Step by step
The question
Question: find the triple integral of xyz over the box 0 ≤ x ≤ 1, 0 ≤ y ≤ 2, 0 ≤ z ≤ 3.
1 Draw it
The box: x from 0 to 1, y from 0 to 2, z from 0 to 3.
2 Find the limits
Write it as three integrals: dz inside, then dy, then dx outside.
Over a box all the limits are numbers.
Careful!
dz goes with the z limits, dy with the y limits, dx with the x limits.
Inside out.
3 The inner integral
Inside first: integrate in z.
x and y stay fixed, like plain numbers.
Integrate each term: the power goes up by 1, and we divide by the new power.
Put in the top z = 3, then take away the bottom z = 0.
The bottom part is 0.
So it is 9xy/2.
4 The middle integral
Then the middle one: integrate in y.
x stays fixed.
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 9x.
5 The outer integral
Last the outside one, 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 = 0.
The bottom part is 0.
So it is 9/2.
6 Check
Let's check.
f is one term, so the three integrals can be done on their own: 1/2, 2, 9/2.
Multiplied: 9/2.
The same.
Right!
Let's look back.
First: draw it.
Then: find the limits.
Then: the inner integral.
Then: the middle integral.
Then: the outer integral.
Last: check.