Step by step
The question
Question: find the volume when the region under y = 2x from x = 1 to x = 2 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 = 2x.
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 1 to 2: 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 2x.
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 2, then take away the same with 1.
2π times 14/3: the volume is 28π over 3 cubic units.
That is about 29.32.
7 Check
Let's check with a calculator: it adds up 1000 thin shells.
That gives about 29.32, 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.