Applications of integrals: area and volume

The front of a school gate is shaped like a parabola: $4$ m wide at the base and $4$ m high at the top. The whole front of the gate must be painted. What area, in square metres, needs paint?

The rectangle around the gate has area $4\cdot4=16$ m², and the triangle inside the gate has area $\dfrac12\cdot4\cdot4=8$ m². So the answer lies between $8$ and $16$ . An integral gives the exact number.

Area of a plane region

Explore: the parabolic gate

We put the origin at the middle of the base of the gate, with the $y$ -axis pointing up. The parabola passes through $(-2, 0)$ and $(2, 0)$ and has vertex $(0, 4)$ , so its equation is $y=4-x^2$ . The front of the gate is the region under the graph and above the $x$ -axis, for $-2\le x\le2$ .

The front of the gate,

y=4-x^2

The function $4-x^2$ is non-negative on $[-2, 2]$ , so by the geometric meaning of the integral:

$S=\int_{-2}^{2}(4-x^2)\,dx=\Big[4x-\frac{x^3}{3}\Big]_{-2}^{2}=\left(8-\frac83\right)-\left(-8+\frac83\right)=\frac{32}{3}\approx10.67.$

It is indeed between $8$ and $16$ ; it is $\dfrac23$ of the area of the surrounding rectangle.

When part of a graph lies below the $x$ -axis, the integral over that part is negative, while an area is always positive. That is why the general formula has an absolute value.

Why is it true? If $f(x)\ge g(x)$ on $[a, b]$ , we translate both graphs up by the same amount $k$ so that they lie above the $x$ -axis. The region does not change, and its area is the area under $f+k$ minus the area under $g+k$ , that is $\displaystyle\int_a^b(f-g)\,dx$ . Wherever $f<g$ the two functions swap roles, so we use $\lvert f-g\rvert$ .

Method. Solve $f(x)-g(x)=0$ to find the crossing points in $(a, b)$ . On each piece, $f-g$ keeps one sign, so $\displaystyle\int\lvert f-g\rvert\,dx=\left\lvert\int(f-g)\,dx\right\rvert$ . If the lines $x=a$ , $x=b$ are not given, the limits are the $x$ -coordinates of the intersection points.

Example A graph that crosses the

x

-axis

Find the area of the region bounded by the graph of $y=x^2-2x$ , the $x$ -axis and the lines $x=0$ , $x=3$ .

Solution. We find where the graph meets the $x$ -axis: $x^2-2x=0\iff x=0$ or $x=2$ . On $[0, 2]$ the graph lies below the axis, and on $[2, 3]$ above it. We split at $x=2$ :

$\int_0^2(x^2-2x)\,dx=\Big[\frac{x^3}{3}-x^2\Big]_0^2=-\frac43,\qquad\int_2^3(x^2-2x)\,dx=0-\left(-\frac43\right)=\frac43.$

So $S=\left\lvert-\dfrac43\right\rvert+\dfrac43=\dfrac83$ .

Look back. Evaluating $\displaystyle\int_0^3(x^2-2x)\,dx$ in one go gives $0$ , because the negative and positive parts cancel.

The region in the example above

When a region lies between two graphs, the limits are usually the $x$ -coordinates of the intersection points.

The region between

y=x^2

and

y=x+2

Volume of a solid

Cut a solid with a plane perpendicular to the $x$ -axis at the point with coordinate $x$ : we get a cross-section of area $S(x)$ . A thin slice of thickness $h$ has volume about $S(x)\cdot h$ . Adding up all the slices, just as we added small rectangles to get an area, gives the following formula.

Solids of revolution

Rotating the region bounded by $y=f(x)$ , the $x$ -axis, $x=a$ and $x=b$ about the $x$ -axis produces a solid of revolution. The cross-section at $x$ is a disc of radius $\lvert f(x)\rvert$ , with area $S(x)=\pi f^2(x)$ . Putting this into the volume formula:

The formula for solids of revolution also gives the volume of real curved objects.

Watch out Common mistakes

Forgetting the absolute value. Wrong: the area bounded by $y=x^2-2x$ , the $x$ -axis, $x=0$ and $x=3$ is $\displaystyle\int_0^3(x^2-2x)\,dx=0$ . Why it is wrong: the part below the axis gives a negative integral. Right: split at $x=2$ to get $\dfrac83$ .
Subtracting the wrong way round with no absolute value. Wrong: the area is $\displaystyle\int_{-1}^2\big(x^2-(x+2)\big)\,dx=-\frac92$ . Why it is wrong: an area cannot be negative. Right: take the upper function minus the lower one to get $\dfrac92$ .
Forgetting the square or the $\pi$ . Wrong: the volume of the sphere is $\pi\displaystyle\int_{-3}^3\sqrt{9-x^2}\,dx$ . Why it is wrong: each cross-section is a disc of area $\pi r^2=\pi f^2(x)$ . Right: $\pi\displaystyle\int_{-3}^3(9-x^2)\,dx=36\pi$ .
Using the wrong limits. Wrong: finding the area between $y=x^2$ and $y=x+2$ on $[0, 2]$ . Why it is wrong: the region starts at the intersection point $x=-1$ . Right: the limits are $-1$ and $2$ .

Key terms

English	Tiếng Việt
plane region	hình phẳng
area of a region	diện tích hình phẳng
$x$ -coordinate of an intersection point	hoành độ giao điểm
cross-section	thiết diện
solid	vật thể
solid of revolution	khối tròn xoay
rotate about the $x$ -axis	quay quanh trục $Ox$
volume	thể tích

Summary

Area: $S=\displaystyle\int_a^b\lvert f(x)\rvert\,dx$ ; between two graphs $S=\displaystyle\int_a^b\lvert f(x)-g(x)\rvert\,dx$ .
Find the roots of $f-g$ to split the interval; on each piece take the absolute value of the integral.
If $x=a$ and $x=b$ are not given, the limits are the $x$ -coordinates of the intersection points.
Volume from cross-sections: $V=\displaystyle\int_a^bS(x)\,dx$ .
Solid of revolution about the $x$ -axis: $V=\pi\displaystyle\int_a^bf^2(x)\,dx$ (remember the square and the $\pi$ ).

Practice

Each exercise comes with what you need and some steps to start. Try it yourself first, then open the solution.

Exercise 1

Find the area of the region bounded by the graph of $y=x^2+1$ , the $x$ -axis and the lines $x=0$ , $x=3$ .

What you need and how to start

What you need: $f(x)=x^2+1>0$ , so $S=\displaystyle\int_0^3(x^2+1)\,dx$ .

Steps:

Find the antiderivative $\dfrac{x^3}{3}+x$ .
Substitute the limits.

Show solution

$S=\Big[\dfrac{x^3}{3}+x\Big]_0^3=9+3=12$ .

Answer. $12$ .

Exercise 2

Find the area of the region bounded by the graph of $y=x^3$ , the $x$ -axis and the lines $x=-1$ , $x=2$ .

What you need and how to start

What you need: $x^3<0$ when $x<0$ , so we must split at $x=0$ .

Steps:

Evaluate $\displaystyle\int_{-1}^0x^3dx$ and $\displaystyle\int_0^2x^3dx$ .
Take the absolute value of each part and add.

Show solution

$\displaystyle\int_{-1}^0x^3dx=-\frac14$ and $\displaystyle\int_0^2x^3dx=4$ . So $S=\dfrac14+4=\dfrac{17}{4}$ .

Answer. $\dfrac{17}{4}$ .

Exercise 3

Find the area of the region bounded by the parabola $y=x^2$ and the line $y=2x$ .

What you need and how to start

What you need: the limits are the $x$ -coordinates of the intersection points; take the upper function minus the lower one.

Steps:

Solve $x^2=2x$ .
Test $x=1$ to see which graph is on top.

Show solution

$x^2=2x\iff x=0$ or $x=2$ . At $x=1$ : $2x=2>1=x^2$ . $S=\displaystyle\int_0^2(2x-x^2)\,dx=4-\frac83=\frac43$ .

Answer. $\dfrac43$ .

Exercise 4

Rotate the region bounded by the line $y=x$ , the $x$ -axis and the line $x=3$ about the $x$ -axis. Find the volume of the solid of revolution and compare it with the formula for the volume of a cone.

What you need and how to start

What you need: $V=\pi\displaystyle\int_a^bf^2(x)\,dx$ ; a cone has volume $\dfrac13\pi r^2h$ .

Steps:

Write $V=\pi\displaystyle\int_0^3x^2dx$ .
The solid is a cone with $r=3$ and $h=3$ .

Show solution

$V=\pi\Big[\dfrac{x^3}{3}\Big]_0^3=9\pi$ . The cone formula gives $\dfrac13\pi\cdot9\cdot3=9\pi$ , the same.

Answer. $9\pi$ .

Exercise 5

Let $H$ be the region bounded by the graph of $y=x^2-4$ and the $x$ -axis. Is each claim true or false?

a) $H$ lies between the lines $x=-2$ and $x=2$ .

b) The area of $H$ is $\displaystyle\int_{-2}^2(x^2-4)\,dx$ .

c) The area of $H$ is $\dfrac{32}{3}$ .

d) Rotating $H$ about the $x$ -axis gives a solid of revolution of volume $\dfrac{512\pi}{15}$ .

What you need and how to start

What you need: $x^2-4\le0$ on $[-2, 2]$ ; $V=\pi\displaystyle\int f^2dx$ .

Steps:

Find where the graph meets the $x$ -axis.
Find the area using $4-x^2$ , then evaluate $\pi\displaystyle\int_{-2}^2(x^2-4)^2dx$ .

Show solution

a) True: $x^2-4=0\iff x=\pm2$ .

b) False: that integral equals $-\dfrac{32}{3}<0$ , so it cannot be an area.

c) True: $\displaystyle\int_{-2}^2(4-x^2)\,dx=\frac{32}{3}$ .

d) True: $(x^2-4)^2=x^4-8x^2+16$ , and $\pi\displaystyle\int_{-2}^2(x^4-8x^2+16)\,dx=\pi\left(\frac{64}{5}-\frac{128}{3}+64\right)=\frac{512\pi}{15}$ .

Answer. a) True; b) False; c) True; d) True.

Exercise 6

Find the area of the region bounded by the parabola $y=x^2-2x$ and the line $y=x$ (give the result as a decimal).

What you need and how to start

What you need: the limits are the $x$ -coordinates of the intersection points.

Steps:

Solve $x^2-2x=x$ .
Evaluate $\displaystyle\int\big(x-(x^2-2x)\big)\,dx$ between the two roots.

Show solution

$x^2-3x=0\iff x=0$ or $x=3$ . $S=\displaystyle\int_0^3(3x-x^2)\,dx=\frac{27}{2}-9=\frac92$ .

Answer. $4.5$ .

Exercise 7

A window has the shape of the region bounded by the parabola $y=1.5-1.5x^2$ and the $x$ -axis (units in metres): it is $2$ m wide and $1.5$ m high. The glass costs $800$ thousand đồng per square metre. Find the cost of the glass.

What you need and how to start

What you need: the area of the window is the integral of a non-negative function between its two crossings of the $x$ -axis.

Steps:

The crossings with the $x$ -axis are $x=\pm1$ .
Evaluate $S=\displaystyle\int_{-1}^1(1.5-1.5x^2)\,dx$ , then multiply by the price.

Show solution

$S=\Big[1.5x-0.5x^3\Big]_{-1}^{1}=1-(-1)=2$ (m²). The glass costs $2\cdot800=1600$ thousand đồng.

Answer. $1\,600$ thousand đồng ( $1.6$ million đồng).

Exercise 8

A solid lies between the planes $x=0$ and $x=4$ . Its cross-section perpendicular to the $x$ -axis at $x$ $(0\le x\le4)$ is a rectangle with sides $x$ and $4-x$ . Find the volume of the solid.

What you need and how to start

What you need: $V=\displaystyle\int_a^bS(x)\,dx$ .

Steps:

$S(x)=x(4-x)=4x-x^2$ .
Evaluate $\displaystyle\int_0^4(4x-x^2)\,dx$ .

Show solution

$V=\Big[2x^2-\dfrac{x^3}{3}\Big]_0^4=32-\dfrac{64}{3}=\dfrac{32}{3}$ .

Answer. $\dfrac{32}{3}$ .

Exercise 9 Review

Find the area of the region bounded by the graph of $y=\sin x$ , the $x$ -axis and the lines $x=0$ , $x=\pi$ .

What you need and how to start

What you need: $\sin x\ge0$ on $[0, \pi]$ ; integrals of trigonometric functions (see definite integrals).

Steps:

$S=\displaystyle\int_0^{\pi}\sin x\,dx$ .
Use the antiderivative $-\cos x$ .

Show solution

$S=\Big[-\cos x\Big]_0^{\pi}=1-(-1)=2$ .

Answer. $2$ .

Exercise 10 Challenge

Find the area of the region bounded by the graphs of $y=x^3-x$ and $y=3x$ .

What you need and how to start

What you need: the graphs cross three times, so the upper function changes.

Steps:

Solve $x^3-x=3x\iff x(x^2-4)=0$ .
Evaluate $\displaystyle\int\lvert x^3-4x\rvert\,dx$ on $[-2, 0]$ and $[0, 2]$ .

Show solution

The intersections are $x=-2$ , $0$ , $2$ . $\displaystyle\int_{-2}^0(x^3-4x)\,dx=\Big[\frac{x^4}{4}-2x^2\Big]_{-2}^0=0-(4-8)=4$ ; $\displaystyle\int_0^2(4x-x^3)\,dx=8-4=4$ . So $S=8$ .

Answer. $8$ .

Area of a plane region

Explore: the parabolic gate

Volume of a solid

Solids of revolution

Key terms

Summary

Practice

Mini test

Part I. Multiple choice

Part II. True or false

Part III. Short answer

a) The parabola meets the $x$ -axis at $x=0$ and $x=4$ .	True	False
b) The area of $H$ is $\dfrac{32}{3}$ .	True	False
c) Rotating $H$ about the $x$ -axis gives a solid of revolution of volume $\pi\displaystyle\int_0^4(4x-x^2)\,dx$ .	True	False
d) The line $x=2$ divides $H$ into two parts of equal area.	True	False