Step by step
The question
Question: find the double integral of 3(x² + y²) over the first quarter of the disk x² + y² ≤ 16.
1 Draw the region
The region on the floor is the first quarter of a disk: radius 4.
Above it is the surface z = 3(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.
Why the extra r?
The small pieces far from the center are wider.
Their width grows with r.
r goes from 0 to 4.
θ goes from 0 to π/2: the first quarter.
Write it in r and θ.
3(x² + y²) becomes 3r².
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: 3r³.
Integrate each term: the power goes up by 1, and we divide by the new power.
Put in the top r = 4, then take away the bottom r = 0.
The bottom part is 0.
So it is 192.
4 The outer integral
Now θ.
192 is a plain number: its integral is 192 times θ.
Top minus bottom: π/2 minus 0 is π/2.
So the answer is 96π.
That is about 301.59.
5 Check
Let's check if it makes sense.
The region's area is 4π.
On it, f goes from 0 to 48.
So the answer must be between 0 and 603.19.
301.59 is.
Right!
Let's look back.
First: draw the region.
Then: change to polar.
Then: the inner integral.
Then: the outer integral.
Last: check.