Post on 17-Mar-2020
WeBWorK #4
Evaluate the iterated integral I =
∫ 1
0
∫ 1+x
1−x(18x2 + 2y) dy dx .
I =
∫ 1
0
[18x2y + y2
]1+x
1−xdx
=
∫ 1
018x2(1 + x) + (1 + x)2 − 18x2(1− x)− (1− x)2 dx
=
∫ 1
018x2 + 18x3 + 1 + 2x + x2 − 18x2 + 18x3 − 1 + 2x − x2 dx
=
∫ 1
036x3 + 4x dx
=
[9x4 + 2x2
]10
= (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 justarea
I Similarly, double-integration also has many applications
I Today, I’m going to give you a quick overview of a couple, and thenfocus on one in particular
I Quick mentions:
I Area
I Volume between two surfaces
I 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 regionR, then the volume between thetwo surfaces is
Volume =
∫∫Rf (x , y)−g(x , y) dA
I Notice that this is not SignedVolume: as long as you’vecorrectly identified which surfaceis above and which is below, thiswill always be positive.
Math 236-Multi (Sklensky) 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 youare.
R
Where should we put a support so that the plate can balance perfectlyhorizontally?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 regionR ⊂ 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 either using traditional calc 1 means, or
using a double integral A =
∫∫RdA.)
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 variesthroughout the plate? That is, what if density=ρ(x , y)?
Math 236-Multi (Sklensky) In-Class Work April 19, 2013 6 / 25
Mass, in 2 dimensions:
Partition R into smaller regions Ri (only using rectangles that liecompletely within R), and pick (ui , vi ) ∈ Ri .
R
R1
R2 R3 R4
(u1,v1)
(u2,v2)
(u3,v3)
(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. Wewill use the density at (ui , vi ).
Math 236-Multi (Sklensky) In-Class Work April 19, 2013 7 / 25
Mass in 2 dimensions
R
R1
R2 R3 R4
(u1,v1)
(u2,v2)
(u3,v3)
(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∑
i=1
mi =n∑
i=1
ρ(ui , vi )∆Ai .
Math 236-Multi (Sklensky) In-Class Work April 19, 2013 8 / 25
Mass in 2 dimensions
R
R1
R2 R3 R4
(u1,v1)
(u2,v2)
(u3,v3)
(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∑
i=1
mi =n∑
i=1
ρ(ui , vi )∆Ai .
Math 236-Multi (Sklensky) In-Class Work April 19, 2013 8 / 25
Mass, in 2 dimensions
R
R1
R2 R3 R4
(u1,v1)
(u2,v2)
(u3,v3)
(u4,v4)
Since
total mass of lamina ≈n∑
i=1
mi =n∑
i=1
ρ(ui , vi )∆Ai ,
=⇒ m = limi→∞
∑ρ(ui , vi ) ∆Ai
=⇒ m =
∫∫Rρ(x , y) dA
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 youare.
R
Where should we put a support so that the plate can balance perfectlyhorizontally? 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 aregion R:
First, subdivide the region Rinto rectangles Ri , and for eachrectangle, choose one of the cor-ner 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 aregion R:
For each i , take the bit of the tan-gent 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 “tan-gent parallelogram” will of coursevary, depending on which rectangleRi 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 aregion R:
Plan:Find the area of each parallelo-gram, and add them all up.
Then take the limit as the diagonalsof our rectangles approach 0
This amounts to ∆A → 0, and sowe 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. Thentake the limit.
Remember: the cross-product of two vectors gives the area of theparallelogram they span!
Area = −→ai ×−→bi
Why?
area = (base)(height)
= ‖−→ai ‖‖−→bi‖ sin θ = −→ai ×
−→bi
Math 236-Multi (Sklensky) In-Class Work April 19, 2013 14 / 25
Determining the Surface Area of z = f (x , y) over aregion R:
I For each Ri , let −→a i ‖ xz-plane,−→b i ‖ yz-plane.
I −→a i =⟨∆x , 0, fx(ui , vi )∆x
⟩and−→b i =
⟨0,∆y , fy (ui , vi )∆y
⟩.
(Why? Think about it!)
I
SAi ≈ ‖−→a i ×−→b i‖
...
≈ ‖⟨− fx(ui , vi )∆x∆y ,−fy (ui , vi )∆x∆y ,∆x∆y
⟩‖
≈√fx(ui , vi )2 + fy (ui , vi )2 + 1∆x∆y
SA ≈∑i
√fx(ui , vi )2 + fy (ui , vi )2 + 1∆x∆y
Math 236-Multi (Sklensky) In-Class Work April 19, 2013 15 / 25
Double Integrals: Not just for Volume anymore
I Volume between 2 surfaces =
∫∫R
top surface− bottom surface dA
I Area:
∫∫R
dA = the area of the region R.
I Mass=
∫∫Rρ(x , y) dA, where ρ(x , y) = density of 2d plate at (x , y)
I Center of Mass= (x , y), where
x =
∫∫Rxρ(x , y) dA∫∫
Rρ(x , y) dA
y =
∫∫Ryρ(x , y) dA∫∫
Rρ(x , y) dA
I Surface Area=
∫∫R
√fx(x , y)2 + fy (x , y)2 + 1
Math 236-Multi (Sklensky) In-Class Work April 19, 2013 16 / 25
Double Integrals: Not just for Volume anymore
I Volume between 2 surfaces =
∫∫R
top surface− bottom surface dA
I Area:
∫∫R
dA = the area of the region R.
I Mass=
∫∫Rρ(x , y) dA, where ρ(x , y) = density of 2d plate at (x , y)
I Center of Mass= (x , y), where
x =
∫∫Rxρ(x , y) dA∫∫
Rρ(x , y) dA
y =
∫∫Ryρ(x , y) dA∫∫
Rρ(x , y) dA
I Surface Area=
∫∫R
√fx(x , y)2 + fy (x , y)2 + 1
Math 236-Multi (Sklensky) In-Class Work April 19, 2013 16 / 25
Double Integrals: Not just for Volume anymore
I Volume between 2 surfaces =
∫∫R
top surface− bottom surface dA
I Area:
∫∫R
dA = the area of the region R.
I Mass=
∫∫Rρ(x , y) dA, where ρ(x , y) = density of 2d plate at (x , y)
I Center of Mass= (x , y), where
x =
∫∫Rxρ(x , y) dA∫∫
Rρ(x , y) dA
y =
∫∫Ryρ(x , y) dA∫∫
Rρ(x , y) dA
I Surface Area=
∫∫R
√fx(x , y)2 + fy (x , y)2 + 1
Math 236-Multi (Sklensky) In-Class Work April 19, 2013 16 / 25
Double Integrals: Not just for Volume anymore
I Volume between 2 surfaces =
∫∫R
top surface− bottom surface dA
I Area:
∫∫R
dA = the area of the region R.
I Mass=
∫∫Rρ(x , y) dA, where ρ(x , y) = density of 2d plate at (x , y)
I Center of Mass= (x , y), where
x =
∫∫Rxρ(x , y) dA∫∫
Rρ(x , y) dA
y =
∫∫Ryρ(x , y) dA∫∫
Rρ(x , y) dA
I Surface Area=
∫∫R
√fx(x , y)2 + fy (x , y)2 + 1
Math 236-Multi (Sklensky) In-Class Work April 19, 2013 16 / 25
Double Integrals: Not just for Volume anymore
I Volume between 2 surfaces =
∫∫R
top surface− bottom surface dA
I Area:
∫∫R
dA = the area of the region R.
I Mass=
∫∫Rρ(x , y) dA, where ρ(x , y) = density of 2d plate at (x , y)
I Center of Mass= (x , y), where
x =
∫∫Rxρ(x , y) dA∫∫
Rρ(x , y) dA
y =
∫∫Ryρ(x , y) dA∫∫
Rρ(x , y) dA
I Surface Area=
∫∫R
√fx(x , y)2 + fy (x , y)2 + 1
Math 236-Multi (Sklensky) In-Class Work April 19, 2013 16 / 25
Double Integrals: Not just for Volume anymore
I Volume between 2 surfaces =
∫∫R
top surface− bottom surface dA
I Area:
∫∫R
dA = the area of the region R.
I Mass=
∫∫Rρ(x , y) dA, where ρ(x , y) = density of 2d plate at (x , y)
I Center of Mass= (x , y), where
x =
∫∫Rxρ(x , y) dA∫∫
Rρ(x , y) dA
y =
∫∫Ryρ(x , y) dA∫∫
Rρ(x , y) dA
I Surface Area=
∫∫R
√fx(x , y)2 + fy (x , y)2 + 1
Math 236-Multi (Sklensky) In-Class Work April 19, 2013 16 / 25
In Class Work
1. Set up the formulas for the center of mass of the laminaR =
{0 ≤ x ≤ 1, 0 ≤ y ≤ x2
}with density function
ρ(x , y) = 1 + x2 + y2.
Do not worry (yet) about whether you’ll need to reverse the order ofintegration.
2. Set up the surface area of the portion of z = x2 + y2 betweenx = 4− y2 and x = 1
3. (If time) Go back and evaluate these integrals, or determine whetheryou’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,∫ √3−√3
∫ 4−y2
1
√4x2 + 4y2 + 1 dx dy
I As we found, this integral gives the surface area of z = x2 + y2 overthe region between x = 4− y2 and x = 1.
I It also gives the volume below z =√
4x2 + 4y2 + 1 and above thatsame region
I And it gives the mass of a parabolic lamina with density functionρ(x , y) =
√4x2 + 4y2 + 1 dx dy .
The same integral can have different interpretations depending upon thecontext.
Math 236-Multi (Sklensky) In-Class Work April 19, 2013 18 / 25
One Double Integral, Many Stories
Consider the integral we found in #2,∫ √3−√3
∫ 4−y2
1
√4x2 + 4y2 + 1 dx dy
I As we found, this integral gives the surface area of z = x2 + y2 overthe region between x = 4− y2 and x = 1.
I It also gives the volume below z =√
4x2 + 4y2 + 1 and above thatsame region
I And it gives the mass of a parabolic lamina with density functionρ(x , y) =
√4x2 + 4y2 + 1 dx dy .
The same integral can have different interpretations depending upon thecontext.
Math 236-Multi (Sklensky) In-Class Work April 19, 2013 18 / 25
One Double Integral, Many Stories
Consider the integral we found in #2,∫ √3−√3
∫ 4−y2
1
√4x2 + 4y2 + 1 dx dy
I As we found, this integral gives the surface area of z = x2 + y2 overthe region between x = 4− y2 and x = 1.
I It also gives the volume below z =√
4x2 + 4y2 + 1 and above thatsame region
I And it gives the mass of a parabolic lamina with density functionρ(x , y) =
√4x2 + 4y2 + 1 dx dy .
The same integral can have different interpretations depending upon thecontext.
Math 236-Multi (Sklensky) In-Class Work April 19, 2013 18 / 25
One Double Integral, Many Stories
Consider the integral we found in #2,∫ √3−√3
∫ 4−y2
1
√4x2 + 4y2 + 1 dx dy
I As we found, this integral gives the surface area of z = x2 + y2 overthe region between x = 4− y2 and x = 1.
I It also gives the volume below z =√
4x2 + 4y2 + 1 and above thatsame region
I And it gives the mass of a parabolic lamina with density functionρ(x , y) =
√4x2 + 4y2 + 1 dx dy .
The same integral can have different interpretations depending upon thecontext.
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 ≤ x2}
with density function ρ(x , y) = 1 + x2 + y2.
x =
∫∫Rxρ(x , y) dA∫∫
Rρ(x , y) dA
y =
∫∫Ryρ(x , y) dA∫∫
Rρ(x , y) dA
I will find the double integrals separately, and then put it all together tofind my center of mass.
Math 236-Multi (Sklensky) In-Class Work April 19, 2013 19 / 25
Solutions1. Find the center of mass of the lamina R =
{0 ≤ x ≤ 1, 0 ≤ y ≤ x2
}with density function ρ(x , y) = 1 + x2 + y2.First, I’ll find the mass:
m =
∫∫Rρ(x , y) dA
=
∫ 1
0
(∫ x2
01 + x2 + y2 dy
)dx
=
∫ 1
0
(y + x2y +
1
3y3∣∣∣∣x20
)dx
=
∫ 1
0x2 + x4 +
1
3x6 dx
=...
=61
105≈ 0.581
Math 236-Multi (Sklensky) In-Class Work April 19, 2013 20 / 25
Solutions1. Find the center of mass of the lamina R =
{0 ≤ x ≤ 1, 0 ≤ y ≤ x2
}with density function ρ(x , y) = 1 + x2 + y2.
m =61
105≈ 0.581.
Next, I’ll find the numerator of x .∫∫Rxρ(x , y) dA =
∫ 1
0
(∫ x2
0x(1 + x2 + y2) dy
)dx
=
∫ 1
0
(xy + x3y +
1
3xy3∣∣∣∣x20
)dx
=
∫ 1
0x3 + x5 +
1
3x7 dx
= . . . =11
24≈ 0.458
Math 236-Multi (Sklensky) In-Class Work April 19, 2013 21 / 25
Solutions1. Find the center of mass of the lamina R =
{0 ≤ x ≤ 1, 0 ≤ y ≤ x2
}with density function ρ(x , y) = 1 + x2 + y2.
m =61
105≈ 0.581 My =
11
24≈ 0.458.
Finally, I’ll find the numerator of y∫∫Ryρ(x , y) dA =
∫ 1
0
(∫ x2
0y(1 + x2 + y2) dy
)dx
=
∫ 1
0
(1
2y2 +
1
2x2y2 +
1
4y4∣∣∣∣x20
)dx
=
∫ 1
0
1
2x4 +
1
2x6 +
1
4x8 dx
= . . . =251
1260≈ 0.199
Math 236-Multi (Sklensky) In-Class Work April 19, 2013 22 / 25
Solutions1. Find the center of mass of the lamina R =
{0 ≤ x ≤ 1, 0 ≤ y ≤ x2
}with density function ρ(x , y) = 1 + x2 + y2.
m =61
105≈ 0.581 numx =
11
24≈ 0.458 numy =
251
1260≈ 0.199.
And now I’m ready to find the center of mass:
x =
11
2461
105
=11
24
105
61=
385
488≈ 0.78893
y =
251
126061
105
=251
1260
105
61=
251
732≈ .342896
Math 236-Multi (Sklensky) In-Class Work April 19, 2013 23 / 25
Solutions2. Find the surface area of the portion of z = x2 + y2 between x = 4− y2
and x = 1
1.5
x
1.0
0.5
4.00.0
−0.5
3.6
−1.0
−1.5
3.22.82.42.01.61.20.80.40.0
SA =
∫∫R
√fx(x , y)2 + fy (x , y)2 + 1 dA
=
∫ √3−√3
∫ 4−y2
1
√fx(x , y)2 + fy (x , y)2 + 1 dx dy
=
∫ √3−√3
∫ 4−y2
1
√4x2 + 4y2 + 1 dx dy
Math 236-Multi (Sklensky) In-Class Work April 19, 2013 24 / 25
2. Find the surface area of the portion of z = x2 + y2 between x = 4− y2
and x = 1
1.5
x
1.0
0.5
4.00.0
−0.5
3.6
−1.0
−1.5
3.22.82.42.01.61.20.80.40.0
SA =
∫ √3−√3
∫ 4−y2
1
√4x2 + 4y2 + 1 dx dy
Can’t antidifferentiate. Could try reversing the order of integration, butthat 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