Steply

Volume of the solid under the plane 2x + 3y + z = 6 in the first…

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 2x + 3y + z = 6 in the first octant.

V = ∫ ∫ ∫ 1 dV, under 2x + 3y + z = 6

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 = 2 and z = 6.

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 6 − 2x − 3y

On the floor, z = 0, so the edge is 2x + 3y = 6.

Solve for y: y goes from 0 to 2 − 2x/3.

y = 0 to 2 − 2x3

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 2 − 2x over 3 of ∫ from 0 to 6 − 2x − 3y 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 6 − 2x − 3y of 1 dz = z from 0 to 6 − 2x − 3y

Put in the top z = 6 − 2x − 3y, then take away the bottom z = 0.

∫ from 0 to 6 − 2x − 3y of 1 dz = (6 − 2x − 3y) − (0)

Work out each part.

The bottom part is 0.

∫ from 0 to 6 − 2x − 3y of 1 dz = 6 − 2x − 3y − 0

So it is 6 − 2x − 3y.

∫ from 0 to 6 − 2x − 3y of 1 dz = 6 − 2x − 3y

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 2 − 2x over 3 of (6 − 2x − 3y) dy = [6y − 2xy − 3y22] from 0 to 2 − 2x over 3

Put in the top y = 2 − 2x/3, then take away the bottom y = 0.

∫ from 0 to 2 − 2x over 3 of (6 − 2x − 3y) dy = (6(2 − 2x3) − 2x(2 − 2x3) − 3(2 − 2x3)22) − (6(0) − 2x(0) − 3(0)22)

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

∫ from 0 to 2 − 2x over 3 of (6 − 2x − 3y) dy = ((12 − 4x) − (4x − 4x23) − (6 − 4x + 2x23)) − (0)

Add the like terms in each part.

The bottom part is 0.

∫ from 0 to 2 − 2x over 3 of (6 − 2x − 3y) dy = (6 − 4x + 2x23) − (0)

So it is 6 − 4x + 2x² over 3.

∫ from 0 to 2 − 2x over 3 of (6 − 2x − 3y) dy = 6 − 4x + 2x23

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 (6 − 4x + 2x23) dx = [6x − 2x2 + 2x39] from 0 to 3

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

∫ from 0 to 3 of (6 − 4x + 2x23) dx = (6(3) − 2(3)2 + 2(3)39) − (6(0) − 2(0)2 + 2(0)39)

The bottom part is 0.

So it is 6.

∫ from 0 to 3 of (6 − 4x + 2x23) dx = 6

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 2.

We get 6.

Right!

pyramid: ⅓ × (½ × 3 × 2) × 6 = 6

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 (6 − 4x + 2x23) dx = 6

Your turn

Try this one yourself, then check:

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

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