math 19520/51 class 12math.uchicago.edu/~mqt/math/teaching/math-195/math-195-51_class-12.pdf ·...

MATH 19520/51 Class 12

Minh-Tam Trinh

University of Chicago

2017-10-25

Minh-Tam Trinh MATH 19520/51 Class 12

1 Review double integrals.2 Double integrals over weird regions in the plane.

Review of Double Integrals

What is the integral of f (x, y) =√4 – x2 over the rectangle

[–2, 2] × [0, 5]?

Direct way: It’s∫ 50

∫ 2–2

√4 – x2 dx dy.

Geometric way: It’s the half the volume of a cylinder of radius 2and height 5.

[–2, 2] × [0, 5]?

∫ 2–2

√4 – x2 dx dy.

[–2, 2] × [0, 5]?

∫ 2–2

√4 – x2 dx dy.

Answer?

12 · π · 2

2 · 5 = 10π.

Answer? 12 · π · 2

2 · 5 = 10π.

What is the integral of f (x, y) = 1 –��y�� over the rectangle

[–1, 1] × [–1, 1]?

It’s a triangular prism. Its base is a 45◦-45◦-90◦ triangle ofhypotenuse 2, and its height is 2.

What is the integral of f (x, y) = 1 –��y�� over the rectangle

[–1, 1] × [–1, 1]?

It’s a triangular prism. Its base is a 45◦-45◦-90◦ triangle ofhypotenuse 2, and its height is 2.

Answer?

The base has area 12 ·√2 ·√2 = 1, so the volume

underneath the prism is 1 · 2 = 2.

Answer? The base has area 12 ·√2 ·√2 = 1, so the volume

underneath the prism is 1 · 2 = 2.

Double Integrals over Weird Regions

In the previous two cases, we were integrating over a rectangularregion of the xy-plane.

But integration makes sense over other shapes as well.

The disk in the xy-plane is given by x2 + y2 ≤ 9. How would youintegrate f (x, y) =

√9 – x2 – y2 over this disk?

(Stewart would write∬x2+y2≤9 f (x, y) dA.)

Double Integrals over Weird Regions

In the previous two cases, we were integrating over a rectangularregion of the xy-plane.

But integration makes sense over other shapes as well.

The disk in the xy-plane is given by x2 + y2 ≤ 9. How would youintegrate f (x, y) =

√9 – x2 – y2 over this disk?

(Stewart would write∬x2+y2≤9 f (x, y) dA.)

Again, we can use clever geometry.

It’s half the volume of a sphere of radius 3, i.e., 12(

43π · 3

2) = 6π.

Again, we can use clever geometry.

It’s half the volume of a sphere of radius 3, i.e., 12(

43π · 3

2) = 6π.

Integrating f (x, y) = 1√9–x2

over the disk x2 + y2 ≤ 9 looks. . . harder.

What kind of Riemann sum would you use?

Idea: In the xy-plane, the lower part of the disk is bounded byy = –√9 – x2 and the upper part is bounded by y =

√9 – x2.

Now look at ∫ 3

∫ √9–x2

–√9–x2

f (x, y) dy dx.(1)

After we compute the y-integral, we’ll have a function of x that stillmakes sense inside the x-integral!

Exercise: Go back and figure out the answer.

√9 – x2.

Now look at ∫ 3

∫ √9–x2

–√9–x2

f (x, y) dy dx.(1)

√9 – x2.

Now look at ∫ 3

∫ √9–x2

–√9–x2

f (x, y) dy dx.(1)

General idea: Suppose you want to integrate f (x, y) over the region

{(x, y) ∈ R2 : a ≤ x ≤ b and c(x) ≤ y ≤ d(x)}.(2)

Then you use ∫ b

∫ d(x)

c(x)f (x, y) dy dx.(3)

Warning!

For the double integral to give a number, you need the x-integral onthe outside and the y-integral on the inside. The order matters!

Example

Compute∬D y dA, where D = {0 ≤ x ≤ π4 and sin x ≤ y ≤ cos x}.

First, ∫ π/4

∫ cos x

sin xy dy dx =

∫ π/4

0( 12y

2)��cos xy=sin x dx

∫ π/4

0(cos2 x – sin2 x) dx.

Double-angle formula to the rescue: cos(2x) = cos2 x – sin2 x. Sothe last expression equals

∫ π/4

0cos(2x) dx =

(12sin(2x)

)��π/4x=0

Example

Compute∬D y dA, where D = {0 ≤ x ≤ π4 and sin x ≤ y ≤ cos x}.

First, ∫ π/4

∫ cos x

sin xy dy dx =

∫ π/4

0( 12y

2)��cos xy=sin x dx

∫ π/4

0(cos2 x – sin2 x) dx.

Double-angle formula to the rescue: cos(2x) = cos2 x – sin2 x. Sothe last expression equals

∫ π/4

0cos(2x) dx =

(12sin(2x)

)��π/4x=0

The green region is where 0 ≤ x ≤ π4 and the blue region is wheresin x ≤ y ≤ cos x. We integrated f (x, y) = y on their overlap.

Example (Stewart §15.2, Example 1)

The parabolas y = 2x2 and y = 1 + x2 bound a region D. Find∬D(x + 2y) dA.

We aren’t given the bounds on x, so we need to find them.

The parabolas intersect at (–1, 2) and (1, 2), so we infer D isbounded by –1 ≤ x ≤ 1.

Example (Stewart §15.2, Example 1, cont.)

Now, ∬D(x + 2y) dA

=∫ 1

∫ 1+x2

2x2(x + 2y) dy dx

=∫ 1

–1(xy + y2)

��1+x2y=2x2 dx

=∫ 1

–1((x(1 + x2) + (1 + x2)2) – (x(2x2) – (2x2)2)) dx

· · ·

See Stewart.

Very similar: Suppose you want to integrate f (x, y) over the region

{(x, y) ∈ R2 : a(y) ≤ x ≤ b(y) and c ≤ y ≤ d}.(7)

Then you use ∫ d

∫ b(y)

a(y)f (x, y) dx dy.(8)

Example

Integrate the function f (x, y) = 1 over the region where–1 ≤ x ≤

√1 – y2 and

��y�� ≤ 1.

Note that��y�� ≤ 1 is the same as –1 ≤ y ≤ 1. Plugging in,∫ 1

∫ √1–y2–1

1 dx dy =∫ 1

(√1 – y2 – (–1)

=∫ 1

√1 – y2 dy +

–11 dy.

In the final expression, the first integral equals half the area of adisk of radius 1, i.e., π2 ; and the second integral equals 2. So theanswr is π2 + 2.

Example

Integrate the function f (x, y) = 1 over the region where–1 ≤ x ≤

√1 – y2 and

��y�� ≤ 1.

Note that��y�� ≤ 1 is the same as –1 ≤ y ≤ 1. Plugging in,∫ 1

∫ √1–y2–1

1 dx dy =∫ 1

(√1 – y2 – (–1)

=∫ 1

√1 – y2 dy +

–11 dy.

In the final expression, the first integral equals half the area of adisk of radius 1, i.e., π2 ; and the second integral equals 2. So theanswr is π2 + 2.

The region where –1 ≤ x ≤√1 – y2.

math 19520/51 class 12math.uchicago.edu/~mqt/math/teaching/math-195/math-195-51_class-12.pdf ·...

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