Steply

Volume of the solid under the plane x + y + 2z = 4 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 + 2z = 4 in the first octant.

V = ∫ ∫ ∫ 1 dV, under x + y + 2z = 4

1 Draw it

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

The plane cuts the axes at x = 4, y = 4 and z = 2.

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 2 − x2 − y2

On the floor, z = 0, so the edge is x + y = 4.

Solve for y: y goes from 0 to 4 − x.

y = 0 to 4 − x

And x goes from 0 to 4, where that edge meets the x-axis.

x = 0 to 4

Put the limits in: the most changing ones inside.

V = ∫ from 0 to 4 of ∫ from 0 to 4 − x of ∫ from 0 to 2 − x over 2 − y over 2 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 2 − x over 2 − y over 2 of 1 dz = z from 0 to 2 − x over 2 − y over 2

Put in the top z = 2 − x/2 − y/2, then take away the bottom z = 0.

∫ from 0 to 2 − x over 2 − y over 2 of 1 dz = (2 − x2 − y2) − (0)

Work out each part.

The bottom part is 0.

∫ from 0 to 2 − x over 2 − y over 2 of 1 dz = 2 − x2 − y2 − 0

So it is 2 − x/2 − y/2.

∫ from 0 to 2 − x over 2 − y over 2 of 1 dz = 2 − x2 − y2

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 4 − x of (2 − x2 − y2) dy = [2y − xy2 − y24] from 0 to 4 − x

Put in the top y = 4 − x, then take away the bottom y = 0.

∫ from 0 to 4 − x of (2 − x2 − y2) dy = (2(4 − x) − x(4 − x)2 − (4 − x)24) − (2(0) − x(0)2 − (0)24)

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

∫ from 0 to 4 − x of (2 − x2 − y2) dy = ((8 − 2x) − (2x − x22) − (4 − 2x + x24)) − (0)

Add the like terms in each part.

The bottom part is 0.

∫ from 0 to 4 − x of (2 − x2 − y2) dy = (4 − 2x + x24) − (0)

So it is 4 − 2x + x² over 4.

∫ from 0 to 4 − x of (2 − x2 − y2) dy = 4 − 2x + x24

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 4 of (4 − 2x + x24) dx = [4x − x2 + x312] from 0 to 4

Put in the top x = 4, then take away the bottom x = 0.

∫ from 0 to 4 of (4 − 2x + x24) dx = (4(4) − (4)2 + (4)312) − (4(0) − (0)2 + (0)312)

The bottom part is 0.

So it is 16/3.

∫ from 0 to 4 of (4 − 2x + x24) dx = 163

6 Check

Let's check.

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

The base triangle is half of 4 times 4.

We get 16/3.

Right!

pyramid: ⅓ × (½ × 4 × 4) × 2 = 16/3

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 4 of (4 − 2x + x24) dx = 163

Your turn

Try this one yourself, then check:

V = ∫ ∫ ∫ 1 dV, under 2x + 3y + 2z = 6
Show the answer
3

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