WeBWorK #4
Z
1 Z 1+x
Evaluate the iterated integral I =
0
Z
I
1
=
18x 2 y + y 2
1+x
0
Z
(18x 2 + 2y ) dy dx.
1−x
dx
1−x
1
18x 2 (1 + x) + (1 + x)2 − 18x 2 (1 − x) − (1 − x)2 dx
=
0
Z
1
18x 2 + 18x 3 + 1 + 2x + x 2 − 18x 2 + 18x 3 − 1 + 2x − x 2 dx
=
0
Z
=
1
36x 3 + 4x dx
0
=
4
9x + 2x
2
1
0
= (9 + 2) − (0)
= 11
Math 236-Multi (Sklensky)
In-Class Work
April 19, 2013
1 / 25
Where we’re going:
I
In Calc 2, you saw that integration has many applications beyond just
area
I
Similarly, double-integration also has many applications
I
Today, I’m going to give you a quick overview of a couple, and then
focus on one in particular
I
I
Quick mentions:
I
Area
I
Volume between two surfaces
A bit longer on:
I
Mass and Center of mass
I
Surface area
Math 236-Multi (Sklensky)
In-Class Work
April 19, 2013
2 / 25
Volume Between 2 Surfaces
I
If f (x, y ) ≥ g (x, y ) over a region
R, then the volume between the
two surfaces is
ZZ
Volume =
f (x, y )−g (x, y ) dA
R
I
Math 236-Multi (Sklensky)
Notice that this is not Signed
Volume: as long as you’ve
correctly identified which surface
is above and which is below, this
will always be positive.
In-Class Work
April 19, 2013
3 / 25
Question:
Suppose we have a thin plate, or lamina in the shape of a region R ⊆ R2 ,
and the density of this lamina varies depending on where on the plate you
are.
R
Where should we put a support so that the plate can balance perfectly
horizontally?
That is, where is the center of mass?
Math 236-Multi (Sklensky)
In-Class Work
April 19, 2013
4 / 25
Mass, in 2 dimensions:
Suppose we’re given a thin plate, or lamina, in the shape of a region
R ⊂ R2 .
R
If the density of the lamina is a constant, that is if density= ρ g/cm2 , then
mass = (density)(area of lamina) =⇒ m = ρA.
(We can of course find the area
Z Z either using traditional calc 1 means, or
using a double integral A =
dA.)
R
Math 236-Multi (Sklensky)
In-Class Work
April 19, 2013
5 / 25
Mass, in 2 dimensions:
R
But ... what if the density of the lamina is not constant? What if it varies
throughout the plate? That is, what if density=ρ(x, y )?
Math 236-Multi (Sklensky)
In-Class Work
April 19, 2013
6 / 25
R
Mass, in 2 dimensions:
Partition R into smaller regions Ri (only using rectangles that lie
completely within R), and pick (ui , vi ) ∈ Ri .
(u1,v1)
R1
(u3,v3)
R2 R3
(u2,v2)
R4
(u4,v4)
If the regions are small enough, the density won’t vary much over them,
and so we can treat the density over any sub-region as if it’s constant. We
will use the density at (ui , vi ).
Math 236-Multi (Sklensky)
In-Class Work
April 19, 2013
7 / 25
Mass in 2 dimensions
(u1,v1)
R1
(u3,v3)
R2 R3
(u2,v2)
R4
(u4,v4)
Over the region Ri , we approximate the density with ρ(ui , vi ); the area is
∆Ai . Thus the mass of the ith bit of lamina is approximated by
mi ≈ ρ(ui , vi )∆Ai ,
Math 236-Multi (Sklensky)
In-Class Work
April 19, 2013
8 / 25
Mass in 2 dimensions
(u1,v1)
R1
(u3,v3)
R2 R3
(u2,v2)
R4
(u4,v4)
Over the region Ri , we approximate the density with ρ(ui , vi ); the area is
∆Ai . Thus the mass of the ith bit of lamina is approximated by
mi ≈ ρ(ui , vi )∆Ai ,
and so
total mass of lamina ≈
n
X
mi =
i=1
Math 236-Multi (Sklensky)
In-Class Work
n
X
ρ(ui , vi )∆Ai .
i=1
April 19, 2013
8 / 25
Mass, in 2 dimensions
(u1,v1)
R1
(u3,v3)
R2 R3
(u2,v2)
Since
total mass of lamina ≈
n
X
R4
(u4,v4)
mi =
i=1
=⇒ m =
lim
ZZ
X
i→∞
=⇒ m =
n
X
ρ(ui , vi )∆Ai ,
i=1
ρ(ui , vi ) ∆Ai
ρ(x, y ) dA
R
Math 236-Multi (Sklensky)
In-Class Work
April 19, 2013
9 / 25
Center of Mass, in 2 dimensions
Back to our Original Question:
Suppose we have a thin plate, or lamina in the shape of a region R ⊆ R2 ,
and the density of this lamina varies depending on where on the plate you
are.
R
Where should we put a support so that the plate can balance perfectly
horizontally? That is, where is the center of mass?
Math 236-Multi (Sklensky)
In-Class Work
April 19, 2013
10 / 25
Determining the Surface Area of z = f (x, y ) over a
region R:
First, subdivide the region R
into rectangles Ri , and for each
rectangle, choose one of the corner and label it (ui , vi ).
Math 236-Multi (Sklensky)
In-Class Work
April 19, 2013
11 / 25
Determining the Surface Area of z = f (x, y ) over a
region R:
For each i, take the bit of the tangent plane at ui , vi , f (ui , vi ) that
lies over the ith rectangle Ri .
This will be a parallelogram.
The size and shape of each “tangent parallelogram” will of course
vary, depending on which rectangle
Ri we’re looking at.
Math 236-Multi (Sklensky)
In-Class Work
April 19, 2013
12 / 25
Determining the Surface Area of z = f (x, y ) over a
region R:
Plan:
Find the area of each parallelogram, and add them all up.
Then take the limit as the diagonals
of our rectangles approach 0
This amounts to ∆A → 0, and so
we get a double integral.
Math 236-Multi (Sklensky)
In-Class Work
April 19, 2013
13 / 25
Plan: Find the area of each parallelogram, and add them all up. Then
take the limit.
Remember: the cross-product of two vectors gives the area of the
parallelogram they span!
Why?
→
−
−
Area = →
ai × bi
Math 236-Multi (Sklensky)
area = (base)(height)
→
−
→
−
−
−
= k→
ai kk bi k sin θ = →
ai × bi
In-Class Work
April 19, 2013
14 / 25
Determining the Surface Area of z = f (x, y ) over a
region R:
I
I
→
−
−
For each Ri , let →
a i k xz-plane, b i k yz-plane.
→
−
→
−
a i = ∆x, 0, fx (ui , vi )∆x and b i = 0, ∆y , fy (ui , vi )∆y .
(Why? Think about it!)
I
SAi
→
−
−
≈ k→
a i × b ik
..
.
≈ k − fx (ui , vi )∆x∆y , −fy (ui , vi )∆x∆y , ∆x∆y k
q
≈
fx (ui , vi )2 + fy (ui , vi )2 + 1∆x∆y
SA ≈
Xq
fx (ui , vi )2 + fy (ui , vi )2 + 1∆x∆y
i
Math 236-Multi (Sklensky)
In-Class Work
April 19, 2013
15 / 25
Double Integrals: Not just for Volume anymore
Math 236-Multi (Sklensky)
In-Class Work
April 19, 2013
16 / 25
Double Integrals: Not just for Volume anymore
ZZ
I
top surface − bottom surface dA
Volume between 2 surfaces =
R
Math 236-Multi (Sklensky)
In-Class Work
April 19, 2013
16 / 25
Double Integrals: Not just for Volume anymore
ZZ
I
I
Volume between 2 surfaces =
top surface − bottom surface dA
R
ZZ
Area:
dA = the area of the region R.
R
Math 236-Multi (Sklensky)
In-Class Work
April 19, 2013
16 / 25
Double Integrals: Not just for Volume anymore
ZZ
I
I
I
Volume between 2 surfaces =
top surface − bottom surface dA
R
ZZ
Area:
dA = the area of the region R.
R
ZZ
Mass=
ρ(x, y ) dA, where ρ(x, y ) = density of 2d plate at (x, y )
R
Math 236-Multi (Sklensky)
In-Class Work
April 19, 2013
16 / 25
Double Integrals: Not just for Volume anymore
ZZ
I
I
I
Volume between 2 surfaces =
top surface − bottom surface dA
R
ZZ
Area:
dA = the area of the region R.
R
ZZ
Mass=
ρ(x, y ) dA, where ρ(x, y ) = density of 2d plate at (x, y )
R
I
Center of Mass= (x̄, ȳ ), where
ZZ
xρ(x, y ) dA
R
Z
Z
x̄ =
ρ(x, y ) dA
R
Math 236-Multi (Sklensky)
ZZ
y ρ(x, y ) dA
ȳ = Z ZR
ρ(x, y ) dA
R
In-Class Work
April 19, 2013
16 / 25
Double Integrals: Not just for Volume anymore
ZZ
I
I
I
Volume between 2 surfaces =
top surface − bottom surface dA
R
ZZ
Area:
dA = the area of the region R.
R
ZZ
Mass=
ρ(x, y ) dA, where ρ(x, y ) = density of 2d plate at (x, y )
R
I
Center of Mass= (x̄, ȳ ), where
ZZ
xρ(x, y ) dA
R
Z
Z
x̄ =
ρ(x, y ) dA
R
I
ZZ
y ρ(x, y ) dA
ȳ = Z ZR
ρ(x, y ) dA
R
ZZ q
Surface Area=
fx (x, y )2 + fy (x, y )2 + 1
R
Math 236-Multi (Sklensky)
In-Class Work
April 19, 2013
16 / 25
In Class Work
1. Set up
of mass of the lamina
the formulas for the 2center
R = 0 ≤ x ≤ 1, 0 ≤ y ≤ x with density function
ρ(x, y ) = 1 + x 2 + y 2 .
Do not worry (yet) about whether you’ll need to reverse the order of
integration.
2. Set up the surface area of the portion of z = x 2 + y 2 between
x = 4 − y 2 and x = 1
3. (If time) Go back and evaluate these integrals, or determine whether
you’d need to use technology.
Math 236-Multi (Sklensky)
In-Class Work
April 19, 2013
17 / 25
One Double Integral, Many Stories
Consider the integral we found in #2,
√
Z
3
√
− 3
I
Z
4−y 2
p
4x 2 + 4y 2 + 1 dx dy
1
As we found, this integral gives the surface area of z = x 2 + y 2 over
the region between x = 4 − y 2 and x = 1.
Math 236-Multi (Sklensky)
In-Class Work
April 19, 2013
18 / 25
One Double Integral, Many Stories
Consider the integral we found in #2,
√
Z
3
√
− 3
I
I
Z
4−y 2
p
4x 2 + 4y 2 + 1 dx dy
1
As we found, this integral gives the surface area of z = x 2 + y 2 over
the region between x = 4 − y 2 and x = 1.
p
It also gives the volume below z = 4x 2 + 4y 2 + 1 and above that
same region
Math 236-Multi (Sklensky)
In-Class Work
April 19, 2013
18 / 25
One Double Integral, Many Stories
Consider the integral we found in #2,
√
Z
3
√
− 3
I
I
I
Z
4−y 2
p
4x 2 + 4y 2 + 1 dx dy
1
As we found, this integral gives the surface area of z = x 2 + y 2 over
the region between x = 4 − y 2 and x = 1.
p
It also gives the volume below z = 4x 2 + 4y 2 + 1 and above that
same region
And it gives
p the mass of a parabolic lamina with density function
ρ(x, y ) = 4x 2 + 4y 2 + 1 dx dy .
Math 236-Multi (Sklensky)
In-Class Work
April 19, 2013
18 / 25
One Double Integral, Many Stories
Consider the integral we found in #2,
√
Z
3
√
− 3
I
I
I
Z
4−y 2
p
4x 2 + 4y 2 + 1 dx dy
1
As we found, this integral gives the surface area of z = x 2 + y 2 over
the region between x = 4 − y 2 and x = 1.
p
It also gives the volume below z = 4x 2 + 4y 2 + 1 and above that
same region
And it gives
p the mass of a parabolic lamina with density function
ρ(x, y ) = 4x 2 + 4y 2 + 1 dx dy .
The same integral can have different interpretations depending upon the
context.
Math 236-Multi (Sklensky)
In-Class Work
April 19, 2013
18 / 25
Solutions
1. Find the center of mass of the lamina R = 0 ≤ x ≤ 1, 0 ≤ y ≤ x 2
with density function ρ(x, y ) = 1 + x 2 + y 2 .
ZZ
ZZ
xρ(x, y ) dA
y ρ(x, y ) dA
x̄ = Z ZR
ȳ = Z ZR
ρ(x, y ) dA
ρ(x, y ) dA
R
R
I will find the double integrals separately, and then put it all together to
find my center of mass.
Math 236-Multi (Sklensky)
In-Class Work
April 19, 2013
19 / 25
Solutions
1. Find the center of mass of the lamina R = 0 ≤ x ≤ 1, 0 ≤ y ≤ x 2
with density function ρ(x, y ) = 1 + x 2 + y 2 .
First, I’ll find the mass:
ZZ
m =
ρ(x, y ) dA
R
!
Z
Z 2
1
x
1 + x 2 + y 2 dy
=
0
Z
1
=
0
Z
=
0
dx
0
1
x 2 !
1
dx
y + x 2 y + y 3 3 0
1
x 2 + x 4 + x 6 dx
3
.
= ..
=
Math 236-Multi (Sklensky)
61
≈ 0.581
105
In-Class Work
April 19, 2013
20 / 25
Solutions
1. Find the center of mass of the lamina R = 0 ≤ x ≤ 1, 0 ≤ y ≤ x 2
with density function ρ(x, y ) = 1 + x 2 + y 2 .
61
≈ 0.581.
105
m=
Next, I’ll find the numerator of x̄.
ZZ
Z 1
xρ(x, y ) dA =
R
1
=
0
Z
!
x2
x(1 + x 2 + y 2 ) dy
dx
0
0
Z
Z
x 2 !
1
xy + x 3 y + xy 3 dx
3
0
1
1
x 3 + x 5 + x 7 dx
3
0
11
= ... =
≈ 0.458
24
=
Math 236-Multi (Sklensky)
In-Class Work
April 19, 2013
21 / 25
Solutions
1. Find the center of mass of the lamina R = 0 ≤ x ≤ 1, 0 ≤ y ≤ x 2
with density function ρ(x, y ) = 1 + x 2 + y 2 .
m=
61
≈ 0.581
105
Finally, I’ll find the numerator of ȳ
ZZ
Z 1
y ρ(x, y ) dA =
R
0
=
0
Z
Z
11
≈ 0.458.
24
!
x2
y (1 + x 2 + y 2 ) dy
dx
0
1
Z
My =
2!
1 2 1 2 2 1 4 x
y + x y + y dx
2
2
4 0
1
1 4 1 6 1 8
x + x + x dx
2
4
0 2
251
= ... =
≈ 0.199
1260
=
Math 236-Multi (Sklensky)
In-Class Work
April 19, 2013
22 / 25
Solutions
1. Find the center of mass of the lamina R = 0 ≤ x ≤ 1, 0 ≤ y ≤ x 2
with density function ρ(x, y ) = 1 + x 2 + y 2 .
m=
61
≈ 0.581
105
numx =
11
≈ 0.458
24
numy =
251
≈ 0.199.
1260
And now I’m ready to find the center of mass:
x̄
ȳ
=
11
24 = 11 105 = 385 ≈ 0.78893
61
24 61
488
105
=
251
1260 = 251 105 = 251 ≈ .342896
61
1260 61
732
105
Math 236-Multi (Sklensky)
In-Class Work
April 19, 2013
23 / 25
Solutions
2. Find the surface area of the portion of z = x 2 + y 2 between x = 4 − y 2
and x = 1
1.5
1.0
0.5
0.0
0.0
0.4
0.8
1.2
1.6
2.0
2.4
2.8
3.2
3.6
4.0
x
−0.5
−1.0
−1.5
ZZ q
SA =
fx (x, y )2 + fy (x, y )2 + 1 dA
Z
=
=
Math 236-Multi (Sklensky)
R
√
3
Z
4−y 2
√
− 3 1
√
Z 3 Z 4−y 2
√
− 3
q
fx (x, y )2 + fy (x, y )2 + 1 dx dy
p
4x 2 + 4y 2 + 1 dx dy
1
In-Class Work
April 19, 2013
24 / 25
2. Find the surface area of the portion of z = x 2 + y 2 between x = 4 − y 2
and x = 1
1.5
1.0
0.5
0.0
0.0
0.4
0.8
1.2
1.6
2.0
2.4
2.8
3.2
3.6
4.0
x
−0.5
−1.0
−1.5
√
Z
SA =
3
√
− 3
Z
4−y 2
p
4x 2 + 4y 2 + 1 dx dy
1
Can’t antidifferentiate. Could try reversing the order of integration, but
that square root is still intractable.
In fact, Maple can’t do it exactly either.
So . . . approximate!
SA ≈ 33.47.
Math 236-Multi (Sklensky)
In-Class Work
April 19, 2013
25 / 25