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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


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.


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?


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.)


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)?


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 ).


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 .


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


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?


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 ).



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.



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.


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



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


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


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.


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.


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.


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



{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



{0 ≤ x ≤ 1, 0 ≤ y ≤ x2


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



{0 ≤ x ≤ 1, 0 ≤ y ≤ x2


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


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


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.


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