Steply

Double integral of x + y over 0 ≤ x ≤ 1, 0 ≤ y ≤ 2

Step by step, with the reason for every step · Double integrals over rectangles · Calculus 3

Step by step

The question

Question: find the double integral of x + y/0 ≤ x ≤ 1, 0 ≤ y ≤ 2.

∫ ∫ (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 = 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.

∫ ∫ (x + y) dA = ∫ from 0 to 1 of ∫ from 0 to 2 of (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 (x + y) dy = [xy + y22] from 0 to 2

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

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

The bottom part is 0.

So it is 2x + 2.

∫ from 0 to 2 of (x + y) dy = 2x + 2

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 (2x + 2) dx = [x2 + 2x] from 0 to 1

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

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

The bottom part is 0.

So it is 3.

∫ from 0 to 1 of (2x + 2) dx = 3

5 Check

Let's check in the other order.

x first: 1/2 + y.

∫ from 0 to 1 of (x + y) dx = 1 over 2 + y

Then y: 3.

The same answer.

Right!

∫ from 0 to 2 of (12 + y) dy = 3

Let's look back.

First: draw it.

Then: find the limits.

Then: the inner integral.

Then: the outer integral.

Last: check.

Answer
∫ from 0 to 1 of (2x + 2) dx = 3

Your turn

Try this one yourself, then check:

Find the double integral of ∫ ∫ (3x + 3y) dA, 0 ≤ x ≤ 1, 0 ≤ y ≤ 1
Show the answer
3

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