Surface Integrals of Vector Fields: Definition We learned how to find surface integrals of scalar fields. Let’s now learn how to find surface integrals of vector fields. Setup: • Let F(x, y, z) be a vector field on R3 . • Let S ⊂ R3 be an orientable surface with orientation n. Def: The surface integral (or flux) of F(x, y, z) on an orientable surface S with orientation n(x, y, z) is: ¨ ¨ Flux = F · dS = F · n dS. S S So: The surface integral of the vector field F is defined to be the surface integral of the scalar field F · n. Warning: The orientation of the surface S matters. Using the wrong orientation will often result in the wrong answer. Surface Integrals of Vector Fields: Parametric Surfaces Setup: • Let F(x, y, z) be a vector field on R3 . • Let S ⊂ R3 be an orientable surface with orientation n. Suppose that S is a parametric surface r(u, v) = hx(u, v), y(u, v), z(u, v)i with parameter domain D. The orientation n is a normal vector field to S of unit length, so it must be: ru × rv n=± (1) kru × rv k The ± accounts for the fact that there are two possible orientations for S. In practice, one has to figure out which sign (+ or −) is appropriate. Recall also the formula dS = kru × rv k du dv. (2) Putting the formulas (1) and (2) together, we observe that: ¨ ¨ ¨ ru × rv (kru × rv k du dv) F · dS = F · n dS = F· ± kru × rv k S S D ¨ =± F · (ru × rv ) du dv. D Conclusion: If an orientable surface S is given as a parametric surface r(u, v) with parameter domain D, then the surface integral of F across S is: ¨ ¨ F · dS = ± F · (ru × rv ) du dv. (?) S D ¨ F · dS on parametric surfaces S, one typi- Evaluation: To find the flux S cally needs to: (1) Find a parametrization r of S. (2) Find the parameter domain D. (3) Compute ru × rv and determine + or − in Formula (?). (4) Compute F · (ru × rv ) and apply Formula (?). Surface Integrals of Vector Fields: Graphical Surfaces The formula ¨ ¨ F · dS = ± S F · (ru × rv ) du dv. (?) D works for parametric surfaces r : D → R3 . In the special case where S is a graphical surface z = g(x, y), the formula (?) simplifies. Let’s see how. Setup: • Let F(x, y, z) = hP (x, y, z), Q(x, y, z), R(x, y, z)i be a vector field on R3 . • Let S ⊂ R3 be a graphical surface z = g(x, y) over a region D in the xy-plane, with a choice of orientation (either upwards or downwards). In this case, we can choose our parametrization to be r(x, y) = hx, y, g(x, y)i so that rx × ry = h−gx , −gy , 1i. Therefore: F · (rx × ry ) = hP, Q, Ri · h−gx , −gy , 1i = −P gx − Qgy + R. Finally, notice that rx × ry is always an upward normal vector field to S. Plugging into (?), we have proved: Conclusion: If S is given as a graphical surface z = g(x, y) over the region D ⊂ R2 , then the surface integral of F = hP, Q, Ri across S is: ¨ ¨ (−P gx − Qgy + R) dx dy. (∗) F · dS = ± S D Here, we take the + sign if S is oriented upwards, and take the − sign if S is oriented downwards. ¨ Evaluation: To find the flux F · dS on graphical surfaces S: S (1) Choose + or − depending on orientation of S (upwards or downwards). (2) Determine bounds for the region D in the xy-plane. (3) Use the formula (∗). Exercise: Surface Integrals of Vector Fields 1. Let S be the part of the plane 3x = z − 2 lying within the cylinder x2 + y 2 = 4, oriented upward. Find the flux of F(x, y, z) = h0, y, zi through S. Solution: Our surface S = {3x = z − 2} can be expressed as a graphical surface {z = 3x + 2}. Therefore, we can use formula (∗), meaning that the flux of F across S is: ¨ ¨ Flux = F · dS = ± (−P gx − Qgy + R) dx dy S ¨ D = (−P gx − Qgy + R) dx dy D We chose the + sign because S is oriented upwards. Our vector field is F = hP, Q, Ri = h0, y, zi Our surface is S = {z = 3x + 2}, so we have g(x, y) = 3x + 2, so that gx = 3 gy = 0. Therefore, −P gx − Qgy + R = −0 · 3 − y · 0 + z = z and hence: ¨ ¨ F · dS = Flux = ¨D S = (−P gx − Qgy + R) dx dy z dx dy ¨D (3x + 2) dx dy = D using the fact that z = 3x + 2 on S. We need bounds for the region D, which is the projection of S into the xy-plane. To see the situation clearly, draw a picture. The region D is the disk x2 + y 2 ≤ 4, so we can express D in polar coordinates via 0≤r≤2 0 ≤ θ ≤ 2π. Therefore: ¨ ¨ F · dS = Flux = S (3x + 2) dx dy ˆ ˆ 2 = (3r cos θ + 2) r dr dθ ˆ0 2π ˆ0 2 = 3r2 cos θ + 2r dr dθ ˆ0 2π 0 2 = r3 cos θ + r2 0 dθ ˆ0 2π (8 cos θ + 4) dθ = D 2π 0 = 0 + 4(2π) = 8π ♦ Exercise: Surface Integrals of Vector Fields 2. Find the flux of the vector field F(x, y, z) = hx, y, 3i out of the closed surface bounded by z = x2 + y 2 and z = 4. Solution: Let S be the closed surface in question. It has two pieces: the surface S1 consisting of the plane z = 4, and the surface S2 consisting of the paraboloid z = x2 + y 2 . So, our flux is: ‹ ¨ ¨ Flux = F · dS = F · dS + F · dS. S S1 S2 We need to evaluate both surface integrals. (1) The surface S1 is the graphical surface S1 = {z = 4}. Therefore, ¨ ¨ F · dS = + (−P gx − Qgy + R) dx dy, S1 D1 where D1 is the projection of S1 into the xy-plane. We have chosen the + sign because S1 is oriented upwards. Our vector field is F = hP, Q, Ri = hx, y, 3i Our surface S1 = {z = 4}, so g(x, y) = 4, so gx = 0, gy = 0. Therefore, −P gx − Qgy + R = −0 − 0 + 3 = 3, so that ¨ ¨ F · dS = (−P gx − Qgy + R) dx dy ¨ S1 D1 = 3 dx dy D1 = 3 Area(D1 ) Finally, note that D1 is the disk x2 + y 2 ≤ 4 (draw a picture to see this), so that: ¨ F · dS = 3 Area(D1 ) = 3(πr2 ) = 3(4π) = 12π. S1 (2) The surface S2 is the graphical surface S2 = {z = x2 + y 2 }. Therefore, ¨ ¨ F · dS = − (−P gx − Qgy + R) dx dy S2 D2 ¨ = (P gx + Qgy − R) dx dy D2 where D2 is the projection of S2 into the xy-plane. We have chosen the − sign because S2 is oriented downwards. Our vector field is F = hP, Q, Ri = hx, y, 3i Our surface S1 = {z = x2 + y 2 }, so g(x, y) = x2 + y 2 , so gx = 2x, gy = 2x. Therefore, P gx + Qgy − R = x(2x) + y(2y) − 3 = 2x2 + 2y 2 − 3, so that ¨ ¨ F · dS = (P gx + Qgy − R) dx dy ¨ S2 D2 2x2 + 2y 2 − 3 dx dy. = D2 Finally, note that D2 is the disk x2 + y 2 ≤ 4 (draw a picture to see this). (Yes, in this problem, D2 is the same as D1 . This does not always happen.) In polar coordinates, D2 has bounds 0≤r≤2 Therefore: 0 ≤ θ ≤ 2π. ¨ ˆ 2π ˆ 2 2r2 − 3 r dr dθ F · dS = S2 0 ˆ 0 2π = 0 ˆ 2π = 0 ˆ ˆ 2 2r3 − 3r dr dθ 0 2 4 r2 r −3 dθ 2 2 0 2π (8 − 6) dθ = 0 = 4π. Conclusion: The flux is: ‹ ¨ F · dS = Flux = S ¨ F · dS + S1 F · dS S2 = 12π + 4π = 16π ♦ Outlook: This example was quite tedious. Later in the course, we’ll redo this exact problem with the aid of a powerful theorem: the Divergence Theorem (in R3 ). We will see that the Divergence Theorem will make the solution significantly easier.