Steply

Triple integral of 2x - y over 0 ≤ x ≤ 2, 0 ≤ y ≤ 1, 0 ≤ z ≤ 3

Step by step, with the reason for every step · Triple integrals · Calculus 3

Step by step

The question

Question: find the triple integral of 2x − y over the box 0 ≤ x ≤ 2, 0 ≤ y ≤ 1, 0 ≤ z ≤ 3.

∫ ∫ ∫ (2x − y) dV, 0 ≤ x ≤ 2, 0 ≤ y ≤ 1, 0 ≤ z ≤ 3

1 Draw it

The box: x from 0 to 2, y from 0 to 1, z from 0 to 3.

2 Find the limits

Write it as three integrals: dz inside, then dy, then dx outside.

Over a box all the limits are numbers.

∫ ∫ ∫ (2x − y) dV = ∫ from 0 to 2 of ∫ from 0 to 1 of ∫ from 0 to 3 of (2x − y) dz dy dx

Careful!

dz goes with the z limits, dy with the y limits, dx with the x limits.

Inside out.

Careful: each d goes with its own limits

3 The inner integral

Inside first: integrate in z.

x and y stay fixed, like plain numbers.

Integrate each term: the power goes up by 1, and we divide by the new power.

∫ from 0 to 3 of (2x − y) dz = [2xz − yz] from 0 to 3

Put in the top z = 3, then take away the bottom z = 0.

∫ from 0 to 3 of (2x − y) dz = (2x(3) − y(3)) − (2x(0) − y(0))

The bottom part is 0.

So it is 6x − 3y.

∫ from 0 to 3 of (2x − y) dz = 6x − 3y

4 The middle integral

Then the middle one: integrate in y.

x stays fixed.

Integrate each term: the power goes up by 1, and we divide by the new power.

∫ from 0 to 1 of (6x − 3y) dy = [6xy − 3y22] from 0 to 1

Put in the top y = 1, then take away the bottom y = 0.

∫ from 0 to 1 of (6x − 3y) dy = (6x(1) − 3(1)22) − (6x(0) − 3(0)22)

The bottom part is 0.

So it is 6x − 3/2.

∫ from 0 to 1 of (6x − 3y) dy = 6x − 32

5 The outer integral

Last the outside one, in x.

Its limits are numbers.

Integrate each term: the power goes up by 1, and we divide by the new power.

∫ from 0 to 2 of (6x − 32) dx = [3x2 − 3x2] from 0 to 2

Put in the top x = 2, then take away the bottom x = 0.

∫ from 0 to 2 of (6x − 32) dx = (3(2)2 − 3(2)2) − (3(0)2 − 3(0)2)

The bottom part is 0.

So it is 9.

∫ from 0 to 2 of (6x − 32) dx = 9

6 Check

Let's check if it makes sense.

The box's volume is 6.

In it, f goes from −1 to 4.

volume = 6, f from −1 to 4

So the answer must be between −6 and 24.

It is.

Right!

−6 ≤ 9 ≤ 24

Let's look back.

First: draw it.

Then: find the limits.

Then: the inner integral.

Then: the middle integral.

Then: the outer integral.

Last: check.

Answer
∫ from 0 to 2 of (6x − 32) dx = 9

Your turn

Try this one yourself, then check:

Find the triple integral of ∫ ∫ ∫ (3x − y) dV, 0 ≤ x ≤ 2, 0 ≤ y ≤ 1, 0 ≤ z ≤ 3
Show the answer
15

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