Worked examples
The question
Question: find the double integral of x² + y² over the disk x² + y² ≤ 4.
∫ ∫ (x² + y²) dA, x² + y² ≤ 4
1 Draw the region
The region on the floor is a disk: radius 2, all the way around.
Above it is the surface z = x² + y².
The integral is the volume of this solid.
2 Change to polar
In polar form, x² + y² = r², and the small area is dA = r dr dθ.
x² + y² = r², dA = r dr dθ
Why the extra r?
The small pieces far from the center are wider.
Their width grows with r.
r goes from 0 to 2.
θ goes from 0 to 2π: once all the way around.
= r: 0 to 2, θ: 0 to 2π
Write it in r and θ.
x² + y² becomes r².
Then times r dr dθ.
∫ ∫ (x² + y²) dA = ∫ from 0 to 2π of ∫ from 0 to 2 of r2 × r dr dθ
Careful!
Without the extra r, the answer is wrong.
It is the most common mistake here.
Careful: dA = r dr dθ: do not forget the r
3 The inner integral
Multiply by r first: r³.
∫ ∫ (x² + y²) dA = ∫ from 0 to 2π of ∫ from 0 to 2 of r3 dr dθ
Integrate each term: the power goes up by 1, and we divide by the new power.
∫ from 0 to 2 of r3 dr = r44 from 0 to 2
Put in the top r = 2, then take away the bottom r = 0.
∫ from 0 to 2 of r3 dr = (2)44 − (0)44
The bottom part is 0.
So it is 4.
∫ from 0 to 2 of r3 dr = 4
4 The outer integral
Now θ.
4 is a plain number: its integral is 4 times θ.
∫ from 0 to 2π of 4 dθ = 4θ from 0 to 2π
Top minus bottom: 2π minus 0 is 2π.
∫ from 0 to 2π of 4 dθ = 4 × 2π
So the answer is 8π.
∫ from 0 to 2π of 4 dθ = 8π
That is about 25.13.
∫ from 0 to 2π of 4 dθ ≈ 25.13
5 Check
Let's check if it makes sense.
The region's area is 4π.
On it, f goes from 0 to 4.
area = 4π ≈ 12.57, f from 0 to 4
So the answer must be between 0 and 50.27.
25.13 is.
Right!
0 ≤ 25.13 ≤ 50.27
Let's look back.
First: draw the region.
Then: change to polar.
Then: the inner integral.
Then: the outer integral.
Last: check.
Answer ∫ from 0 to 2π of 4 dθ = 8π
▶ Watch it solved
The question
Question: find the double integral of 4 over the ring 1 ≤ x² + y² ≤ 4.
∫ ∫ 4 dA, 1 ≤ x² + y² ≤ 4
1 Draw the region
The region on the floor is a ring: between radius 1 and radius 2, all the way around.
Above it is the surface z = 4.
The integral is the volume of this solid.
2 Change to polar
In polar form, x² + y² = r², and the small area is dA = r dr dθ.
x² + y² = r², dA = r dr dθ
Why the extra r?
The small pieces far from the center are wider.
Their width grows with r.
r goes from 1 to 2.
θ goes from 0 to 2π: once all the way around.
= r: 1 to 2, θ: 0 to 2π
Write it in r and θ.
Then times r dr dθ.
∫ ∫ 4 dA = ∫ from 0 to 2π of ∫ from 1 to 2 of 4 × r dr dθ
Careful!
Without the extra r, the answer is wrong.
It is the most common mistake here.
Careful: dA = r dr dθ: do not forget the r
3 The inner integral
Multiply by r first: 4r.
∫ ∫ 4 dA = ∫ from 0 to 2π of ∫ from 1 to 2 of 4r dr dθ
Integrate each term: the power goes up by 1, and we divide by the new power.
∫ from 1 to 2 of 4r dr = 2r2 from 1 to 2
Put in the top r = 2, then take away the bottom r = 1.
∫ from 1 to 2 of 4r dr = 2(2)2 − 2(1)2
So it is 6.
∫ from 1 to 2 of 4r dr = 6
4 The outer integral
Now θ.
6 is a plain number: its integral is 6 times θ.
∫ from 0 to 2π of 6 dθ = 6θ from 0 to 2π
Top minus bottom: 2π minus 0 is 2π.
∫ from 0 to 2π of 6 dθ = 6 × 2π
So the answer is 12π.
∫ from 0 to 2π of 6 dθ = 12π
That is about 37.70.
∫ from 0 to 2π of 6 dθ ≈ 37.70
5 Check
Let's check if it makes sense.
The region's area is 3π.
On it, f goes from 4 to 4.
area = 3π ≈ 9.42, f from 4 to 4
f is the same everywhere, so the answer is f times the area: 37.7.
Right!
37.7 ≤ 37.7 ≤ 37.7
Let's look back.
First: draw the region.
Then: change to polar.
Then: the inner integral.
Then: the outer integral.
Last: check.
Answer ∫ from 0 to 2π of 6 dθ = 12π
▶ Watch it solved
The question
Question: find the double integral of (x² + y²)² over the disk x² + y² ≤ 1.
∫ ∫ ((x² + y²)²) dA, x² + y² ≤ 1
1 Draw the region
The region on the floor is a disk: radius 1, all the way around.
Above it is the surface z = (x² + y²)².
The integral is the volume of this solid.
2 Change to polar
In polar form, x² + y² = r², and the small area is dA = r dr dθ.
x² + y² = r², dA = r dr dθ
Why the extra r?
The small pieces far from the center are wider.
Their width grows with r.
r goes from 0 to 1.
θ goes from 0 to 2π: once all the way around.
= r: 0 to 1, θ: 0 to 2π
Write it in r and θ.
(x² + y²)² becomes r⁴.
Then times r dr dθ.
∫ ∫ ((x² + y²)²) dA = ∫ from 0 to 2π of ∫ from 0 to 1 of r4 × r dr dθ
Careful!
Without the extra r, the answer is wrong.
It is the most common mistake here.
Careful: dA = r dr dθ: do not forget the r
3 The inner integral
Multiply by r first: r⁵.
∫ ∫ ((x² + y²)²) dA = ∫ from 0 to 2π of ∫ from 0 to 1 of r5 dr dθ
Integrate each term: the power goes up by 1, and we divide by the new power.
∫ from 0 to 1 of r5 dr = r66 from 0 to 1
Put in the top r = 1, then take away the bottom r = 0.
∫ from 0 to 1 of r5 dr = (1)66 − (0)66
The bottom part is 0.
So it is 1/6.
∫ from 0 to 1 of r5 dr = 16
4 The outer integral
Now θ.
1/6 is a plain number: its integral is 1/6 times θ.
∫ from 0 to 2π of 16 dθ = 16θ from 0 to 2π
Top minus bottom: 2π minus 0 is 2π.
∫ from 0 to 2π of 16 dθ = 16 × 2π
So the answer is π/3.
∫ from 0 to 2π of 16 dθ = π3
That is about 1.05.
∫ from 0 to 2π of 16 dθ ≈ 1.05
5 Check
Let's check if it makes sense.
The region's area is π.
On it, f goes from 0 to 1.
area = π ≈ 3.14, f from 0 to 1
So the answer must be between 0 and 3.14.
1.05 is.
Right!
0 ≤ 1.05 ≤ 3.14
Let's look back.
First: draw the region.
Then: change to polar.
Then: the inner integral.
Then: the outer integral.
Last: check.
Answer ∫ from 0 to 2π of 16 dθ = π3
▶ Watch it solved
