Volume between y = 3 and y = 1 turned around the x-axis from x = 0… Step by Step

Step by step

The question

Question: find the volume when the region between y = 3 and y = 1 from x = 0 to x = 2 turns around the x-axis.

y = 3

y = 1

1 Draw and turn it

Draw the curves first.

The region between them, from x = 0 to x = 2, is shaded.

2 One slice

Cut the solid into thin slices.

Each slice is a washer: a flat disk with a hole.

The outside radius is the curve further from the axis: R = 3.

R = 3

The inside radius, the hole, is the other one: r = 1.

r = 1

A washer is the big circle minus the hole: π R² minus π r², times its thickness dx.

slice = π(R² − r²) dx

Which curve is outside?

Test a number between: 3 is bigger there.

test x = 1: 3 > 1

3 Write the integral

Add up all the slices from 0 to 2: that is an integral.

π is just a number, so it waits in front.

V = π∫ from 0 to 2 of (3² − 1²) dx

Careful!

Not R minus r, squared!

The hole is a whole circle: take away π r squared.

Careful: not π(R − r)² ✗

4 Square it

Square both radii, then take away.

V = π∫ from 0 to 2 of 8 dx

Square the number and the power: 9.

3² = 9

And the hole: 1.

1² = 1

5 The antiderivative

The antiderivative: add 1 to each power, divide by the new power.

V = π[ 8x ] from 0 to 2

6 Top minus bottom

Put in 2, then take away the same with 0.

= π[(8(2)) − (8(0))]

= π[16 − 0]

So the volume is 16π cubic units.

That is about 50.27.

V = 16π ≈ 50.27

7 Check

Let's check with a calculator: it adds up 1000 thin washers.

That gives about 50.27, the same as our answer.

Right!

1000 thin washers: ≈ 50.27

Let's look back.

First: draw and turn it.

Then: one slice.

Then: write the integral.

Then: square it.

Then: the antiderivative.

Then: top minus bottom.

Last: check.

Answer