Math 425 Notes 9

Multiple Integrals: Defintion

We state a rough version of the fundamental theorem of calculus. Almost all calcula-tions involving integrals lead back to this.

Definition 1. Let f : R→ R and let [a, b] be a closed interval. Let

a = x0 < x1 < · · · < xn = b

be a division of the interval [a, b]. For each i let x∗i ∈ [xi, xi−1]. A sum of the form

i=n∑i=1

f(x∗i )(xi − xi−1)

is called a Riemann sum.

Notation: Keep the notation from the definition above. Define∫ b

a

f(x)dx = limi=n∑i=1

f(x∗i )(xi − xi−1).

The limit is taken over partitions of [a, b] as the length of the pieces all go to zero.

Theorem 1. Fundamental Theorem of Calculus Let f : R→ R be continuous. Then∫ b

a

f(x)dx = F (b)− F (a)

wheredF

dx= f.

Notation: Let xi, xi−1 ∈ R, then ∆xi = xi − xi−1 and similarly for ∆yj.

Definition 2. Let R be the rectangle [a, b]× [c, d] ⊆ R2. Let f : R→ R be a function onR. Let xi = a+ i

n(b− a), yj = c+ j

n(d− c). Then

∫ ∫R

fdxdy = limn→∞

i,j=n∑i,j=1

f(xi, yj)∆xi∆yj

provided the limit exists. The limit is taken over all partitions of [a, b] and [c, d] such thatthe lengths of all the subintervals go to zero.

Theorem 2. Assume that f is continuous on a rectangle R except on points in the unionof a finite number of graphs. Then the limit in the definition above exists.

1

Assume that z = f(x, y) is the graph of a continuous (except possibly on points in thefinite union of graphs) non-negative function. Then

∫ ∫Rfdxdy is the volume of the solid

above the z = 0 plane and below the graph of f.Here is a fancier case. How do we find the volume under the function f(x, y), above

the plane z = 0 and within the triangle with vertices (0, 0), (1, 0), (1, 1). Apparently this isnot covered by the above theorems. We extend the function f by making it equal to zerooutside the triangle and using the rectangle [0, 1] × [0, 1]. The singularities are all on theline y = x and hence are allowable.

How do we compute these double integrals. These integrals are the limits of sumsin a rectangular array. Here are two ways of organizing the summing of numbers in arectangular array. The first method is to add the numbers in each column and thenadding these sums. A second method is to add the numbers in each row and then addingthese sums.

Formally we have

i,j=n∑i,j=1

f(xi, yj)∆xi∆yj =

j=n∑j=1

(i=n∑i=1

f(xi, yj)∆xi

)∆yj

=i=n∑i=1

(j=n∑j=1

f(xi, yj)∆yj

)∆xi.

If we take the limit as n→∞, then the three sums above have the same limit providedf is bounded, and continuous outside a finite union of graphs. This is a delicate result.

Theorem 3. Assume that f is nice on a rectangle R = [a, b] × [c, d]. The first sumconverges to ∫ ∫

R

fdxdy.

The second converges to ∫ d

c

(∫ b

a

fdx

)dy.

The third converges to ∫ b

a

(∫ d

c

fdx

)dy.

These are all equal.

Example 4. Let R = [−1, 2]× [3, 4], f(x, y) = 2x2 + xy + y2. Then∫ ∫R

f(x, y)dxdy =

∫ 4

3

(

∫ 2

−1f(x, y)dx)dy.

We separate out the inside integral to get∫ 2

−1(2x2 + xy + y2)dx = x3 + (x2y)/2 + xy2 |x=2

x=−1 = 5y2 + (3/2)y + 9.

2

Figure 1: Tetrahedron y − z + x = 1

and the whole integral is ∫ y=4

y=3

(5y2 + (3/2)y + 9)dy = LTS.

We can also integrate with respect to y first.

Example 5. We find the volume of the tetrahedron bounded by the planes

x = 0, y = 0, z = 0, y − z + x = 1.

The first step is to sketch this. We sketch the intersection of the plane y − z + x = 1with each of the planes x = 0, y = 0, z = 0. We find that the tetrahedron lies belowthe z = 0 plane. Indeed, it lies below the triangle T in the z = 0 plane with vertices(0, 0), (1, 0), (0, 1).

Up to this point we have performed double integrals only over rectangles. We deal withintegrating over a triangle theoretically by extending the integrand by the value 0. We areonly able to anti-differentiate functions given by a single formula such as ex, x2 − 7x + 1.Here is how we get out of this difficulty. Assume that f(x) given by a formula from x = a

to x = b and is given by zero from x = b to x = c. Then∫ cafdx =

∫ bafdx.

Let f(x, y) = 1−x−y. The function f gives the height of the tetrahedron. We integratewith respect to y first and then x. We get∫ x=1

x=0

(∫ y=1−x

y=0

fdy

)dx.

We could also integrate with respect to x first and then with respect to y. In this casewe get ∫ y=1

y=0

(∫ x=1−y

x=0

f(x)dx

)dy.

3

Figure 2: Limits of Integration

We state Fubini’s Theroerm. Let R be the rectangle [a, b] × [c, d]. Assume that f isbounded everywhere in R and continuous except at points in a finite union of graphs. Theintegrals below are all equal. ∫

R

f(x, y)dxdx∫ y=d

y=c

(∫ x=b

x=a

f(x, y)dx

)dy∫ x=b

x=a

(∫ y=d

y=c

f(x, y)dy

)dx.

We can calculate integrals over more general regions than rectangles. If a region canbe describe as

{(x, y)|φ1(x) ≤ y ≤ φ2(x), a ≤ x ≤ b}We say it is y-simple. If a region can be describe as

{(x, y)|φ1(y) ≤ x ≤ φ2(y), a ≤ y ≤ b}

We say it is x-simple. If a region is either of these we say it is elementary.We can integrate over a y-simple region using the formula∫ x=b

x=a

(

∫ y=φ2(x)

y=φ1(x)

f(x, y)dy)dx.

Example 6. How do you integrate over the annulus A = {(x, y)|a2 ≤ x2 + y2 ≤ b}. Youbreak it up into a number of elementary regions, integrate over each, and then sum theresults. In this case you could use the regions (there are many ways of doing this):

R1 = {(x, y) ∈ A|a ≤ x ≤ b, −√a2 − x2 ≤ y ≤

√a2 − x2}

R2 = {(x, y) ∈ A| − a ≤ x ≤ a, y ≥ 0 ,√a2 − x2 ≤ y ≤

√b2 − x2

R3 = {(x, y) ∈ A| − b ≤ x ≤ −a, · · · }R4 = {(x, y) ∈ A| − a ≤ x ≤ a, y ≤ 0, · · · }

4

Figure 3: Annulus

Example 7. We set up an integral to calculate the volume of a right cone of radius Rand height H. We center the cone so that its central axis is the z axis. First we find aformula for the height of the cone over the point (x, y). By symmetry this only dependson r =

√x2 + y2 so we express h(x, y) in terms of r and then express r in terms of (x, y).

Make a vertical slice through the center of the cone and the point (x, y). We get two similartriangles. One has sides of length H and R and the other has sides h and R−r. We obtain

h

R− r=H

R.

From this we can get h = (R−r)HR

.To obtain the volume of the cone we integrate∫ ∫

S

h(x, y)dxdy,

where S is the disk of radius R centered at the origin. The volume is∫ x=R

x=−R

∫ y=√R2−x2

y=−√R2−x2

h(x, y)dxdy.

Example 8. We find the volume of the solid inside the cylinder x2+4y2 = 4, above z = 0,and below the plane x+ 2y + 2z = 2.

The first step is to sketch the cylinder x2 + 4y2 = 4, and, in particular, see where itintersects the x and y axis. The second step is to sketch the plane, and, in particular, tosee how it intersects the plane z = 0, and also see the three points where it intersects thethree axes.

Let h(x, y) be the height of the plane above a point (x, y, 0), for those points where theplane is above the x− y plane. We have h(x, y) = 2−x−2y

2. This means that the volume is∫ ∫

C

h(x, y)dxdy

where C is the region in the plane z = 0 of points (x, y, 0) such that

5

Figure 4: Cone

• the plane x+ 2y + 2z = 2 is above the z = 0 plane, and

• (x, y) is inside the ellipse x2 + 4y2 = 4.

We can integrate with respect to x first or we can integrate with respect to y first.Either choice requires breaking the region into two pieces. We choose to integrate withrespcet to x first. We just give the limits of integration of the two regions. The totalvolume is the sum of these two volumes. The first is∫ y=1

y=0

∫ x=2−2y

x=−√

4−4y2h(x, y)dxdy.

The second is ∫ y=1

y=−1

∫ x=√

4−4y2

x=−√

4−4y2h(x, y)dxdy.

Example 9. Let f(x, y) = (√x) y2. We set up the limits of integration in order to find

the integral of f over the region x > 0, y > x2, y < 10− x2 in two ways.The first step is to sketch the region.We integrate with respect to y first. We obtain∫ x=

√5

x=0

∫ y=10−x2

y=x2f(x, y)dydx.

We can also integrate with respect to x first. We need to break the region into twoparts in this case, namely the region above the line y = 5 and the region below y = 5. Forthe region above y = 5 we obtain∫ y=10

y=5

∫ x=√10−y

x=0

f(x, y)dxdy.

For the region below y = 5 we obtain∫ y=5

y=0

∫ x=√y

x=0

f(x, y)dxdy.

The integral is the sum of the two integrals above.

6

Figure 5: Cylinder

Figure 6: Two Parabolas

7

Figure 7: Change of Variables: One Variable

Change of Variables

The One Variable Case

Here is how we change variables while integrating in the one variable case. We start witha function f(x) and an interval [a, b]. Let x = g(u) be a ’change of variable’ and assumethat g maps the interval [A,B] to [a = g(A), b = g(B)] in a one-to-one onto fashion. Wehave

dx

du=dg

du= g′(u)

so we can write dx = g′(u)du. We get∫ b

a

f(x)dx =

∫ B

A

f(g(u))g′(u)du.

We look at this with more care. An integral is the limit of sums. We look with morecare at a single summand in one of the sums. Let u and x be corresponding points sog(u) = x. Write ∆u for a small interval around u and and ∆x for image of that intervalwith respect to the map g. We have that

∆x ≈ ∆u · g′(u).

We also have that f(x) = f ◦ g(u). Hence

f(x)∆x ≈ f ◦ g(u)g′(u)∆u.

Taking sums and taking the limit we conclude that∫ b

a

f(x)dx =

∫ B

A

f(g(u))g′(u)du.

Example with Plus and Minus: We integrate∫ x=3

x=1

(3x+ 2)dx

by changing variables. Letx = 3− u

8

sox = 1 =⇒ u = 2, x = 3 =⇒ u = 0, and dx = −du.

We get that ∫ x=3

x=1

(3x+ 2)dx =

∫ u=0

u=2

(3(3− u) + 2)(−du).

Note the limits of integration.

Geometry of Maps R2 → R2

Definition 3. Let f : R2 → R2 and let D∗ ⊂ R2. We say that f(D∗) = {f(P )|P ∈ D∗} isthe image of D∗ under f. We say that f is one to one on D∗ provided f(P ) = f(Q), P,Q ∈D∗ implies P = Q. We say that f : D∗ → D is onto provided S ∈ D implies there is aP ∈ D∗ so that f(P ) = S.

Example 10. I claim that a linear map R2 → R2 which is one-to-one maps parallelogramsto parallelograms. To see this the first step is to describe a parallelogram as

{p ∈ R2|p = b+ su+ tv, b, u, v ∈ R2, s, t ∈ [0, 1]}.

Let A be a one-to-one linear map R2 → R2. Apply A to p and we get

A(p) = A(b) + sA(u) + tA(v),

a parallelogram with one vertices at A(b), A(b)+A(u), A(b)+A(v), A(b)+A(u+v). Wheredo we use ’one-to-one’ in this argument?

We see that A multiplies area by the absolute value of det(A). To see this we first verifyit for the unit square with vertex at the orgin. Second we verify it for an arbitray square.Third we fill up a parallelogram with many small squares and use approximation.

Note that if a linear map f is not one-to-one, then f maps a parallelogram to a pointor to a line segment.

Example 11. We consider the map

p : R2 → R2, p : (r, θ) 7→ (x = r cos(θ), y = r sin(θ)).

• This map is not one-to-one. Indeed p(r, θ) = p(r, θ + 2π).

• The map p maps (0 ≤ r <∞)× (0 ≤ θ < 2π) in a one-to-one fashion onto R2.

• It maps 1 ≤ r ≤ 2 in a many-to-one way onto the annulus with inner radius 1 andouter radius 2.

• The map p takes 0 ≤ θ ≤ π/4 onto a sector corvering 1/8-th of the plane.

Let D be a subset of R2 and D∗ be subset of another copy of R2. Let T be a one-to-onedifferentiable map T : D∗ → D. Let f : D → R. How are the integrals of f and f ◦ Trelated?

9

Figure 8: Change of Variables: Two Variables, Points

Example 12. Let f(x) ≡ 1, and let

T : R2 → R2,

(xy

)7→(

3xy

).

Let D∗ = [0, 1]× [0, 1] and D = [0, 3]× [0, 1] so T is a one-to-one map from D∗ to D. Wehave ∫ ∫

D∗T ◦ f = area of D∗ = 1,

while ∫ ∫D

f = area of D = 3.

To see what is happening use the definition of the integral. To find∫ ∫

D∗ f we breakup D∗ into many small squares S, pick a point P ∈ S, and we take the sum

(f ◦ T )(P )( area of S)

over all the squares S.To find

∫ ∫Df we break up D into the squares T (S), take the point Q = T (P ) ∈ T (S),

and then we add up the summands

f(Q)( area of T (S)).

Note that in the two summands, the terms f ◦T (P ) and f(Q) are equal, but the areasare not the same. Indeed, we have the area of D ≈ |detDT (P )|( area of D∗). We concludethat

f(Q)( area of T (S)) = f ◦ T (P )( area of S)|det(DT (P )|.This gives ∫ ∫

D

f =

∫ ∫D∗f ◦ T |det(DT (P ))|.

This is the Change of Variables Formula.

Definition 4. Let D,D∗ be subsets of Rn and let T : D∗ → D be one-to-one and differ-entiable. The Jacobian of T is the determinant of DT.

10

Figure 9: Change of Variables: Two Variables, Regions

Figure 10: Polar Coordinates

Example 13. Let

p : R2 → R2,

(rθ

)7→(r cos(θ)r sin(θ)

).

We have

Dp(r, θ) =

(cos(θ) −r sin(θ)sin(θ) r cos(θ)

),

so the Jacobian is r.We find the area inside the curves r = sin(θ), θ = 0, θ = π, r = 0 Let D∗ denote the

region of the r, θ plane so that 0 ≤ r ≤ sin(θ). We are asking to find the area of D = p(D∗).To find the area of a region we integrate the function 1 over the region. By the change ofvariable formula this is equal to∫ ∫

D

1 =

∫ ∫D∗|detDp|1 =

∫ θ=π

θ=0

∫ r=sin(θ

r=0

1 · rdrdθ = π/4

We solve this problem in a different way. Multiply r = sin(θ) by r and use r2 = x2 +y2, y = sin(θ) to get x2+y2 = y. Completing the squares gives us x2+(y−(1/2))2 = (1/2)2.This is a circle of radius 1/2 which has an area of π/4.

Example 14. We evaluate the integral

A =

∫ ∞−∞

e−x2

dx.

11

We have

A2 =

∫ ∞−∞

e−x2

dx

∫ ∞−∞

e−y2

dy = lima→∞

∫ ∫Da

e−x2−y2dxdy

where Da is the disk with center at the origin and radius a.We change variables to polar coordinates. This last integral is∫ ∫

Da

e−r2

rdrdθ =

∫ θ=2π

θ=0

∫ r=a

r=0

e−r2

rdrdθ = ∫ 2π

0

−e−r2

2|r=ar=0dθ = 2π(

1

2− e−a

2

2). (1)

Taking the limit we obtainA2 = π =⇒ A =

√π.

Implicitly we have used the definition of integrals of unbounded regions. We just takeit to be the integral of bounded regions which eventually fill up the unbounded region.

Example 15. Let

T : R2 → R2,

(uv

)7→(x = u− uvy = uv

).

Let D be the region bounded by x = 0, y = 0, x + y = 1, x + y = 4. We calculate theintegral ∫

D

1

x+ ydxdy

using change of variables. Write f(x, y) = 1/(x + y). First we sketch the region D. Nextwe find the region D∗ in the u, v plane so that T (D∗) = D. Note that change of variablesallows us to compute integrals over D using integrals over the region D∗.

The region D∗ is bounded by u = 1, u = 4, v = 0, v = 1. To see this note that x+ y =(u−uv)+uv = u. Since u 6= 0, we can solve x = 0 ⇐⇒ u−uv = 0 = u(1−v) ⇐⇒ v = 1.Similarly we see that y = 0 =⇒ v = 0.

The derivative of T is

(1− v −uv u

), so the Jacobian of T is u. Note that u 6= 0 in the

region D∗. Is the map T one-to-one on D∗? This is the same as asking if the equationsx = u − uv, y = uv have a unique solution for (x, y) ∈ D. We find that u = x + y, v =y/u = y/(x+ y). Since x+ y 6= 0 for (x, y) ∈ D, there is a unique solution.

The change of variables formula gives∫ ∫D∗f ◦ T |det(DT )|dudv =

∫ ∫D

fdxdy∫ ∫D∗

1

u· |u|dvdu =

∫ ∫D

1

x+ ydxdy∫ u=4

u=1

∫ v=1

v=0

dvdu =

∫ ∫D

1

x+ y.

We use that |u| = u in D∗.

12

Figure 11: Trapaziods

13