Worked examples
The question
Question: find the double integral of xy over 0 ≤ x ≤ 2, 0 ≤ y ≤ 1.
∫ ∫ xy dA, 0 ≤ x ≤ 2, 0 ≤ y ≤ 1
1 Draw it
The rectangle lies on the floor: x from 0 to 2, y from 0 to 1.
Above it is the surface z = xy.
The double integral is the volume of this solid.
2 Find the limits
Write it as two integrals.
Inside: y from 0 to 1.
Outside: x from 0 to 2.
∫ ∫ xy dA = ∫ from 0 to 2 of ∫ from 0 to 1 of xy dy dx
Careful!
The inside integral is dy, so its limits are the y limits.
They may have x in them.
The outside limits are only numbers.
Careful: dy inside: the y limits go inside
3 The inner integral
Inside first: integrate in y.
x stays fixed, like a plain number.
Integrate each term: the power goes up by 1, and we divide by the new power.
∫ from 0 to 1 of xy dy = xy22 from 0 to 1
Put in the top y = 1, then take away the bottom y = 0.
∫ from 0 to 1 of xy dy = x(1)22 − x(0)22
The bottom part is 0.
So it is x/2.
∫ from 0 to 1 of xy dy = x2
4 The outer integral
Now the outside integral, 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 x2 dx = x24 from 0 to 2
Put in the top x = 2, then take away the bottom x = 0.
∫ from 0 to 2 of x2 dx = (2)24 − (0)24
The bottom part is 0.
So it is 1.
∫ from 0 to 2 of x2 dx = 1
5 Check
Let's check in the other order.
x first: 2y.
∫ from 0 to 2 of xy dx = 2y
Then y: 1.
The same answer.
Right!
∫ from 0 to 1 of 2y dy = 1
Let's look back.
First: draw it.
Then: find the limits.
Then: the inner integral.
Then: the outer integral.
Last: check.
Answer ∫ from 0 to 2 of x2 dx = 1
▶ Watch it solved
The question
Question: find the double integral of y/0 ≤ x ≤ 3, 0 ≤ y ≤ 2.
∫ ∫ y dA, 0 ≤ x ≤ 3, 0 ≤ y ≤ 2
1 Draw it
The rectangle lies on the floor: x from 0 to 3, y from 0 to 2.
Above it is the surface z = y.
The double integral is the volume of this solid.
2 Find the limits
Write it as two integrals.
Inside: y from 0 to 2.
Outside: x from 0 to 3.
∫ ∫ y dA = ∫ from 0 to 3 of ∫ from 0 to 2 of y dy dx
Careful!
The inside integral is dy, so its limits are the y limits.
They may have x in them.
The outside limits are only numbers.
Careful: dy inside: the y limits go inside
3 The inner integral
Inside first: integrate in y.
x stays fixed, like a plain number.
Integrate each term: the power goes up by 1, and we divide by the new power.
∫ from 0 to 2 of y dy = y22 from 0 to 2
Put in the top y = 2, then take away the bottom y = 0.
∫ from 0 to 2 of y dy = (2)22 − (0)22
The bottom part is 0.
So it is 2.
∫ from 0 to 2 of y dy = 2
4 The outer integral
Now the outside integral, 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 3 of 2 dx = 2x from 0 to 3
Put in the top x = 3, then take away the bottom x = 0.
∫ from 0 to 3 of 2 dx = 2(3) − 2(0)
The bottom part is 0.
So it is 6.
∫ from 0 to 3 of 2 dx = 6
5 Check
Let's check in the other order.
x first: 3y.
∫ from 0 to 3 of y dx = 3y
Then y: 6.
The same answer.
Right!
∫ from 0 to 2 of 3y dy = 6
Let's look back.
First: draw it.
Then: find the limits.
Then: the inner integral.
Then: the outer integral.
Last: check.
Answer ∫ from 0 to 3 of 2 dx = 6
▶ Watch it solved
The question
Question: find the volume under z = 6 − 2x − y/0 ≤ x ≤ 1, 0 ≤ y ≤ 2.
∫ ∫ (6 − 2x − y) dA, 0 ≤ x ≤ 1, 0 ≤ y ≤ 2
1 Draw it
The rectangle lies on the floor: x from 0 to 1, y from 0 to 2.
Above it is the surface z = 6 − 2x − y.
The double integral is the volume of this solid.
2 Find the limits
Write it as two integrals.
Inside: y from 0 to 2.
Outside: x from 0 to 1.
∫ ∫ (6 − 2x − y) dA = ∫ from 0 to 1 of ∫ from 0 to 2 of (6 − 2x − y) dy dx
Careful!
The inside integral is dy, so its limits are the y limits.
They may have x in them.
The outside limits are only numbers.
Careful: dy inside: the y limits go inside
3 The inner integral
Inside first: integrate in y.
x stays fixed, like a plain number.
Integrate each term: the power goes up by 1, and we divide by the new power.
∫ from 0 to 2 of (6 − 2x − y) dy = [6y − 2xy − y22] from 0 to 2
Put in the top y = 2, then take away the bottom y = 0.
∫ from 0 to 2 of (6 − 2x − y) dy = (6(2) − 2x(2) − (2)22) − (6(0) − 2x(0) − (0)22)
The bottom part is 0.
So it is 10 − 4x.
∫ from 0 to 2 of (6 − 2x − y) dy = 10 − 4x
4 The outer integral
Now the outside integral, 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 (10 − 4x) dx = [10x − 2x2] from 0 to 1
Put in the top x = 1, then take away the bottom x = 0.
∫ from 0 to 1 of (10 − 4x) dx = (10(1) − 2(1)2) − (10(0) − 2(0)2)
The bottom part is 0.
So it is 8.
∫ from 0 to 1 of (10 − 4x) dx = 8
5 Check
Let's check in the other order.
x first: 5 − y.
∫ from 0 to 1 of (6 − 2x − y) dx = 5 − y
Then y: 8.
The same answer.
Right!
∫ from 0 to 2 of (5 − y) dy = 8
Let's look back.
First: draw it.
Then: find the limits.
Then: the inner integral.
Then: the outer integral.
Last: check.
Answer ∫ from 0 to 1 of (10 − 4x) dx = 8
▶ Watch it solved
