Step by step
The question
Question: find the double integral of 3x²y² over −1 ≤ x ≤ 1, 0 ≤ y ≤ 2.
1 Draw it
The rectangle lies on the floor: x from −1 to 1, y from 0 to 2.
Above it is the surface z = 3x²y².
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 −1 to 1.
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.
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.
Put in the top y = 2, then take away the bottom y = 0.
The bottom part is 0.
So it is 8x².
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.
Put in the top x = 1, then take away the bottom x = −1.
So it is 16/3.
5 Check
Let's check in the other order.
x first: 2y².
Then y: 16/3.
The same answer.
Right!
Let's look back.
First: draw it.
Then: find the limits.
Then: the inner integral.
Then: the outer integral.
Last: check.