Step by step
The question
Question: find the volume when the region under y = x³ from x = 0 to x = 1 turns around the y-axis.
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.
Its height is the height of the curve: h = x³.
Its thickness is a tiny dx.
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.
3 Write the integral
Add up all the shells from 0 to 1: that is an integral.
2π is just a number, so it waits in front.
Careful!
The radius is not the height of the curve.
It is the distance to the y-axis: x.
4 Multiply
Multiply x into the bracket: x times x³.
Add the powers of x.
5 The antiderivative
The antiderivative: add 1 to each power, divide by the new power.
6 Top minus bottom
Put in 1, then take away the same with 0.
2π times 1/5: the volume is 2π over 5 cubic units.
That is about 1.26.
7 Check
Let's check with a calculator: it adds up 1000 thin shells.
That gives about 1.26, the same as our answer.
Right!
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.