Chapter 8 Change of Variables, Parametrizations, Surface Integrals

Chapter 8

Change of Variables, Parametrizations, Surface Integrals

8.1 The transformation formula

In evaluating any integral, if the integral depends on an auxiliary function of the variables involved, it is often a good idea to change variables and try to simplify the integral. The formula which allows one to pass from the original integral to the new one is called the transformation formula (or change of variables formula). It should be noted that certain conditions need to be met before one can achieve this, and we begin by reviewing the one variable situation. Let D be an open interval, say (a, b), in R , and let ϕ : D → R be a 1-1 , C1 mapping (function) such that ϕ0 6= 0 on D. Put D∗ = ϕ(D). By the hypothesis on ϕ, it’s either increasing or decreasing everywhere on D. In the former case D∗ = (ϕ(a), ϕ(b)), and in the latter case, D∗ = (ϕ(b), ϕ(a)). Now suppose we have to evaluate the integral

b Z I = f(ϕ(u))ϕ0(u) du,

a for a nice function f. Now put x = ϕ(u), so that dx = ϕ0(u) du. This change of variable allows us to express the integral as

ϕ(b) Z Z I = f(x) dx = sgn(ϕ0) f(x) dx,

ϕ(a) D∗ where sgn(ϕ0) denotes the sign of ϕ0 on D. We then get the transformation formula Z Z f(ϕ(u))|ϕ0(u)| du = f(x) dx

D D∗ This generalizes to higher dimensions as follows:

1 Theorem 1. Let D be a bounded open set in Rn, ϕ : D → Rn a C1, 1-1 mapping whose Jacobian determinant det(Dϕ) is everywhere non-vanishing on D, D∗ = ϕ(D), and f an integrable function on D∗. Then we have the transformation formula Z Z Z Z ··· f(ϕ(u))| det Dϕ(u)| du1... dun = ··· f(x) dx1... dxn. D D∗

Of course, when n = 1, det Dϕ(u) is simply ϕ0(u), and we recover the old formula. This theorem is quite hard to prove, and we will discuss the 2-dimensional case in detail in the next section. In any case, this is one of the most useful things one can learn in Calculus, and one should feel free to (properly) use it as much as possible.

8.2 The formula in the plane

Let D be an open set in R2. We will call a mapping ϕ : D → R2 as above primitive if it is either of the form

(P 1)g ˜ :(u, v) 7→ (u, g(u, v)) or of the form

(P 2) h˜ :(u, v) 7→ (h(u, v), v), with g, h continuously diﬀerentiable (i.e. C1) and ∂g/∂v, ∂h/∂u nowhere vanishing on D. The condition ∂g/∂v 6= 0 ensures thatg ˜ is bijective and similarly for h˜. If ϕ is a linear map then it can be written as the composite of primitive linear maps. Indeed a linear map of the form (P1) looks like

1 a (u, v) 7→ (u, v) 0 b and a linear map of the form (P2) like

a0 0 (u, v) 7→ (u, v) b0 1 and we have 1 a a0 0 a0 + ab0 a a0 0 1 a a0 aa0 · = , · = . 0 b b0 1 bb0 b b0 1 0 b b0 b + ab0

Now given any 2 × 2 matrix M one can always solve for a, b, a0, b0 for at least one order of the factors. Hence M will be a product of primitive linear maps.

2 We will now prove the transformation formula when ϕ is a composition of two primitive transformations, one of type (P1) and the other of type (P2), i.e.

φ =g ˜ ◦ h˜ and for f a continuous function. For simplicity, let us assume that the functions ∂g/∂v(u, v) and ∂h/∂u(u, v) are always positive. (If either of them is negative, it is elementary to modify the argu- ∗ ment.) Put D1 = {(h(u, v), v)|(u, v) ∈ D} and D = {(x, g(x, v))|(x, v) ∈ D1}. so that we have bijective maps h˜ g˜ ∗ D −→D1 −→D whose composite is φ. Enclose D1 in a closed rectangle R, and look at the intersection P of D1 with a partition of R, which is bounded by the lines x = xm, m = 1, 2, ..., and v = vr, r = ∗ 1, 2, ..., with the subrectangles Rmr being of sides 4x = l and 4v = k. Let R , ∗ respectively Rmr, denote the image of R, respectively Rmr, under (u, v) → (u, g(u, v)). ∗ Then each Rmr is bounded by the parallel lines x = xm and x = xm + l and by the arcs of the two curves y = g(x, vr) and y = g(x, vr + k) (it is a region of type I in previous terminology). Then we have

x +l Zm ∗ area(Rmr) = (g(x, vr + k) − g(x, vr)) dx.

xm By the mean value theorem of 1-variable integral calculus and continuity of g, we can write

∗ 0 0 area(Rmr) = l[g(xm, vr + k) − g(xm, vr)], 0 for some point xm ∈ (xm, xm +l). By the mean value theorem of 1-variable diﬀerential calculus, we get ∂g area(R∗ ) = lk (x0 , v0 ), mr ∂v m r 0 ∗ for some vr ∈ (vr, vr + k). Now for any continuous function f on D we have ZZ ZZ X 0 0 0 ∗ f(x, y) dx dy = lim sP dx dy = lim f(xm, g(xm, vr))area(Rmr) P P D∗ D∗ m,r

0 0 0 ∗ where sP is the step function with constant value f(xm, g(xm, vr)) on Rmr. Indeed, given > 0, by the small span theorem applied to the continuous function (x, v) 7→ f(x, g(x, v)) we can ﬁnd a partition P so that |f(x0, g(x0, v0)) − f(x, g(x, v))| < area(D∗)

3 0 0 for (x, v), (x , v ) contained in the same subrectangle Rmr of P . But then

ZZ ZZ X 0 0 0 (f(x, y) − sP ) dx dy ≤ |f(x, y) − f(x , g(x , v ))| dx dy < . m m r ∗ m,r ∗ D Rmr Hence we obtain ZZ X 0 0 0 ∂g 0 0 f(x, y) dx dy = lim klf(xm, g(xm, vr)) (xm, vr). P ∂v D∗ m,r

RR ∂g The expression on the right tends to the integral f(x, g(x, v)) ∂v (x, v) dx dv. Thus D1 we get the identity

RR RR ∂g f(x, y) dx dy = f(x, g(x, v)) ∂v (x, v) dx dv ∗ D D1

By applying the same argument, with the roles of x, y (respectively g, h) switched, we obtain

RR ∂g RR ∂g ∂h f(x, g(x, v)) ∂v (x, v) dx dy = f(h(u, v), g(h(u, v), v)) ∂v (h(u, v), v) ∂u (u, v) du dv D1 D

Since ϕ =g ˜ ◦ h˜ we get by the chain rule that

∂g ∂h ˜ 1 ∂u (h(u, v), v) ∂u (u, v) 0 Dϕ(u, v) = Dg˜(h(u, v), v) · Dh(u, v) = ∂g · ∂h 0 ∂v (h(u, v), v) ∂v (u, v) 1 and ∂g ∂h det Dϕ(u, v) = (h(u, v), v) · (u, v) ∂v ∂u which is by hypothesis > 0. Thus we get

RR f(x, y) dx dy = RR f(ϕ(u, v))| det Dϕ(u, v)| du dv D∗ D as asserted in the Theorem. How to do the general case of ϕ? The fact is, we can subdivide D into a ﬁnite union of subregions, on each of which ϕ can be realized as a composition of primitive transformations. We refer the interested reader to chapter 3, volume 2, of ”Introduction to Calculus and Analysis” by R. Courant and F. John.

4 8.3 Examples

(1) Let Φ = {(x, y) ∈ R2|x2 +y2 ≤ a2}, with a > 0. We know that A = area(Φ) = πa2. But let us now do this by using polar coordinates. Put

2 D = {(r, θ) ∈ R |0 ≤ r ≤ a, 0 ≤ θ < 2π}, and deﬁne ϕ : D → Φ by ϕ(r, θ) = (r cos θ, r sin θ). Then it is easy to check that ϕ is 1-1, onto, and moreover, ∂ϕ ∂ϕ = (cos θ, sin θ) and = (−r sin θ, r cos θ). ∂r ∂θ Hence

cos θ −r sin θ det(Dϕ) = det = r, sin θ r cos θ which is non-negative, and is non-zero iﬀ r > 0. The integral will not change if we integrate over {r ≥ 0, 0 ≤ θ ≤ 2π} or over {r > 0, 0 ≤ θ ≤ 2π} as the set {r = 0, 0 ≤ θ ≤ 2π} has content zero. By the transformation formula, we get

a 2π ZZ ZZ Z Z A = dx dy = | det Dϕ(r, θ)| dr dθ = r dr dθ =

Φ D 0 0

a 2π a Z Z Z r2 a = r dr dθ = 2π r dr = 2π = πa2. 2 0 0 0 0 (Note that D is a rectangular strip, which is transformed by ϕ into a circular region.) RR (2) Compute the integral I = R xdxdy where R is the region {(r, φ) | 1 ≤ r ≤ 2, 0 ≤ φ ≤ π/4}. 2 √ √ R 2 R π/4 π/4 r3 i 2 7 2 We have I = 1 0 r cos(φ)rdrdφ = sin(φ)]0 · 3 = 2 (8/3 − 1/3) = 6 . 1 (3) Let Φ be the region inside the parallelogram bounded by y = 2x, y = 2x−2, y = x and y = x + 1. Evaluate I = RR xy dx dy. Φ The parallelogram is spanned by the vectors (1, 2) and (2, 2), so it seems reasonable to make the change of variable

x 1 2 u = y 2 2 v

5 since then we’ll have to integrate over the box [0, 1] × [0, 1] in the u − v-region. The Jacobi matrix of this transformation is of course constant and has determinant −2. So we get

Z 1 Z 1 Z 1 Z 1 I = (u + 2v)(2u + 2v)| − 2|dudv = 2 (2u2 + 6uv + 4v2)dudv 0 0 0 0 Z 1 Z 1 3 2 2 1 2 = 2 2u /3 + 3u v + 4v u 0 dv = 2 (2/3 + 3v + 4v )dv 0 0 2 3 1 = 2(2v/3 + 3v /2 + 4v /3 0) = 2(2/3 + 3/2 + 4/3) = 7.

(4) Find the volume of the cylindrical region in R3 deﬁned by

W = {(x, y, z)|x2 + y2 ≤ a2, 0 ≤ z ≤ h}, where a, h are positive real numbers. We need to compute ZZZ I = vol(W ) = dx dy dz.

W It is convenient here to introduce the cylindrical coordinates given by the transformation

3 ϕ : D → R given by

ϕ(r, θ, z) = (r cos θ, r sin θ, z), where D = {0 ≤ r ≤ a2, 0 ≤ θ ≤ 2π, 0 ≤ z ≤ h}.

It is easy to see ϕ is 1-1, C1 and onto W. Moreover,

cos θ −r sin θ 0 | det Dϕ| = | det sin θ r cos θ 0 | = r 0 0 1 Again, the set {(0, θ, z)} where | det Dϕ| = 0 has content 0, and the integral will be unchanged if this set is included. By the transformation formula,

h a 2π Z Z Z I = r dr dθ dz = πa2h,

0 0 0 as expected.

3 (5) Let W be the unit ball B0(1) in R with center at the origin and radius 1. Evaluate

6 ZZZ I = e(x2+y2+z2)3/2 dx dy dz.

W Here it is best to use the spherical coordinates given by the transformation

3 ψ : D → R , (ρ, θ, φ) → (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ), where D = {0 ≤ ρ ≤ 1, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π}. Then ψ is C1, 1-1 and onto W. Moreover

sin φ cos θ −ρ sin φ sin θ ρ cos φ cos θ det(Dψ) = det sin φ sin θ ρ sin φ cos θ ρ cos φ sin θ = cos φ 0 −ρ sin φ

−ρ sin φ sin θ ρ cos φ cos θ sin φ cos θ −ρ sin φ sin θ = cos φ det − ρ sin φ det = ρ sin φ cos θ ρ cos φ sin θ sin φ sin θ ρ sin φ cos θ

= −ρ2(cos φ)2 sin φ − ρ2(sin φ)3 = −ρ2 sin φ. Note that sin φ is ≥ 0 on [0, π]. Hence | det(Dψ)| = ρ2 sin φ, and we get (by the transformation formula):

1 π 2π 1 π 2π Z Z Z Z Z Z I = eρ3 ρ2 sin φ dρ dφ dθ = eρ3 ρ2 dρ sin φ dφ dθ =

0 0 0 0 0 0

1 1 Z 3 4π Z = 2π eρ ρ2 dρ (− cos φ)|π = eu du, 0 3 0 0 where u = ρ3

4π 1 4π = eu = (e − 1). 3 0 3 8.4 Parametrizations

Let n, k be positive integers with k ≤ n. A subset Φ of Rn is called a parametrized k-fold iﬀ there exist a bounded, connected region T in Rk together with a C1 injective mapping

n ϕ : T → R , u → (x1(u), x2(u), ..., xn(u)), such that ϕ(T ) = Φ. It is called a parametrized surface when k = 2, and a parametrized curve when k = 1.

7 Example: We restrict the spherical coordinates of the above example to the sphere S0(a) of radius a > 0, i.e. we set ρ = a. Then the parametrizing domain is

2 T = {(θ, φ) ∈ R |0 ≤ θ < 2π, 0 < φ < π} = [0, 2π) × (0, π) and the parametrization is given by

(θ, φ) → (a sin φ cos θ, a sin φ sin θ, a cos φ).

Note that we omitted the endpoints θ = 2π and φ = 0, π so that the parametrization is injective. This means we are really parametrizing the unit sphere minus its ”north” and ”south” pole (corresponding to φ = 0 and φ = π and arbitrary θ). If we want to integrate over the sphere, removing those two points doesn’t make a diﬀerence because they form a set of content zero.

8.5 Surface integrals in R3 8.5.1 Integrals of scalar valued functions

Let Φ be a parametrized surface in R3, given by a C1 injective mapping

3 2 ϕ : T → R ,T ⊂ R , ϕ(u, v) = (x(u, v), y(u, v), z(u, v)). When we say that ϕ is C1 we mean that it is so on an open set containing T. Let ξ = ϕ(u, v) = (x, y, z) be a point on Φ. Consider the curve C1 on Φ passing through ξ on which v is constant. Then the tangent vector to C1 at ξ is simply given by ∂ϕ ∂u (u, v). Similarly, we may consider the curve C2 on Φ passing through ξ on which u ∂ϕ is constant. Then the tangent vector to C2 at ξ is given by ∂v (u, v). So the surface Φ ∂ϕ ∂ϕ has a tangent plane JΦ(ξ) at ξ iff ∂u and ∂v are linearly independent there. From now on we will assume this to be the case (see also the discussion of tangent spaces in Ch. 3 of the class notes). When this happens at every point, we call Φ smooth. ∂ϕ ∂ϕ In fact, for integration purposes, it suffices to know that ∂u and ∂v are independent except at a set {ϕ(u, v)} of content zero. By the definition of the cross product, there is a natural choice for a normal vector to Φ at ξ = ϕ(u, v) given by: ∂ϕ ∂ϕ (u, v) × (u, v). ∂u ∂v

Deﬁnition : Let f be a bounded scalar ﬁeld on the parametrized surface Φ. The surface integral of f over Φ, denoted RR f dS, is given by the formula Φ ZZ ZZ ∂ϕ ∂ϕ f dS = f(ϕ(u, v)) (u, v) × (u, v) du dv. ∂u ∂v Φ T

8 ∂ϕ ∂ϕ We say that f is integrable on Φ if f(u, v) ∂u (u, v) × ∂v (u, v) is integrable on T . An important special case is when f = 1. In this case, we get

RR ∂ϕ ∂ϕ area(Φ) = || ∂u (u, v) × ∂v (u, v)|| du dv T

Note that this formula is similiar to that for a line integral in that we have to put in a scaling factor which measures how the parametrization changes the (inﬁnitesimal) length of the curve, resp. area of the surface. Note also that unlike the case of curves this formula only covers the case of surfaces in R3 rather than in a general Rn (more about the general case in the last section).

Note further that the deﬁnition of a surface integral is independent of the 1 parametrization. Indeed, suppose α : T1 → T is a bijective C -map whose total derivative Dα is everywhere bijective, i.e. det(Dα)(u1, v1) 6= 0 for all (u1, v1) ∈ T1. Write α(u1, v1) = (u(u1, v1), v(u1, v1)) and ϕ1(u1, v1) := ϕ(α(u1, v1)). Then the chain rule implies ∂ϕ1 ∂ϕ1 ∂ϕ ∂u ∂ϕ ∂v (u1, v1) × (u1, v1) = (α(u1, v1)) (u1, v1) + (α(u1, v1)) (u1, v1) ∂u1 ∂v1 ∂u ∂u1 ∂v ∂u1 ∂ϕ ∂u ∂ϕ ∂v × (α(u1, v1)) (u1, v1) + (α(u1, v1)) (u1, v1) ∂u ∂v1 ∂v ∂v1 ∂ϕ ∂ϕ ∂u ∂v ∂v ∂u = (α(u1, v1)) × (α(u1, v1)) · (u1, v1) (u1, v1) − (u1, v1) (u1, v1) ∂u ∂v ∂u1 ∂v1 ∂u1 ∂v1 ∂ϕ ∂ϕ = (α(u , v )) × (α(u , v )) · det(Dα)(u , v ) ∂u 1 1 ∂v 1 1 1 1 (8.1) and the transformation formula of Theorem 1 implies ZZ ZZ ∂ϕ ∂ϕ f dS = f(u, v) (u, v) × (u, v) du dv ∂u ∂v Φ T ZZ ∂ϕ1 ∂ϕ1 = f(α(u1, v1)) (u1, v1) × (u1, v1) du1 dv1. ∂u1 ∂v1 T1

3 Example: Find the area of the standard sphere S = S0(a) in R given by x2 + y2 + z2 = a2 with a > 0. Recall the parametrization of S from above given by

2 T = {(θ, φ) ∈ R |0 ≤ θ < 2π, 0 < φ < π} = [0, 2π) × (0, π)

9 and

ϕ(θ, φ) = (x(θ, φ), y(θ, φ), z(θ, φ)) = (a sin φ cos θ, a sin φ sin θ, a cos φ).

So we have ∂ϕ = (−a sin φ sin θ, a sin φ cos θ, 0) ∂θ and ∂ϕ = (a cos φ cos θ, a cos φ sin θ, −a sin φ). ∂φ Hence  i j k  ∂ϕ ∂ϕ × = det −a sin φ sin θ a sin φ cos θ 0 ∂θ ∂φ   a cos φ cos θ a cos φ sin θ −a sin φ a sin φ cos θ 0 −a sin φ sin θ 0 = det i − det j+ a cos φ sin θ −a sin φ a cos φ cos θ −a sin φ −a sin φ sin θ a sin φ cos θ + det k a cos φ cos θ a cos φ sin θ = − a2 cos θ sin2 φi − a2 sin θ sin2 φj − a2(sin2 θ sin φ cos φ + cos2 θ sin φ cos φ)k = − a2 cos θ sin2 φi − a2 sin θ sin2 φj − a2 sin φ cos φk.

π Note that for a point in the upper hemisphere z > 0 we have 0 < φ < 2 with our parametrisation, hence sin φ > 0 and cos φ > 0 which means that the k-component ∂ϕ ∂ϕ of ∂θ × ∂φ is negative. So what we’ve computed here is an inward pointing normal vector to the sphere. Moreover ∂ϕ ∂ϕ || × || =a2(cos2 θ sin4 φ + sin2 θ sin4 φ + sin2 φ cos2 φ)1/2 ∂θ ∂φ =a2(sin4 φ + sin2 φ cos2 φ)1/2 = a2| sin φ|.

So we ﬁnd

2π π π Z Z Z area(S) = a2 dθ | sin φ| dφ = 2πa2 sin φ dφ = 4πa2.

0 0 0

Example: Find the area of the cylinder Φ = ϕ(T ) where T = [0, 2π] × [0, 1] and ϕ(u, v) = (cos(u), sin(u), v). We have ∂ϕ ∂ϕ ∂ϕ ∂ϕ = (− sin(u), cos(u), 0), = (0, 0, 1), × = (cos(u), sin(u), 0) ∂u ∂v ∂u ∂v and ∂ϕ ∂ϕ × = 1. ∂u ∂v

10 Hence Z 1 Z 2π area(Φ) = dudv = 2π. 0 0 Assume now that we are trying to approximate the area of a parametrized surface Φ by a sum over areas of little triangles with vertices lying on Φ, in the same way that we approximated the length of a parametrized curve by a sum over lengths of little line segments with vertices on the curve. This corresponds to a subdivision (partition) of the parameter space T into triangles. Continuing the example of the cylinder we subdivide T = [0, 2π]×[0, 1] into the 2km congruent triangles with vertex sets 2jπ i (2j + 2)π i (2j + 1)π i + 1 S := , , , , , ij m k m k m k and (2j + 1)π i + 1 (2j + 3)π i + 1 (2j + 2)π i S0 := , , , , , ij m k m k m k for 0 ≤ j ≤ m − 1 and 0 ≤ i ≤ k − 1 (and taking the ﬁrst coordinate modulo 2π). Of course, the areas of these 2km triangles sum up to the area of T which is (also) 2π. Now let’s look at the triangle ∆ spanned by ϕ(Sij) which has vertices

2jπ 2jπ i (2j + 2)π (2j + 2)π i cos , sin , , cos , sin , , m m k m m k (2j + 1)π (2j + 1)π i + 1 cos , sin , m m k

π The baseline of ∆ has length b := 2 sin( m ) and the height of ∆ is r 1 π 2 h := + 1 − cos , k2 m so the area of ∆ is r 1 π 1 π 2 area(∆) = · b · h = sin( ) · + 1 − cos . 2 m k2 m Summing over all 2km triangles, our approximation to the area of the cylinder then becomes r π 1 π 2 A := 2 · k · m · sin( ) · + 1 − cos . k,m m k2 m The first thing to notice is that this set of numbers is unbounded as can be seen from fixing m and letting k tend to infinity. This is in contrast to the curve case where the length of a polygon inscribed into a curve is always bounded by the length of the curve (essentially because of the triangle inequality). The second thing to notice is that if we pick a sequence of subdivisions of T whose diameter tends to zero and so that the corresponding sequence of areas Ak,m does have a limit (for example

11 k = m2 → ∞), this limit need not equal the area of Φ. Indeed, setting x = 1 = √1 m k we have

q 2 2 sin(πx) x4 + (1 − cos(πx)) lim Am2,m = lim · m→∞ x→0 x x2 s s x4 + (1 − cos(πx))2 (1 − cos(πx))2 =2π · lim = 2π · 1 + lim x→0 x4 x→0 x4 r π4 =2π · 1 + . 4 Moreover, this limit will usually depend on the particular sequence of subdivisions we choose. For example, for k = m → ∞ the limit is 2π. So the idea of approximating areas by ﬁner and ﬁner subdivisions into triangles fails in all respects.

Here is a useful result:

Proposition 1. Let Φ be a surface in R3 parametrized by a C1, 1-1 function

3 ϕ : T → R of the form ϕ(u, v) = (u, v, h(u, v)). In other words, Φ is the graph of z = h(x, y). Then for any integrable scalar ﬁeld f on Φ, we have r ZZ ZZ ∂h ∂h f dS = f(u, v, h(u, v)) ( )2 + ( )2 + 1 du dv. ∂u ∂v Φ T Proof. ∂ϕ ∂h ∂ϕ ∂h = (1, 0, ) and = (0, 1, ). ∂u ∂u ∂v ∂v i j k  ∂ϕ ∂ϕ ∂h ∂h ∂h ⇒ × = det 1 0 ∂u  L = − i − j + k. ∂u ∂v ∂h ∂u ∂v 0 1 ∂v r ∂ϕ ∂ϕ ∂h ∂h ⇒ || × || = ( )2 + ( )2 + 1. ∂u ∂v ∂u ∂v Now the assertion follows by the deﬁnition of RR f dS. Φ

Example. Let Φ be the surface in R3 bounded by the triangle with vertices (1,0,0), (0,1,0) and (0,0,1). Evaluate the surface integral RR x dS. Φ Note that Φ is a triangular piece of the plane x + y + z = 1. Hence Φ is parametrized by 3 ϕ : T → R , ϕ(u, v) = (u, v, h(u, v)),

12 where h(u, v) = 1 − u − v, and T = {0 ≤ v ≤ 1 − u, 0 ≤ u ≤ 1}. r ∂h ∂h ∂h ∂h √ √ = −1 , = −1, and ( )2 + ( )2 + 1 = 1 + 1 + 1 = 3. ∂u ∂v ∂u ∂v By the Proposition above, we have:

1 1−u 1 √ ZZ √ Z Z √ Z 3 x dS = 3 u du dv = 3 u(1 − u) du = . 6 Φ 0 0 0

8.5.2 Integrals of vector valued functions There is also a notion of an integral of a vector field over a surface. As in the case of line integrals this is in fact obtained by integrating a suitable projection of the vector field (which is then a scalar field) over the surface. Whereas for curves the natural direction to project on is the tangent direction, for a surface in R3 one uses the normal direction to the surface. Note that a unit normal vector to Φ at ξ = ϕ(u, v) is given by

∂ϕ ∂ϕ ∂u (u, v) × ∂v (u, v) n = ∂ϕ ∂ϕ || ∂u (u, v) × ∂v (u, v)|| and that n = n(u, v) varies with (u, v) ∈ T . This defines a unit vector field on Φ called the unit normal field. Definition: Let F be a vector field on Φ. Then the surface integral of F over Φ, denoted RR F · n dS, is defined by Φ ZZ ZZ ∂ϕ ∂ϕ F · n dS = F (ϕ(u, v)) · n(u, v) × du dv ∂u ∂v Φ T ZZ ∂ϕ ∂ϕ = F (ϕ(u, v)) · ( × ) du dv. ∂u ∂v T

∂ϕ ∂ϕ Again, we say that F is integrable over Φ if F (ϕ(u, v))·n(u, v)|| ∂u × ∂v || is integrable over T . It is clear that this deﬁnition is independent of the parametrization if we choose a coordinate change map α : T1 → T with det(Dα)(u1, v1) > 0 for all (u1, v1) ∈ T1. Then we have

∂ϕ1 ∂ϕ1 ∂ϕ ∂ϕ (u1, v1) × (u1, v1) (α(u , v )) × (α(u , v )) ∂u ∂v1 = ∂u 1 1 ∂v 1 1 ∂ϕ1 ∂ϕ1 ∂ϕ ∂ϕ || (u1, v1) × (u1, v1)|| || (α(u1, v1)) × (α(u1, v1))|| ∂u1 ∂v1 ∂u ∂v using (8.1) and det(Dα)(u1, v1) = | det(Dα)(u1, v1)|. So the surface integral we have deﬁned associates a number to a vector ﬁeld and an oriented surface. A change of orientation changes the sign of the integral, just as is the case for line integrals.

13 The notation RR F · n dS does not make reference to any coordinates whereas Φ the deﬁnition of the surface integral explicitly refers to a parametrization. As in the case of line integrals there is a third notation for this integral which makes explicit reference to coordinates (x, y, z) of the ambient space. If F = (P, Q, R) we write ZZ ZZ F · n dS = P dy ∧ dz + Qdz ∧ dx + Rdx ∧ dy. (8.2)

Φ Φ Here the notation v ∧ w indicates a product of vectors which is bilinear (i.e. (λ1v1 + λ2v2) ∧ w = λ1(v1 ∧ w) + λ2(v1 ∧ w) and similarly in the other variable) and antisymmetric (i.e. v ∧ w = −w ∧ v, in particular v ∧ v = 0). In this sense ∧ is similar to × on R3 except that v ∧ w does not lie in the same space where v and w lie. On the positive side v ∧ w can be deﬁned for vectors v, w in any vector space. After these lengthy remarks let’s see how this formalism works out in practice. If the surface Φ is ∂y ∂y parametrized by a function ϕ(u, v) = (x(u, v), y(u, v), z(u, v)) then dy = ∂u du + ∂v dv and by the properties of ∧ outlined above we have ∂y ∂z ∂y ∂z dy ∧ dz = ( − )du ∧ dv ∂u ∂v ∂v ∂u ∂ϕ ∂ϕ which is the x-coordinate of ∂u × ∂v . So the equation (8.2) does indeed hold if we R 2 interpret an integral fdu ∧ dv (over some region T in R ) as an ordinary double R T R R integral T fdudv whereas T fdv∧du equals − T fdudv. All of this suggests that one should give meaning to dx, dy, dz as elements of some vector space instead of being purely formal notation as in our deﬁnition of multiple integrals. This can be done and is in fact necessary to a complete development of integration in higher dimensions and on spaces more general than Rn.

8.6 Integrals over k-folds in Rn

If T is a connected open region of Rk with boundary of content zero, 1 ≤ k ≤ n and ϕ is a parametrization

n ϕ : T → R , u → (x1(u), x2(u), ..., xn(u)), such that Φ := ϕ(T ) has a well deﬁned tangent space at every point ϕ(u), i.e. so that for any u the vectors (Dϕ)(u)(ei) are linearly independent for i = 1, . . . , k, one might wonder what the correct way to integrate over Φ is. In particular, how do we compute the k-dimensional volume of Φ? For that we need a digression on linear algebra. n If v1, . . . , vn are linearly independent vectors in R then the volume of the parallelepiped P spanned by v1, . . . , vn is

vol(P ) = | det(vij)| where vij, j = 1, . . . , n are the coordinates of vi. Here the notion of volume is related to the standard inner product <, > in Rn. In fact, all geometric notions that have to

14 do with measuring lengths of vectors, angles between vectors, areas and volumes can be expressed in terms of the inner product (and would correspondingly change if we would put a different positive definite bilinear form on Rn as the inner product). For t the volume the relation is simply the following. If V is the matrix (vij) then V · V is Pn the symmetric matrix with entries ι=1 viιvjι =< vi, vj >=< vj, vi >. But since V and the transpose matrix V t have the same determinant we get by multiplicativity of the determinant q vol(P ) = | det(V )| = det(< vi, vj >) and this last expression visibly only depends on the inner product. The length of a √ vector v has a similar expression ||v|| = < v, v >. If we now have k linearly inde- n pendent vectors v1, . . . , vk ∈ R then the volume of the k-dimensional parallelepiped P = P (v1, . . . , vk) spanned by v1, . . . , vk is given by the same formula q vol(P ) = det(< vi, vj >) where the determinant is that of a k × k-matrix. One could say that we endow the k-dimensional space spanned by the vi with the inner product of the ambient space Rn. Coming back to our k-fold Φ ⊆ Rn the definition of an integral of a scalar valued function f over Φ is then ZZ ZZ q f dV = f(u) det(< Dϕ(u)(ei), Dϕ(u)(ej) >) du1 ··· duk Φ T s ZZ ∂ϕ ∂ϕ = f(u) det < (u), (u) > du1 ··· duk. (8.3) T ∂ui ∂uj

When discussing the cross product in R3 we have remarked that the length of v1 × v2 is the area of the parallelogram spanned by v1 and v2. Hence for n = 3 we have s < v1, v1 > < v1, v2 > p 2 2 2 kv1 × v2k = det = ||v1|| · ||v2|| − < v1, v2 > , < v2, v1 > < v2, v2 > a formula that can be checked directly. So the case k = 2, n = 3 of formula (8.3) recovers the deﬁnition of surface integrals in R3 given in section 8.5. The case k = 1 of formula (8.3) recovers our deﬁnition of line integrals and the case k = n of formula (8.3) recovers the transformation formula of Theorem 1.