Volume of y = 3 turned around the x-axis from x = 0 to x = 2 Step by Step

Step by step

The question

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

y = 3

1 Draw and turn it

Draw the curve first.

The region under the curve, from x = 0 to x = 2, is shaded.

2 One slice

Cut the solid into thin slices.

Each slice is a disk: a very flat cylinder.

Its radius is the height of the curve: r = 3.

Its thickness is a tiny dx.

r = 3

A cylinder is its circle's area times its thickness.

So one slice is π r² dx.

slice = πr² dx

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 dx

Careful!

Square the radius!

A circle's area is π r squared, not π r.

Careful: not V = π∫ from 0 to 2 of 3 dx ✗

4 Square it

Square the radius.

V = π∫ from 0 to 2 of 9 dx

Square the number and the power: 9.

3² = 9

5 The antiderivative

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

V = π[ 9x ] from 0 to 2

6 Top minus bottom

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

= π[(9(2)) − (9(0))]

= π[18 − 0]

So the volume is 18π cubic units.

That is about 56.55.

V = 18π ≈ 56.55

7 Check

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

That gives about 56.55, the same as our answer.

Right!

1000 thin disks: ≈ 56.55

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