Worked examples
The question
Question: find the volume when the region under y = x² from x = 0 to x = 2 turns around the y-axis.
y = x2
1 Draw and turn it
Draw the curve first.
The region under the curve is shaded.
2 One shell
Cut the solid into thin tubes, one inside the other.
Each one is a shell: like a can with no top and no bottom.
A shell at x is x away from the y-axis.
That is its radius.
radius = x
Its height is the height of the curve: h = x².
Its thickness is a tiny dx.
h = x2
Cut a shell open and lay it flat.
It is a thin slab: 2πx long, the distance around a circle of radius x.
h tall, and dx thick.
shell = 2πx × h × dx
3 Write the integral
Add up all the shells from 0 to 2: that is an integral.
2π is just a number, so it waits in front.
V = 2π∫ from 0 to 2 of xx2 dx
Careful!
The radius is not the height of the curve.
It is the distance to the y-axis: x.
Careful: not radius = x2 ✗
4 Multiply
Multiply x into the bracket: x times x².
V = 2π∫ from 0 to 2 of x3 dx
Add the powers of x.
xx2 = x3
5 The antiderivative
The antiderivative: add 1 to each power, divide by the new power.
V = 2π[ x44 ] from 0 to 2
6 Top minus bottom
Put in 2, then take away the same with 0.
= 2π[(244) − (044)]
= 2π[4 − 0]
= 2π × 4
2π times 4: the volume is 8π cubic units.
That is about 25.13.
V = 8π ≈ 25.13
7 Check
Let's check with a calculator: it adds up 1000 thin shells.
That gives about 25.13, the same as our answer.
Right!
1000 thin shells: ≈ 25.13
Let's look back.
First: draw and turn it.
Then: one shell.
Then: write the integral.
Then: multiply.
Then: the antiderivative.
Then: top minus bottom.
Last: check.
▶ Watch it solved
The question
Question: find the volume when the region under y = 2x − x² turns around the y-axis.
y = 2x − x2
1 Draw and turn it
Draw the curve first.
2 Where they meet
The region ends where the curve meets the x-axis: where y is 0.
2x − x2 = 0
Times −1 on both sides: every sign changes, and 0 stays 0.
Now x² comes first with a plus.
x2 − 2x = 0
Every term has an x in it: take it out.
x(x − 2) = 0
A product is 0 when one of its parts is 0.
So x = 0, or x = 2.
x = 0 or x − 2 = 0 → x = 2
The region under the curve is shaded.
3 One shell
Cut the solid into thin tubes, one inside the other.
Each one is a shell: like a can with no top and no bottom.
A shell at x is x away from the y-axis.
That is its radius.
radius = x
Its height is the height of the curve: h = 2x − x².
Its thickness is a tiny dx.
h = 2x − x2
Cut a shell open and lay it flat.
It is a thin slab: 2πx long, the distance around a circle of radius x.
h tall, and dx thick.
shell = 2πx × h × dx
4 Write the integral
Add up all the shells from 0 to 2: that is an integral.
2π is just a number, so it waits in front.
V = 2π∫ from 0 to 2 of x(2x − x2) dx
Careful!
The radius is not the height of the curve.
It is the distance to the y-axis: x.
Careful: not radius = 2x − x2 ✗
5 Multiply
Multiply x into the bracket: x times 2x − x².
V = 2π∫ from 0 to 2 of (2x2 − x3) dx
x times every part.
x(2x − x2) = 2x2 − x3
6 The antiderivative
The antiderivative: add 1 to each power, divide by the new power.
V = 2π[ 2x33 − x44 ] from 0 to 2
7 Top minus bottom
Put in 2, then take away the same with 0.
= 2π[(2(2)33 − 244) − (2(0)33 − 044)]
= 2π[43 − 0]
= 2π × 43
2π times 4/3: the volume is 8π over 3 cubic units.
That is about 8.38.
V = 8π3 ≈ 8.38
8 Check
Let's check with a calculator: it adds up 1000 thin shells.
That gives about 8.38, the same as our answer.
Right!
1000 thin shells: ≈ 8.38
Let's look back.
First: draw and turn it.
Then: where they meet.
Then: one shell.
Then: write the integral.
Then: multiply.
Then: the antiderivative.
Then: top minus bottom.
Last: check.
▶ Watch it solved
The question
Question: find the volume when the region between y = 4 and y = x² turns around the y-axis.
y = 4
y = x2
1 Draw and turn it
Draw the curves first.
The region between them is shaded.
2 One shell
Cut the solid into thin tubes, one inside the other.
Each one is a shell: like a can with no top and no bottom.
A shell at x is x away from the y-axis.
That is its radius.
radius = x
Its height is the top curve minus the bottom one.
h = (4) − (x2)
Which is on top?
Test a number between: 4 is bigger there.
test x = 1: 4 > 1
So h = 4 − x².
= 4 − x2
Cut a shell open and lay it flat.
It is a thin slab: 2πx long, the distance around a circle of radius x.
h tall, and dx thick.
shell = 2πx × h × dx
3 Write the integral
Add up all the shells from 0 to 2: that is an integral.
2π is just a number, so it waits in front.
V = 2π∫ from 0 to 2 of x(4 − x2) dx
Careful!
The radius is not the height of the curve.
It is the distance to the y-axis: x.
Careful: not radius = 4 − x2 ✗
4 Multiply
Multiply x into the bracket: x times 4 − x².
V = 2π∫ from 0 to 2 of (4x − x3) dx
x times every part.
x(4 − x2) = 4x − x3
5 The antiderivative
The antiderivative: add 1 to each power, divide by the new power.
V = 2π[ 2x2 − x44 ] from 0 to 2
6 Top minus bottom
Put in 2, then take away the same with 0.
= 2π[(2(2)2 − 244) − (2(0)2 − 044)]
= 2π[4 − 0]
= 2π × 4
2π times 4: the volume is 8π cubic units.
That is about 25.13.
V = 8π ≈ 25.13
7 Check
Let's check with a calculator: it adds up 1000 thin shells.
That gives about 25.13, the same as our answer.
Right!
1000 thin shells: ≈ 25.13
Let's look back.
First: draw and turn it.
Then: one shell.
Then: write the integral.
Then: multiply.
Then: the antiderivative.
Then: top minus bottom.
Last: check.
▶ Watch it solved
