Steply

Volume of the solid under the plane x + y + z = 3 in the first octant

Step by step, with the reason for every step · Triple integrals · Calculus 3

Step by step

The question

Question: find the volume of the solid under the plane x + y + z = 3 in the first octant.

V = ∫ ∫ ∫ 1 dV, under x + y + z = 3

1 Draw it

First octant means x, y and z are all positive: the corner in front.

The plane cuts the axes at x = 3, y = 3 and z = 3.

The solid is the corner piece under it.

Its floor is the green triangle.

2 Find the limits

z goes from the floor, z = 0, up to the plane: solve the plane for z.

z = 0 to 3 − x − y

On the floor, z = 0, so the edge is x + y = 3.

Solve for y: y goes from 0 to 3 − x.

y = 0 to 3 − x

And x goes from 0 to 3, where that edge meets the x-axis.

x = 0 to 3

Put the limits in: the most changing ones inside.

V = ∫ from 0 to 3 of ∫ from 0 to 3 − x of ∫ from 0 to 3 − x − y of 1 dz dy dx

Careful!

The inside limits may have letters.

The outside limits must be plain numbers.

Careful: outer limits: numbers only

3 The inner integral

Inside first: integrate in z.

x and y stay fixed, like plain numbers.

Integrate each term: the power goes up by 1, and we divide by the new power.

∫ from 0 to 3 − x − y of 1 dz = z from 0 to 3 − x − y

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

∫ from 0 to 3 − x − y of 1 dz = (3 − x − y) − (0)

Work out each part.

The bottom part is 0.

∫ from 0 to 3 − x − y of 1 dz = 3 − x − y − 0

So it is 3 − x − y.

∫ from 0 to 3 − x − y of 1 dz = 3 − x − y

4 The middle integral

Then the middle one: integrate in y.

x stays fixed.

Integrate each term: the power goes up by 1, and we divide by the new power.

∫ from 0 to 3 − x of (3 − x − y) dy = [3y − xy − y22] from 0 to 3 − x

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

∫ from 0 to 3 − x of (3 − x − y) dy = (3(3 − x) − x(3 − x) − (3 − x)22) − (3(0) − x(0) − (0)22)

Work out each term on its own: multiply out the brackets.

∫ from 0 to 3 − x of (3 − x − y) dy = ((9 − 3x) − (3x − x2) − (92 − 3x + x22)) − (0)

Add the like terms in each part.

The bottom part is 0.

∫ from 0 to 3 − x of (3 − x − y) dy = (92 − 3x + x22) − (0)

So it is 9/2 − 3x + x² over 2.

∫ from 0 to 3 − x of (3 − x − y) dy = 92 − 3x + x22

5 The outer integral

Last the outside one, 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 3 of (92 − 3x + x22) dx = [9x2 − 3x22 + x36] from 0 to 3

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

∫ from 0 to 3 of (92 − 3x + x22) dx = (9(3)2 − 3(3)22 + (3)36) − (9(0)2 − 3(0)22 + (0)36)

The bottom part is 0.

So it is 9/2.

∫ from 0 to 3 of (92 − 3x + x22) dx = 92

6 Check

Let's check.

The solid is a pyramid: one third of the base times the height.

The base triangle is half of 3 times 3.

We get 9/2.

Right!

pyramid: ⅓ × (½ × 3 × 3) × 3 = 9/2

Let's look back.

First: draw it.

Then: find the limits.

Then: the inner integral.

Then: the middle integral.

Then: the outer integral.

Last: check.

Answer
∫ from 0 to 3 of (92 − 3x + x22) dx = 92

Your turn

Try this one yourself, then check:

V = ∫ ∫ ∫ 1 dV, under 2x + 3y + 3z = 12
Show the answer
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