Estimating Areas Consider the challenge of estimating the volume of - - PDF document

Estimating Areas

Consider the challenge of estimating the volume of a solid {(x, y, z)|0 ≤ z ≤ f(x, y), (x, y) ∈ R}, where R is a region in the xy-plane. This may be thought of as the solid under the graph of z = f(x, y) and above the plane region R. We could try partitioning R into a large number n of small, compact subregions Ri, i = 1, 2, . . . , n. For each region Ri, we could pick some point Pi = (x∗

i , y∗ i ) ∈ Ri and estimate the volume of the portion of the

region aboue Ri by f(x∗

i , y∗ i )∆Ai, where ∆Ai is the area of that region.

We could then estimate the entire volume by n

i=1 f(x∗ i , y∗ i )∆Ai. This

is reminiscent of a Riemann Sum and, amazingly enough, will be called a Riemann Sum.

Double Integrals

If these Reimann Sums in a reasonable sense approach a limit when the regions get very small in all directions, we call that limit the double integral of the function over the region and denote it by

• R f(x, y) dA
• r
• R f(x, y) dx dy.

Properties of Multiple Integrals

In general, properties of ordinary integrals which make some sense hold for double integrals and for multiple integrals in general.

• The integral of a sum or difference is the sum or difference of

the integrals.

• If one function is bigger than another, then its integral is bigger.
• If a function is smaller than some constant, then its integral

is smaller than that constant times the area of the region over which the integration is done.

• If a region is partitioned into subregions, then the integral over

the entire region is equal to the sum of the integrals over the subregions.

Applications of Double Integrals

• R dA is equal to the area of R.
• If f(x, y) > 0, then
• R f(x, y) dA is equal to the volume of the

region below the graph of z = f(x, y) and above the region R in the plane. We will come up with additional applications in the near future.

Calculation of a Double Integral as an Iterated Integral

1

2

The calculation of most double integrals reduces to repeated ordinary integration. Type I Region: A Type I Region is a plane region bounded on the left and right by vertical lines and above and below by the graphs of functions, which we will generally assume to be continuous. We will for the present describe it as {(x, y)|α(x) ≤ y ≤ β(x), a ≤ x ≤ b}. a < b would be real numbers and α(x) < β(x) would be functions. Suppose we let P = {x0, x1, x2, x3, . . . , xn} be a partition of [a, b] and let Ri be the portion of R for which xi−1 ≤ x ≤ xi. We can then calculate

• R f(x, y) dA = n

i=1

• Ri f(x, y) dA.

Partition and Riemann Sum

For a fixed i, we can estimate

• Ri f(x, y) dA by choosing some

xi−1 ≤ x∗

i ≤ xi, forming a partition {y0, y1, y2, . . . , ym} of the inter-

val [α(x∗

i ), β(x∗ i )], choosing yj−1 ≤ y∗ j ≤ yj for each j, and adding

m

j=1 f(x∗ i , y∗ j) ∆xi ∆yj =

m

j=1 f(x∗ i , y∗ j) ∆yj

• ∆xi.

The expression in the parentheses is actually a Riemann Sum which will be close to β(x∗

i )

α(x∗

i ) f(x∗

i , y) dy.

We can think of this as g(x∗

i ), where g(x) =

β(x)

α(x) f(x, y) dy.

So

• Ri f(x, y) dA ≈ g(x∗

i ) ∆xi, and thus

• R f(x, y) dA ≈ n

i=1 g(x∗ i ) ∆xi.

The Iterated Integral

This is itself a Riemann Sum, which will be close to b

a g(x) dx.

We thus get

• R f(x, y) dA ≈

b

a

β(x)

α(x) f(x, y) dy

• dx, which we usu-

ally write without the parentheses as b

a

β(x)

α(x) f(x, y) dy dx.

This is called an iterated integral. We calculate it by first treating x as a constant and calculating β(x)

α(x) f(x, y) dy. This gives us a function

• f x, which we then integrate with respect to x, from a to b.

Iterated Integrals Over Type II Regions

A region bounded above and below by horizontal lines and bounded

• n the left and right by curves which are the graphs of functions where

y is the independent variable and x is the dependent variable is called a Type II Region. It may be described as {(x, y)|α(y) ≤ x ≤ β(y), a ≤ y ≤ b}.

3

An integral over such a region may also be calculated as an iterated integral, for which we integrate with respect to x first and then y. If R is such a region, then

• R f(x, y) dA =

b

a

β(y)

α(y) f(x, y) dx dy.

Fubini’s Theorem

It’s natural to ask, if a region is both a Type I Region and a Type II Region, whether we will obtain the same result regardless of which

• rder we integrate.

Fubini’s Theorem says we do. We will not state Fubini’s Theorem in any careful manner, but for now will be satisfied with the above vague statement.

Double Integrals Using Polar Coordinates

Suppose we wish to integrate some function over a region best de- scribed using polar coordinates. Such a region might be {(r, θ)|α(θ) ≤ r ≤ β(θ), a ≤ θ ≤ b}. The region might be divided into subregions between two lines going through the origin, making an angle ∆θ with each other, and two circles

• f radius approximately r for which the difference between the radii is

approximately ∆θ. These regions would be almost rectangular, and their areas might be estimated by multiplying together the length of one of their arcs, rep- resenting the base of the rectangle, and ∆r, representing the height of the rectangle. The arcs are arcs in circles of radius ≈ r, subtendend through an angle ∆θ, and thus has length ≈ r∆θ. We can thus approximate the area ∆A of the rectangle by (∆r)·(r∆θ) = r∆r∆θ. This suggests replacing dA in the integral, which is replaced by dx dy when using rectangular coordinates, with r dr dθ. We conclude, although we haven’t actually proven it, that using polar coordinates and iterated integrals,

• α(θ)≤r≤β(θ), a≤θ≤b f(r, θ) dA =

b

a

β(θ)

α(θ) f(r, θ)r dr dθ.

This is sometimes written b

a dθ

β(θ)

α(θ) dr rf(r, θ).

Applications of Double Integrals – Volume

The volume of a solid of the form f(x, y) ≤ z ≤ g(x, y) for (x, y) ∈ R is given by the double integral

4

• R

g(x, y) − f(x, y) dA.

Applications of Double Integrals – Mass

Suppose we, somewhat artificially, associate mass with a plane region R by assigning a density δ = δ(x, y) with each point in the region. If we have a small, compact portion of the region with area ∆A, we might approximate its mass by δ(x∗, y∗) ∆A. using some point (x∗, y∗) in that portion of the region. We may thus estimate the entire mass by a Riemann Sum δ(x∗

i , y∗ i ) ∆Ai, so we conclude the mass is

• R

δ(x, y) dA.

Applications of Double Integrals – Moments

Suppose we have a plane region R with density function δ = δ(x, y) and wish to find its moments about the two axes and its center of mass. First, let’s estimate its moment My about the y−axis. This may later be used to estimate the x−coordinate of its center of mass by taking My M . If we have a small, compact portion of the region with area ∆A, we might approximate its contribution to the total moment by multiplying its mass ∆m by x∗, the signed distance of a point in that portion from the y−axis. Since ∆m ≈ δ(x∗, y∗) ∆A, its contribution to the total moment is ≈ x∗δ(x∗, y∗) ∆A. We may thus estimate My by a Riemann Sum x∗

i δ(x∗ i , y∗ i ) ∆Ai, so

we conclude the moment is My =

• R

xδ(x, y) dA. Similarly, Mx =

• R

yδ(x, y) dA. If the center of mass is (x, y), then x = My M and y = Mx M .

Surface Area

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Consider a surface z = f(x, y), (x, y) ∈ D for some region D ⊂ R2 for which we’d like to find the surface area. If we take a small piece ∆D of the region D with area ∆A and look at the portion of the surface above it, that portion of the surface, if it’s small enough, will be relatively

• flat. It’s surface area ∆S could be estimated by the area of the portion
• f the tangent plane directly above ∆D.

The tangent plane makes some angle γ with the xy−plane. We may view portion of the tangent plane as mirroring ∆D but scaled in one direction by the factor sec γ. It thus seems reasonable that ∆S ≈ (sec γ)∆A. γ will also equal the angle a normal n makes with a vertical line, so n · k = |n||k| cos γ = |n| cos γ and sec γ = |n| n · k. We know < −fx, −fy, 1 > is a normal to the plane, so we may take n =< −fx, −fy, 1 >. Then |n| = f 2

x + f 2 y + 1 and n · k = 1, so

sec γ = f 2

x + f 2 y + 1, and ∆S ≈ f 2 x + f 2 y + 1 ∆A.

If we divide the region D into compact subregions Di with areas ∆Ai, the surface area of the corresponding portion of the surface will be ∆Si ≈ f 2

x + f 2 y + 1 ∆Ai.

We may thus estimate the entire surface area S = n

i=1 ∆Si ≈ n i=1

f 2

x + f 2 y + 1 ∆Ai.

This is a Riemann Sum and we conclude the surface area is S =

• D
• f 2

x + f 2 y + 1 dA.