Worked examples
The question
Question: find the triple integral of xyz over the box 0 ≤ x ≤ 1, 0 ≤ y ≤ 2, 0 ≤ z ≤ 3.
∫ ∫ ∫ xyz dV, 0 ≤ x ≤ 1, 0 ≤ y ≤ 2, 0 ≤ z ≤ 3
1 Draw it
The box: x from 0 to 1, y from 0 to 2, 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.
∫ ∫ ∫ xyz dV = ∫ from 0 to 1 of ∫ from 0 to 2 of ∫ from 0 to 3 of xyz 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 xyz dz = xyz22 from 0 to 3
Put in the top z = 3, then take away the bottom z = 0.
∫ from 0 to 3 of xyz dz = xy(3)22 − xy(0)22
The bottom part is 0.
So it is 9xy/2.
∫ from 0 to 3 of xyz dz = 9xy2
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 2 of 9xy2 dy = 9xy24 from 0 to 2
Put in the top y = 2, then take away the bottom y = 0.
∫ from 0 to 2 of 9xy2 dy = 9x(2)24 − 9x(0)24
The bottom part is 0.
So it is 9x.
∫ from 0 to 2 of 9xy2 dy = 9x
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 1 of 9x dx = 9x22 from 0 to 1
Put in the top x = 1, then take away the bottom x = 0.
∫ from 0 to 1 of 9x dx = 9(1)22 − 9(0)22
The bottom part is 0.
So it is 9/2.
∫ from 0 to 1 of 9x dx = 92
6 Check
Let's check.
f is one term, so the three integrals can be done on their own: 1/2, 2, 9/2.
(∫ x dx = 1/2) × (∫ y dy = 2) × (∫ z dz = 9/2)
Multiplied: 9/2.
The same.
Right!
= 9/2
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 1 of 9x dx = 92
▶ Watch it solved
The question
Question: find the triple integral of 2xz over the box 0 ≤ x ≤ 1, 0 ≤ y ≤ 2, 0 ≤ z ≤ 2.
∫ ∫ ∫ 2xz dV, 0 ≤ x ≤ 1, 0 ≤ y ≤ 2, 0 ≤ z ≤ 2
1 Draw it
The box: x from 0 to 1, y from 0 to 2, z from 0 to 2.
2 Find the limits
Write it as three integrals: dz inside, then dy, then dx outside.
Over a box all the limits are numbers.
∫ ∫ ∫ 2xz dV = ∫ from 0 to 1 of ∫ from 0 to 2 of ∫ from 0 to 2 of 2xz 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 2 of 2xz dz = xz2 from 0 to 2
Put in the top z = 2, then take away the bottom z = 0.
∫ from 0 to 2 of 2xz dz = x(2)2 − x(0)2
The bottom part is 0.
So it is 4x.
∫ from 0 to 2 of 2xz dz = 4x
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 2 of 4x dy = 4xy from 0 to 2
Put in the top y = 2, then take away the bottom y = 0.
∫ from 0 to 2 of 4x dy = 4x(2) − 4x(0)
The bottom part is 0.
So it is 8x.
∫ from 0 to 2 of 4x dy = 8x
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 1 of 8x dx = 4x2 from 0 to 1
Put in the top x = 1, then take away the bottom x = 0.
∫ from 0 to 1 of 8x dx = 4(1)2 − 4(0)2
The bottom part is 0.
So it is 4.
∫ from 0 to 1 of 8x dx = 4
6 Check
Let's check.
f is one term, so the three integrals can be done on their own: 1/2, 2, 2.
2 × (∫ x dx = 1/2) × (∫ 1 dy = 2) × (∫ z dz = 2)
Multiplied: 4.
The same.
Right!
= 4
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 1 of 8x dx = 4
▶ Watch it solved
The question
Question: find the volume of the solid under the plane x + y + 2z = 4 in the first octant.
V = ∫ ∫ ∫ 1 dV, under x + y + 2z = 4
1 Draw it
First octant means x, y and z are all positive: the corner in front.
The plane cuts the axes at x = 4, y = 4 and z = 2.
The solid is the corner piece under it.
Its floor is the green triangle.
2 Find the limits
z goes from the floor, z = 0, up to the plane: solve the plane for z.
z = 0 to 2 − x2 − y2
On the floor, z = 0, so the edge is x + y = 4.
Solve for y: y goes from 0 to 4 − x.
y = 0 to 4 − x
And x goes from 0 to 4, where that edge meets the x-axis.
x = 0 to 4
Put the limits in: the most changing ones inside.
V = ∫ from 0 to 4 of ∫ from 0 to 4 − x of ∫ from 0 to 2 − x over 2 − y over 2 of 1 dz dy dx
Careful!
The inside limits may have letters.
The outside limits must be plain numbers.
Careful: outer limits: numbers only
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 2 − x over 2 − y over 2 of 1 dz = z from 0 to 2 − x over 2 − y over 2
Put in the top z = 2 − x/2 − y/2, then take away the bottom z = 0.
∫ from 0 to 2 − x over 2 − y over 2 of 1 dz = (2 − x2 − y2) − (0)
Work out each part.
The bottom part is 0.
∫ from 0 to 2 − x over 2 − y over 2 of 1 dz = 2 − x2 − y2 − 0
So it is 2 − x/2 − y/2.
∫ from 0 to 2 − x over 2 − y over 2 of 1 dz = 2 − x2 − y2
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 4 − x of (2 − x2 − y2) dy = [2y − xy2 − y24] from 0 to 4 − x
Put in the top y = 4 − x, then take away the bottom y = 0.
∫ from 0 to 4 − x of (2 − x2 − y2) dy = (2(4 − x) − x(4 − x)2 − (4 − x)24) − (2(0) − x(0)2 − (0)24)
Work out each term on its own: multiply out the brackets.
∫ from 0 to 4 − x of (2 − x2 − y2) dy = ((8 − 2x) − (2x − x22) − (4 − 2x + x24)) − (0)
Add the like terms in each part.
The bottom part is 0.
∫ from 0 to 4 − x of (2 − x2 − y2) dy = (4 − 2x + x24) − (0)
So it is 4 − 2x + x² over 4.
∫ from 0 to 4 − x of (2 − x2 − y2) dy = 4 − 2x + x24
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 4 of (4 − 2x + x24) dx = [4x − x2 + x312] from 0 to 4
Put in the top x = 4, then take away the bottom x = 0.
∫ from 0 to 4 of (4 − 2x + x24) dx = (4(4) − (4)2 + (4)312) − (4(0) − (0)2 + (0)312)
The bottom part is 0.
So it is 16/3.
∫ from 0 to 4 of (4 − 2x + x24) dx = 163
6 Check
Let's check.
The solid is a pyramid: one third of the base times the height.
The base triangle is half of 4 times 4.
We get 16/3.
Right!
pyramid: ⅓ × (½ × 4 × 4) × 2 = 16/3
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 4 of (4 − 2x + x24) dx = 163
▶ Watch it solved
