Volume between y = 4 and y = x^2 from x = 0 to x = 2 turned around… Step by Step

Step by step

The question

Question: find the volume when the region between y = 4 and y = x² turns around the y-axis.

y = 4

y = x²

1 Draw and turn it

Draw the curves first.

The region between them 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 top curve minus the bottom one.

h = (4) − (x²)

Which is on top?

Test a number between: 4 is bigger there.

test x = 1: 4 > 1

So h = 4 − x².

= 4 − x²

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 2: that is an integral.

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

V = 2π∫ from 0 to 2 of x(4 − x²) dx

Careful!

The radius is not the height of the curve.

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

Careful: not radius = 4 − x² ✗

4 Multiply

Multiply x into the bracket: x times 4 − x².

V = 2π∫ from 0 to 2 of (4x − x³) dx

x times every part.

x(4 − x²) = 4x − x³

5 The antiderivative

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

V = 2π[ 2x² − x⁴4 ] from 0 to 2

6 Top minus bottom

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

= 2π[(2(2)² − 2⁴4) − (2(0)² − 0⁴4)]

= 2π[4 − 0]

= 2π × 4

2π times 4: the volume is 8π cubic units.

That is about 25.13.

V = 8π ≈ 25.13

7 Check

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

That gives about 25.13, the same as our answer.

Right!

1000 thin shells: ≈ 25.13

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