MATH 19520/51 Class 12math.uchicago.edu/~mqt/math/teaching/math-195/math-195-51_class-12.pdf ·...
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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.
Minh-Tam Trinh MATH 19520/51 Class 12
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.
Minh-Tam Trinh MATH 19520/51 Class 12
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.
Minh-Tam Trinh MATH 19520/51 Class 12
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.
Minh-Tam Trinh MATH 19520/51 Class 12
Answer?
12 · π · 2
2 · 5 = 10π.
Minh-Tam Trinh MATH 19520/51 Class 12
Answer? 12 · π · 2
2 · 5 = 10π.
Minh-Tam Trinh MATH 19520/51 Class 12
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.
Minh-Tam Trinh MATH 19520/51 Class 12
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.
Minh-Tam Trinh MATH 19520/51 Class 12
Answer?
The base has area 12 ·√2 ·√2 = 1, so the volume
underneath the prism is 1 · 2 = 2.
Minh-Tam Trinh MATH 19520/51 Class 12
Answer? The base has area 12 ·√2 ·√2 = 1, so the volume
underneath the prism is 1 · 2 = 2.
Minh-Tam Trinh MATH 19520/51 Class 12
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.)
Minh-Tam Trinh MATH 19520/51 Class 12
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.)
Minh-Tam Trinh MATH 19520/51 Class 12
Again, we can use clever geometry.
It’s half the volume of a sphere of radius 3, i.e., 12(
43π · 3
2) = 6π.
Minh-Tam Trinh MATH 19520/51 Class 12
Again, we can use clever geometry.
It’s half the volume of a sphere of radius 3, i.e., 12(
43π · 3
2) = 6π.
Minh-Tam Trinh MATH 19520/51 Class 12
Integrating f (x, y) = 1√9–x2
over the disk x2 + y2 ≤ 9 looks. . . harder.
What kind of Riemann sum would you use?
Minh-Tam Trinh MATH 19520/51 Class 12
Integrating f (x, y) = 1√9–x2
over the disk x2 + y2 ≤ 9 looks. . . harder.
What kind of Riemann sum would you use?
Minh-Tam Trinh MATH 19520/51 Class 12
Integrating f (x, y) = 1√9–x2
over the disk x2 + y2 ≤ 9 looks. . . harder.
What kind of Riemann sum would you use?
Minh-Tam Trinh MATH 19520/51 Class 12
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
–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.
Minh-Tam Trinh MATH 19520/51 Class 12
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
–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.
Minh-Tam Trinh MATH 19520/51 Class 12
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
–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.
Minh-Tam Trinh MATH 19520/51 Class 12
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
a
∫ 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!
Minh-Tam Trinh MATH 19520/51 Class 12
Example
Compute∬D y dA, where D = {0 ≤ x ≤ π4 and sin x ≤ y ≤ cos x}.
First, ∫ π/4
0
∫ cos x
sin xy dy dx =
∫ π/4
0( 12y
2)��cos xy=sin x dx
=12
∫ π/4
0(cos2 x – sin2 x) dx.
(4)
Double-angle formula to the rescue: cos(2x) = cos2 x – sin2 x. Sothe last expression equals
12
∫ π/4
0cos(2x) dx =
12
(12sin(2x)
)����π/4x=0
=14.
(5)
Minh-Tam Trinh MATH 19520/51 Class 12
Example
Compute∬D y dA, where D = {0 ≤ x ≤ π4 and sin x ≤ y ≤ cos x}.
First, ∫ π/4
0
∫ cos x
sin xy dy dx =
∫ π/4
0( 12y
2)��cos xy=sin x dx
=12
∫ π/4
0(cos2 x – sin2 x) dx.
(4)
Double-angle formula to the rescue: cos(2x) = cos2 x – sin2 x. Sothe last expression equals
12
∫ π/4
0cos(2x) dx =
12
(12sin(2x)
)����π/4x=0
=14.
(5)
Minh-Tam Trinh MATH 19520/51 Class 12
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.
Minh-Tam Trinh MATH 19520/51 Class 12
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.
Minh-Tam Trinh MATH 19520/51 Class 12
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.
Minh-Tam Trinh MATH 19520/51 Class 12
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.
Minh-Tam Trinh MATH 19520/51 Class 12
Example (Stewart §15.2, Example 1, cont.)
Now, ∬D(x + 2y) dA
=∫ 1
–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
· · ·
=3215
.
(6)
See Stewart.
Minh-Tam Trinh MATH 19520/51 Class 12
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
c
∫ b(y)
a(y)f (x, y) dx dy.(8)
Minh-Tam Trinh MATH 19520/51 Class 12
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
∫ √1–y2–1
1 dx dy =∫ 1
–1
(√1 – y2 – (–1)
)dy
=∫ 1
–1
√1 – y2 dy +
∫ 1
–11 dy.
(9)
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.
Minh-Tam Trinh MATH 19520/51 Class 12
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
∫ √1–y2–1
1 dx dy =∫ 1
–1
(√1 – y2 – (–1)
)dy
=∫ 1
–1
√1 – y2 dy +
∫ 1
–11 dy.
(9)
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.
Minh-Tam Trinh MATH 19520/51 Class 12
The region where –1 ≤ x ≤√1 – y2.
Minh-Tam Trinh MATH 19520/51 Class 12