Steply

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

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 xy over 0 ≤ x ≤ 2, 0 ≤ y ≤ 1.

∫ ∫ xy dA, 0 ≤ x ≤ 2, 0 ≤ y ≤ 1

1 Draw it

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

Above it is the surface z = xy.

The double integral is the volume of this solid.

2 Find the limits

Write it as two integrals.

Inside: y from 0 to 1.

Outside: x from 0 to 2.

∫ ∫ xy dA = ∫ from 0 to 2 of ∫ from 0 to 1 of xy 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 1 of xy dy = xy22 from 0 to 1

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

∫ from 0 to 1 of xy dy = x(1)22 − x(0)22

The bottom part is 0.

So it is x/2.

∫ from 0 to 1 of xy dy = x2

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 2 of x2 dx = x24 from 0 to 2

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

∫ from 0 to 2 of x2 dx = (2)24 − (0)24

The bottom part is 0.

So it is 1.

∫ from 0 to 2 of x2 dx = 1

5 Check

Let's check in the other order.

x first: 2y.

∫ from 0 to 2 of xy dx = 2y

Then y: 1.

The same answer.

Right!

∫ from 0 to 1 of 2y dy = 1

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 2 of x2 dx = 1

Your turn

Try this one yourself, then check:

Find the double integral of ∫ ∫ 4xy dA, 0 ≤ x ≤ 1, 0 ≤ y ≤ 1
Show the answer
1

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