Vector Surface Integrals Math 131 Multivariate Calculus N(S, T) N(S, T) = D Joyce, Spring 2014 Kn(S, T)K

Home , Surface integral

The normal vector N doesn’t have to be a unit normal vector. We can make it a unit vector by dividing it by its length. Let n denote the unit normal vector. Vector surface integrals Math 131 Multivariate Calculus N(s, t) n(s, t) = D Joyce, Spring 2014 kN(s, t)k

Summary of the surface diﬀerentials, surface ar- Then N = kNk n, so we can rewrite the vector eas, and scalar surface integrals that we already surface integral as discussed. The surface diﬀerential dS can be written in ZZ ZZ terms of of the normal vector N, or the tangent F · dS = F(X) · N ds dt vectors Ts and Tt, or Jacobians as X D ZZ dS = kNk ds dt = (F(X) · n) kNk ds dt D ZZ = kTs × Ttk ds dt q = (F · n) dS ∂(y,z) 2 ∂(x,z) 2 ∂(x,y) 2 D = ∂(s,t) + ∂(s,t) + ∂(s,t) ds dt

Any of these expressions of dS can be used to find In other words, the vector differential dS is the the total area of the surface, which is the dou- product of the unit normal vector n and the scalar ZZ ble integral dS, where D is the domain of the differential dS. D parametrization X describing the surface. The scalar surface integral for the scalar field f ZZ Example 1. Exercise 2 in section 7.2 has both a is just the double integral f dS. scalar surface integral and a vector surface integral. D The surface is the same for both.

New topics. The deﬁnition of vector surface Let D be the quarter disk integrals, some examples, and the statement of Stokes’ theorem. D = {(s, t)|s2 + t2 ≤ 1, s ≥ 0, t ≥ 0},

Vector surface integrals. Vector surface inte- 3 grals are defined similarly to scalar surface inte- and let the surface parametrization X : D → R grals. Whereas scalar surface integrals are de- be defined by fined in terms of the scalar differential dS, which is kNk ds dt in terms of dS, vector surface inte- X(s, t) = (s + t, s − t, st). grals are defined in terms of the vector differential dS = N ds dt. Precisely, if F is a vector field in R3, then the The surface looks like it’s a quarter disk in the vector surface integral is defined as xy-plane, but lifted up somewhat in the z-direction. ZZ ZZ One edge is on the line y = x in the xy-plane, F · dS = F(X(s, t)) · N(s, t) ds dt. another on the line y = −x. X D

1 du = 2r dr, helps. ZZ 4 dS X ZZ = 4 kN(s, t)k ds dt D ZZ √ = 4 2s2 + 2t2 + 4 ds dt D √ ZZ √ = 4 2 s2 + t2 + 2 ds dt D √ Z π/2 Z 1 √ = 4 2 r2 + 2 r dr dθ 0 0 √ Z π/2 Z 3 √ = 2 2 u du dθ 0 2 √ Z π/2 = 2 2 2 (33/2 − 23/2) dθ For either a scalar surface integral or a vector 3 √ 0 surface integral, we’ll need to calculate the normal = 2 π 2 (33/2 − 23/2) vector N. Let’s use the expression for N in terms 3 Next, let’s evaluate the vector surface integral of Jacobians this time. ZZ F · dS, where F = xi + yj + zk. ∂(y, z) ∂(x, z) ∂(x, y) X N = i − j + k The deﬁnition for vector surface integrals says ∂(s, t) ∂(s, t) ∂(s, t) ZZ ZZ F · dS = F(X) · N ds dt. X D We already have N. So ∂(y, z) ∂y ∂z ∂y ∂z = − = 1s − (−1)t = s + t F(X) · N = (x, y, z) · (s + t, t − s, −2) ∂(s, t) ∂s ∂t ∂t ∂s = (s + t, s − t, st) · (s + t, t − s, −2) ∂(x, z) ∂x ∂z ∂x ∂z = − = 1s − 1t = s − t 2 2 2 2 ∂(s, t) ∂s ∂t ∂t ∂s = s + 2st + t − s + 2st − t − 2st ∂(x, y) ∂x ∂y ∂x ∂y = 2st. = − = 1(−1) − 1(1) = −2 ∂(s, t) ∂s ∂t ∂t ∂s Therefore, the vector surface integral equals ZZ 2st ds dt, Therefore, N = (s + t, t − s, −2). The length of N D is which, in polar coordinates, equals √ Z π/2 Z 1 kNk = p(s + t)2 + (t − s)2 + 4 = 2s2 + 2t2 + 4. 2r2 sin θ cos θ r dr dθ 0 0 Z 1 Z π/2 ! = r2r dr 2 sin θ cos θ dθ First, let’s evaluate the scalar surface integral ZZ 0 0 f dS, where f(x, y, z) = 4. Polar coordinates 1 π/2 1 4 θ 1 X = 4 r sin = 4 help here, and later the substitution, u = r2 + 2, 0 0

2 Vector surface integrals as flux. One nice We’ll verify Stokes’ theorem. physical interpretation of the vector surface inte- You probably recognize the surface S as a 1 -turn RR 4 gral X F · dS is as a measure of the flow of a fluid of a helicoid where the angle t varies from 0 to 90 across the surface S. Suppose a fluid is moving in degrees. space, and F(x, y, z) is the velocity of that fluid at the point (x, y, z). The dot product F · n, where n is the outward unit normal vector to the surface S, gives the component of the velocity of the fluid going out through the surface S. So, if F is perpen- dicular to n, then there is no fluid going through the surface, but if F is in the same direction as n, then the fluid is going through the surface at velocity F. The vector surface integral sums all the velocities (some can be negative indicating inward flow, some positive indicating outward flow) to give the total velocity through the entire surface S. This is called the flux of F across S.

The statement of Stokes’ theorem. Let a surface S in space have a boundary ∂S. This boundary may have one component, but it may have more than one. All the boundaries have to be oriented in the same way. Let F be a vector field in R3. Stokes’ theorem equates a surface integral to a line integral. It says that the curl of F integrated over the surface S equals the vector field F integrated over the boundary ∂S. Let’s evaluate the surface integral RR ∇ × F · dS ZZ I S ∇ × F · dS = F · ds first. We’ll need the normal vector N. S ∂S

i j k In the special case when the surface S lies in ∂x ∂y ∂z N(s, t) = ∂s ∂s ∂s the (x, y)-plane, we’ve already seen this is a conse- ∂x ∂y ∂z ∂t ∂t ∂t quence of Green’s theorem. The general statement i j k of Stokes’ theorem allows the surface S to be in = cos t sin t 0 space. −s sin t s cos t 1 Example 2 (Stokes’ theorem). Let S be the sur- = (sin t, − cos t, s) face parametrized by We’ll also need the curl of F. X(s, t) = (s cos t, s sin t, t)

i j k for 0 ≤ s ≤ 1 and 0 ≤ t ≤ π/2, and let F be the ∂ ∂ ∂ ∇ × F = vector ﬁeld ∂x ∂y ∂z z x y F(x, y, z) = (z, x, y). = (1, 1, 1)

3 Z Z Then 0 F · ds = F(x3(t)) · x3(t) dt ZZ x3 x3 ∇ × F · dS Z 1 S = (π/2, 0, 1 − t) · (0, −1, 0) dt ZZ 0 = ∇ × F · N ds dt Z 1 D = 0 dt = 0 ZZ 0 = (1, 1, 1) · (sin t, − cos t, s) ds dt D Z π/2 Z 1 Z Z 0 = (sin t − cos t + s) ds dt F · ds = F(x4(t)) · x4(t) dt 0 0 x4 x4 Z π/2 Z 1 1 π = (sin t − cos t + 2 ) dt = = (π/2 − t, 0, 0) · (0, 0, −1) dt 0 4 0 Z 1 H Now let’s evaluate the line integral ∂S F·ds. The = 0 dt = 0 boundary ∂S comes in four parts. First, the line 0 segment from the origin to (1, 0, 0) which we can RR Thus, S ∇ × F · dS = π/4, and Stokes’ theorem parametrize by x1(t) = (t, 0, 0) for 0 ≤ t ≤ 1. is veriﬁed. Second, x2: when s = 1, the helix (cos t, sin t, t) for 0 ≤ t ≤ π/2. Third, the line segment from (0, 1, pi/2) back to the z-axis at (0, 0, π/2), which Math 131 Home Page at we can parametrize by x3(t) = (0, 1 − t, π/2). And http://math.clarku.edu/~djoyce/ma131/ fourth, the line segment from (0, 0, π/2 back to the origin, which we can parametrize by x4(t) = (0, 0, π/2 − t) for 0 ≤ t ≤ π/2. We’ll evaluate the integrand over each of the four parts of the boundary, then add the resulting values to get the integral over all of ∂S. Z Z 0 F · ds = F(x1(t)) · x1(t) dt x1 x1 Z 1 = (0, t, 0) · (1, 0, 0) dt 0 Z 1 = 0 dt = 0 0

Z Z 0 F · ds = F(x1(t)) · x2(t) dt x2 x2 Z π/2 = (t, cos t, sin t) · (− sin t, cos t, 1) dt 0 Z π/2 = (−t sin t + cos2 t + sin t) dt 0 = π/4