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Double integral of 6 over 0 ≤ x ≤ 2, 0 ≤ y ≤ 3

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 6/0 ≤ x ≤ 2, 0 ≤ y ≤ 3.

∫ ∫ 6 dA, 0 ≤ x ≤ 2, 0 ≤ y ≤ 3

1 Draw it

The rectangle lies on the floor: x from 0 to 2, y from 0 to 3.

Above it is the surface z = 6.

The double integral is the volume of this solid.

2 Find the limits

Write it as two integrals.

Inside: y from 0 to 3.

Outside: x from 0 to 2.

∫ ∫ 6 dA = ∫ from 0 to 2 of ∫ from 0 to 3 of 6 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 3 of 6 dy = 6y from 0 to 3

Put in the top y = 3, then take away the bottom y = 0.

∫ from 0 to 3 of 6 dy = 6(3) − 6(0)

The bottom part is 0.

So it is 18.

∫ from 0 to 3 of 6 dy = 18

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 18 dx = 18x from 0 to 2

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

∫ from 0 to 2 of 18 dx = 18(2) − 18(0)

The bottom part is 0.

So it is 36.

∫ from 0 to 2 of 18 dx = 36

5 Check

Let's check in the other order.

x first: 12.

∫ from 0 to 2 of 6 dx = 12

Then y: 36.

The same answer.

Right!

∫ from 0 to 3 of 12 dy = 36

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 18 dx = 36

Your turn

Try this one yourself, then check:

Find the double integral of ∫ ∫ 4 dA, 0 ≤ x ≤ 3, 0 ≤ y ≤ 4
Show the answer
48

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