Last time: Greens theorem Let D R 2 be such that D is formed of one - - PowerPoint PPT Presentation

Last time: Green’s theorem

Let D ⊂ R2 be such that ∂D is formed of one or more simple closed curves. Suppose F = ⟨P, Q⟩ is a vector field such that P and Q have continuous first order partial derivatives. Then ∫︂∫︂

D

(Qx − Py)dA = ∫︂

∂D

Pdx + Qdy = ∫︂

∂D

F · dr. Consider the picture on the chalkboard with paths C1, C2, C3. Suppose that Qx = Py on D and also suppose that ∫︁

C1 F · dr = 3

and ∫︁

C2 F · dr = 1. What is

∫︁

C3 F · dr?

(a) 0 (b) 2 (c) 4 (d) We don’t have enough information.

Solution

Fill in the space surrounded by the curves C1, C2, C3 to get a region D with boundary ∂D = (−C1) ∪ (−C2) ∪ C3. So we see that ∫︂

∂D

F · dr = − ∫︂

C1

F · dr − ∫︂

C2

F · dr + ∫︂

C3

F · dr. On the other hand, by Green’s theorem, ∫︂

∂D

F · dr = ∫︂∫︂

D

Qx − PydA, but since Py = Qx this is zero. Combining, we see that 0 = − ∫︂

C1

F · dr − ∫︂

C2

F · dr + ∫︂

C3

F · dr = −3 − 1 + ∫︂

C3

F · dr, so ∫︁

C3 F · dr = 4.

Practice with curl

Find curl⟨ex, z cos y, sin y⟩. How many of the three components are 0? (a) 0 (b) 1 (c) 2 (d) 3 (e) I don’t know. curl⟨ex, z cos y, sin y⟩ = ⟨cos y − cos y, 0 − 0, 0 − 0⟩ = ⟨0, 0, 0⟩

Practice with curl

Find curl⟨P(x, y), Q(x, y), 0⟩. Solution curl⟨P(x, y), Q(x, y), 0⟩ = ⟨∂0 ∂y − ∂Q ∂z , ∂P ∂z − ∂0 ∂x , ∂Q ∂x − ∂P ∂y ⟩ = ⟨0, 0, Qx − Py⟩.

Practice with conservative vector fields and curl

Is F(x, y, z) = ⟨xz, xyz, −y2⟩ conservative? Is G(x, y, z) = ⟨ex, z cos y, sin y⟩ conservative? (a) Yes and yes. (b) Yes and no. (c) No and yes. (d) No and no. (e) Not enough information. curlF ̸= 0, so F is not conservative. curlG = 0 on all of R3, so G is conservative.