Step by step
The question
Question: find the double integral of 2 over the first quarter of the disk x² + y² ≤ 9.
1 Draw the region
The region on the floor is the first quarter of a disk: radius 3.
Above it is the surface z = 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.
Why the extra r?
The small pieces far from the center are wider.
Their width grows with r.
r goes from 0 to 3.
θ goes from 0 to π/2: the first quarter.
Write it in r and θ.
Then times r dr dθ.
Careful!
Without the extra r, the answer is wrong.
It is the most common mistake here.
3 The inner integral
Multiply by r first: 2r.
Integrate each term: the power goes up by 1, and we divide by the new power.
Put in the top r = 3, then take away the bottom r = 0.
The bottom part is 0.
So it is 9.
4 The outer integral
Now θ.
9 is a plain number: its integral is 9 times θ.
Top minus bottom: π/2 minus 0 is π/2.
So the answer is 9π/2.
That is about 14.14.
5 Check
Let's check if it makes sense.
The region's area is 9π/4.
On it, f goes from 2 to 2.
f is the same everywhere, so the answer is f times the area: 14.14.
Right!
Let's look back.
First: draw the region.
Then: change to polar.
Then: the inner integral.
Then: the outer integral.
Last: check.