Volume under z = 2 + x over 0 ≤ x ≤ 2, 0 ≤ y ≤ 3 Step by Step

Step by step

The question

Question: find the volume under z = 2 + x/0 ≤ x ≤ 2, 0 ≤ y ≤ 3.

∫ ∫ (2 + x) 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 = 2 + x.

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.

∫ ∫ (2 + x) dA = ∫ from 0 to 2 of ∫ from 0 to 3 of (2 + x) 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 (2 + x) dy = [2y + xy] from 0 to 3

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

∫ from 0 to 3 of (2 + x) dy = (2(3) + x(3)) − (2(0) + x(0))

The bottom part is 0.

So it is 6 + 3x.

∫ from 0 to 3 of (2 + x) dy = 6 + 3x

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 (6 + 3x) dx = [6x + 3x²2] from 0 to 2

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

∫ from 0 to 2 of (6 + 3x) dx = (6(2) + 3(2)²2) − (6(0) + 3(0)²2)

The bottom part is 0.

So it is 18.

∫ from 0 to 2 of (6 + 3x) dx = 18

5 Check

Let's check in the other order.

x first: 6.

∫ from 0 to 2 of (2 + x) dx = 6

Then y: 18.

The same answer.

Right!

∫ from 0 to 3 of 6 dy = 18

Let's look back.

First: draw it.

Then: find the limits.

Then: the inner integral.

Then: the outer integral.

Last: check.

Answer