Volume under z = 4 - x - y over 0 ≤ x ≤ 1, 0 ≤ y ≤ 2 Step by Step

Step by step

The question

Question: find the volume under z = 4 − x − y/0 ≤ x ≤ 1, 0 ≤ y ≤ 2.

∫ ∫ (4 − x − y) dA, 0 ≤ x ≤ 1, 0 ≤ y ≤ 2

1 Draw it

The rectangle lies on the floor: x from 0 to 1, y from 0 to 2.

Above it is the surface z = 4 − x − y.

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 0 to 1.

∫ ∫ (4 − x − y) dA = ∫ from 0 to 1 of ∫ from 0 to 2 of (4 − x − y) dy dx

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.

Careful: dy inside: the y limits go inside

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.

∫ from 0 to 2 of (4 − x − y) dy = [4y − xy − y²2] from 0 to 2

Put in the top y = 2, then take away the bottom y = 0.

∫ from 0 to 2 of (4 − x − y) dy = (4(2) − x(2) − (2)²2) − (4(0) − x(0) − (0)²2)

The bottom part is 0.

So it is 6 − 2x.

∫ from 0 to 2 of (4 − x − y) dy = 6 − 2x

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.

∫ from 0 to 1 of (6 − 2x) dx = [6x − x²] from 0 to 1

Put in the top x = 1, then take away the bottom x = 0.

∫ from 0 to 1 of (6 − 2x) dx = (6(1) − (1)²) − (6(0) − (0)²)

The bottom part is 0.

So it is 5.

∫ from 0 to 1 of (6 − 2x) dx = 5

5 Check

Let's check in the other order.

x first: 7/2 − y.

∫ from 0 to 1 of (4 − x − y) dx = 7 over 2 − y

Then y: 5.

The same answer.

Right!

∫ from 0 to 2 of (72 − y) dy = 5

Let's look back.

First: draw it.

Then: find the limits.

Then: the inner integral.

Then: the outer integral.

Last: check.

Answer