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

Step by step

The question

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

y = x

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 = x.

Its thickness is a tiny dx.

h = x

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 3: that is an integral.

2π is just a number, so it waits in front.

V = 2π∫ from 0 to 3 of xx dx

Careful!

The radius is not the height of the curve.

It is the distance to the y-axis: x.

Careful: not radius = x ✗

4 Multiply

Multiply x into the bracket: x times x.

V = 2π∫ from 0 to 3 of x² dx

Add the powers of x.

xx = x²

5 The antiderivative

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

V = 2π[ x³3 ] from 0 to 3

6 Top minus bottom

Put in 3, then take away the same with 0.

= 2π[(3³3) − (0³3)]

= 2π[9 − 0]

= 2π × 9

2π times 9: 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 shells.

That gives about 56.55, the same as our answer.

Right!

1000 thin shells: ≈ 56.55

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.

Answer