Double integral of x^2 + y^2 over the upper half of the ring 1 ≤ x^2… Step by Step

Step by step

The question

Question: find the double integral of x² + y² over the upper half of the ring 1 ≤ x² + y² ≤ 9.

∫ ∫ (x² + y²) dA, the upper half, 1 ≤ x² + y² ≤ 9

1 Draw the region

The region on the floor is the upper half of a ring: between radius 1 and radius 3.

Above it is the surface z = x² + y².

The integral is the volume of this solid.

2 Change to polar

In polar form, x² + y² = r², and the small area is dA = r dr dθ.

x² + y² = r², dA = r dr dθ

Why the extra r?

The small pieces far from the center are wider.

Their width grows with r.

r goes from 1 to 3.

θ goes from 0 to π: the upper half.

= r: 1 to 3, θ: 0 to π

Write it in r and θ.

x² + y² becomes r².

Then times r dr dθ.

∫ ∫ (x² + y²) dA = ∫ from 0 to π of ∫ from 1 to 3 of r² × r dr dθ

Careful!

Without the extra r, the answer is wrong.

It is the most common mistake here.

Careful: dA = r dr dθ: do not forget the r

3 The inner integral

Multiply by r first: r³.

∫ ∫ (x² + y²) dA = ∫ from 0 to π of ∫ from 1 to 3 of r³ dr dθ

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

∫ from 1 to 3 of r³ dr = r⁴4 from 1 to 3

Put in the top r = 3, then take away the bottom r = 1.

∫ from 1 to 3 of r³ dr = (3)⁴4 − (1)⁴4

So it is 20.

∫ from 1 to 3 of r³ dr = 20

4 The outer integral

Now θ.

20 is a plain number: its integral is 20 times θ.

∫ from 0 to π of 20 dθ = 20θ from 0 to π

Top minus bottom: π minus 0 is π.

∫ from 0 to π of 20 dθ = 20 × π

So the answer is 20π.

∫ from 0 to π of 20 dθ = 20π

That is about 62.83.

∫ from 0 to π of 20 dθ ≈ 62.83

5 Check

Let's check if it makes sense.

The region's area is 4π.

On it, f goes from 1 to 9.

area = 4π ≈ 12.57, f from 1 to 9

So the answer must be between 12.57 and 113.1.

62.83 is.

Right!

12.57 ≤ 62.83 ≤ 113.1

Let's look back.

First: draw the region.

Then: change to polar.

Then: the inner integral.

Then: the outer integral.

Last: check.

Answer