LemonmathSelf-study sheet

Grade 12 · Lesson 4.3

Applications of integrals: area and volume

Name: Date: Time spent: min

Goals

Find the area of a region bounded by one graph and the $x$ -axis, or by two graphs.
Split the interval where two graphs cross to remove the absolute value.
Find the volume of a solid from the area of its cross-sections $S(x)$ .
Find the volume of the solid of revolution formed by rotating a region about the $x$ -axis.

Remember

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 $\pi$ ).

Exercises

Exercise 1

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

Exercise 2

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

Exercise 3

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

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.

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 $\dfrac{512\pi}{15}$ .

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).

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.

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.

Exercise 9 Review

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

Exercise 10 Challenge

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

Mini test

15 minutes, without looking back at the lesson. Part I: circle one letter; Part II: write T or F for each statement; Part III: write the answer in the box.

Part I. Multiple choice

Question 1. Let

y=f(x)

be continuous on

[a, b]

. The area of the region bounded by the graph of

y=f(x)

, the

x

-axis and the lines

x=a

x=b

\displaystyle\int_a^b f(x)\,dx

.B.

\displaystyle\int_a^b\lvert f(x)\rvert\,dx

.C.

\pi\displaystyle\int_a^b f^2(x)\,dx

.D.

\pi\displaystyle\int_a^b\lvert f(x)\rvert\,dx

Question 2. The area of the region bounded by

y=x^2

and

y=x

\dfrac56

.B.

\dfrac12

.C.

\dfrac16

.D.

-\dfrac16

Question 3. The region bounded by

y=\sqrt x

, the

x

-axis,

x=0

and

x=1

is rotated about the

x

-axis. The volume of the solid of revolution is

\dfrac{\pi}{2}

.B.

\dfrac{2\pi}{3}

.C.

\dfrac12

.D.

\dfrac{\pi^2}{4}

Question 4. A solid lies between the planes

x=0

and

x=2

; its cross-section perpendicular to the

x

-axis at

x

is a square of side

x

. The volume of the solid is

8

.B.

\dfrac83

.C.

\dfrac{8\pi}{3}

.D.

2

Part II. True or false

Question 1. Let

H

be the region bounded by the parabola

y=4x-x^2

and the

x

-axis.

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

Part III. Short answer

Question 1. Find the area of the region bounded by the graphs of

y=x^3

and

y=x

Question 2. The region bounded by

y=\dfrac1x

, the

x

-axis,

x=1

and

x=3

is rotated about the

x

-axis. Find the volume of the solid of revolution. Give your answer to 2 decimal places.

Score: ........ / 10

Hints (only when stuck)

Exercise 1.

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.

Exercise 2.

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.

Exercise 3.

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.

Exercise 4.

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 $h=3$ .

Exercise 5.

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 $\pi\displaystyle\int_{-2}^2(x^2-4)^2dx$ .

Exercise 6.

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.

Exercise 7.

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.

Exercise 8.

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

Steps:

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

Exercise 9.

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 $-\cos x$ .

Exercise 10.

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 $[0, 2]$ .

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