Volume under z = x^2 + y^2 over 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 Step by Step

Step by step

The question

Question: find the volume under z = x² + y² over 0 ≤ x ≤ 1, 0 ≤ y ≤ 1.

∫ ∫ (x² + y²) dA, 0 ≤ x ≤ 1, 0 ≤ y ≤ 1

1 Draw it

The rectangle lies on the floor: x from 0 to 1, y from 0 to 1.

Above it is the surface z = x² + y².

The double integral is the volume of this solid.

2 Find the limits

Write it as two integrals.

Inside: y from 0 to 1.

Outside: x from 0 to 1.

∫ ∫ (x² + y²) dA = ∫ from 0 to 1 of ∫ from 0 to 1 of (x² + y²) 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 1 of (x² + y²) dy = [x²y + y³3] from 0 to 1

Put in the top y = 1, then take away the bottom y = 0.

∫ from 0 to 1 of (x² + y²) dy = (x²(1) + (1)³3) − (x²(0) + (0)³3)

The bottom part is 0.

So it is x² + 1/3.

∫ from 0 to 1 of (x² + y²) dy = x² + 13

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 1 of (x² + 13) dx = [x³3 + x3] from 0 to 1

Put in the top x = 1, then take away the bottom x = 0.

∫ from 0 to 1 of (x² + 13) dx = ((1)³3 + (1)3) − ((0)³3 + (0)3)

The bottom part is 0.

So it is 2/3.

∫ from 0 to 1 of (x² + 13) dx = 23

5 Check

Let's check in the other order.

x first: 1/3 + y².

∫ from 0 to 1 of (x² + y²) dx = 1 over 3 + y²

Then y: 2/3.

The same answer.

Right!

∫ from 0 to 1 of (13 + y²) dy = 2/3

Let's look back.

First: draw it.

Then: find the limits.

Then: the inner integral.

Then: the outer integral.

Last: check.

Answer