $A = \frac{1}{2}\pi (s/2)^2 = \frac{\pi}{8}s^2$ . Using $r = s$ quadruples the answer.
⚠
Adding an unnecessary piSquare and triangle cross sections are not round — do not multiply by $\pi$ . Only disk, washer, and semicircle slices carry a $\pi$ .

Worked Examples

1. Find the intersection points

2. Decide which curve is on top

3. Set up and evaluate

1. Identify top and bottom

2. Set up and evaluate

Sketch the region

Find the limits of integration

Write top minus bottom (or right minus left)

Split wherever the curves cross

Practice — Your Turn

1. Find the y-limits

2. Identify right and left curves

3. Set up and evaluate

1. Find the crossing point

2. Split and order each piece

3. Two integrals

Worked Examples

1. Identify the radius

2. Set up and evaluate

1. Write the radius in terms of y

2. Set up and evaluate

Sketch the region, the axis, and a representative slice

Check for a gap

Write each radius as a distance to the axis

Integrate

Practice — Your Turn

1. Find the limits and radii

2. Set up and evaluate

1. Distances to the axis y = 2

2. Expand the integrand

3. Set up and evaluate

Worked Examples

1. Find the side length at height y

2. Write the square area

3. Integrate over y from 0 to 4

Draw the base and a representative slice

Find the base length s

Plug s into the slice's area formula

Integrate A over the interval

Practice — Your Turn

1. Write the triangle area

2. Integrate

3. Finish

1. Find the radius of a slice

2. Write the semicircle area

3. Integrate from x = -1 to x = 1

Gary Chang