Volume under z = 9 - x^2 over 0 ≤ x ≤ 3, 0 ≤ y ≤ 2 Step by Step

Step by step

The question

Question: find the volume under z = 9 − x² over 0 ≤ x ≤ 3, 0 ≤ y ≤ 2.

∫ ∫ (9 − x²) dA, 0 ≤ x ≤ 3, 0 ≤ y ≤ 2

1 Draw it

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

Above it is the surface z = 9 − x².

The double integral is the volume of this solid.

2 Find the limits

Write it as two integrals.

Inside: y from 0 to 2.

Outside: x from 0 to 3.

∫ ∫ (9 − x²) dA = ∫ from 0 to 3 of ∫ from 0 to 2 of (9 − x²) 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 2 of (9 − x²) dy = [9y − x²y] from 0 to 2

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

∫ from 0 to 2 of (9 − x²) dy = (9(2) − x²(2)) − (9(0) − x²(0))

The bottom part is 0.

So it is 18 − 2x².

∫ from 0 to 2 of (9 − x²) dy = 18 − 2x²

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 3 of (18 − 2x²) dx = [18x − 2x³3] from 0 to 3

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

∫ from 0 to 3 of (18 − 2x²) dx = (18(3) − 2(3)³3) − (18(0) − 2(0)³3)

The bottom part is 0.

So it is 36.

∫ from 0 to 3 of (18 − 2x²) dx = 36

5 Check

Let's check in the other order.

x first: 18.

∫ from 0 to 3 of (9 − x²) dx = 18

Then y: 36.

The same answer.

Right!

∫ from 0 to 2 of 18 dy = 36

Let's look back.

First: draw it.

Then: find the limits.

Then: the inner integral.

Then: the outer integral.

Last: check.

Answer