Day 23

Area And Volume

Area Between Curves Volumes of Revolution: Disk & Washer Method Volume by Known Cross Sections

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

f(x) \ge g(x)

[a, b]

, the area between them is

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

p(y) \ge q(y)

[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.

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 Exercises

Example 01Easy

Find the area of the region bounded by

y = x + 2

and

y = x^2

NEED A HINT?

Set

x + 2 = x^2

to find the intersection points. Between them the line is on top.

SHOW DETAILED EXPLANATION

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

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]

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}

Example 02Easy

Find the area of the region bounded by

y = e^x

y = x

x = 0

, and

x = 1

NEED A HINT?

[0, 1]

e^x

is always above

x

. The vertical lines give the limits.

SHOW DETAILED EXPLANATION

Identify top and bottom

For every

x

[0, 1]

e^x > x

, so top

= e^x

, bottom

= x

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

Example 03Medium

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 DETAILED EXPLANATION

Find the y-limits

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

, so

y = -1

and

y = 2

Identify right and left curves

y = 0

x = y + 2

gives

2

x = y^2

gives

0

. So

x = y + 2

is the right curve.

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}

Example 04Hard

Find the total area between

y = \sin x

and

y = \cos x

from

x = 0

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 DETAILED EXPLANATION

Find the crossing point

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

Split and order each piece

[0, \pi/4]

\cos x \ge \sin x

. On

[\pi/4, \pi/2]

\sin x \ge \cos x

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

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 Exercises

Example 01Easy

The region under

y = \sqrt{x}

from

x = 0

x = 4

is revolved about the

x

-axis. Find the volume.

NEED A HINT?

Each slice perpendicular to the

x

-axis is a disk of radius

R(x) = \sqrt{x}

SHOW DETAILED EXPLANATION

Identify the radius

The axis is the

x

-axis, so

R(x) = \sqrt{x}

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

Example 02Medium

The region bounded by

y = x^3

y = 8

, and

x = 0

is revolved about the

y

-axis. Find the volume.

NEED A HINT?

Slices perpendicular to the

y

-axis are disks. Solve

y = x^3

for

x

to get the radius as a function of

y

SHOW DETAILED EXPLANATION

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

8

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}

Example 03Medium

The region bounded by

y = x

and

y = x^2

is revolved about the

x

-axis. Find the volume.

NEED A HINT?

[0, 1]

y = x

is farther from the

x

-axis (outer) and

y = x^2

is closer (inner).

SHOW DETAILED EXPLANATION

Find the limits and radii

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

. Outer

R(x) = x

, inner

r(x) = x^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}

Example 04Hard

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

[0,1]

y = x

is closer.

SHOW DETAILED EXPLANATION

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

Expand the integrand

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

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

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 Exercises

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.

NEED A HINT?

Slices are perpendicular to the

y

-axis, so work in

y

. At height

y

, the base runs from

x = -\sqrt{y}

x = \sqrt{y}

SHOW DETAILED EXPLANATION

Find the side length at height y

x = \pm\sqrt{y}

, so

s(y) = 2\sqrt{y}

Write the square area

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

Integrate over y from 0 to 4

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

Example 02Hard

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

2

SHOW DETAILED EXPLANATION

Write the triangle area

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

Finish

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

Example 03Hard

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}

y = \sqrt{1 - x^2}

, so the diameter is

2\sqrt{1 - x^2}

SHOW DETAILED EXPLANATION

Find the radius of a slice

Diameter

= 2\sqrt{1 - x^2}

, so radius

= \sqrt{1 - x^2}

Write the semicircle area

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

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 $\pi$ .

Gary Chang

Calculus Educator

5+ Years of Calculus Teaching Experience | AP Calculus Specialist. Dedicated to helping students master calculus through step-by-step logic and clear visualizations.

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Sketch the region

Find the limits of integration

Write top minus bottom (or right minus left)

Split wherever the curves cross

Practice Exercises

Find the intersection points

Decide which curve is on top

Set up and evaluate

Identify top and bottom

Set up and evaluate

Find the y-limits

Identify right and left curves

Set up and evaluate

Find the crossing point

Split and order each piece

Two integrals

Sketch the region, the axis, and a representative slice

Check for a gap

Write each radius as a distance to the axis

Integrate

Practice Exercises

Identify the radius

Set up and evaluate

Write the radius in terms of y

Set up and evaluate

Find the limits and radii

Set up and evaluate

Distances to the axis y = 2

Expand the integrand

Set up and evaluate

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 Exercises

Find the side length at height y

Write the square area

Integrate over y from 0 to 4

Write the triangle area

Integrate

Finish

Find the radius of a slice

Write the semicircle area

Integrate from x = -1 to x = 1

Gary Chang