Lecture 29: Curl, Divergence and Flux

Math 21a: Multivariable calculus Oliver Knill, Fall 2019 Lecture 29: Curl, Divergence and Flux ~ ~ The curl of F = [P; Q] is Qx − Py, a scalar field. The curl of F = [P; Q; R] is curl(P; Q; R) = [Ry − Qz;Pz − Rx;Qx − Py] : We can write curl(F~ ) = r × F~ . Fields of zero curl are called irrotational. 1 The curl of the vector field [x2 + y5; z2; x2 + z2] is [−2z; −2x; −5y4]. If you place a \paddle wheel" pointing into the direction v, its rotation speed F~ · ~v. The direction in which the wheel turns fastest, is the direction of curl(F~ ). The angular velocity is the magnitude of the curl. ~ ~ The divergence of F = [P; Q; R] is div([P; Q; R]) = r · F = Px + Qy + Rz. The ~ ~ divergence of F = [P; Q] is div(P; Q) = r · F = Px + Qy. The divergence measures the \expansion" of a field. Fields of zero divergence are incompressible. ~ ~ ~ ~ With r = [@x;@y;@z], we can write curl(F ) = r × F and div(F ) = r · F . ∆f = div(grad(f)) = fxx + fyy + fzz : is the Laplacian of f. One also writes ∆f = r2f because r · (rf) = div(grad(f). From r · r × F~ = 0 and r × rF~ = ~0, we get div(curl(F~ )) = 0, curl(grad(f)) = ~0. 2 Question: Is there a vector field G~ such that F~ = [x + y; z; y2] = curl(G~ )? Answer: No, because div(F~ ) = 1 is incompatible with div(curl(G~ )) = 0. 3 Show that in simply connected region, every irrotational and incompressible field can be written as a vector field F~ = grad(f) with ∆f = 0. Proof. Since F~ is irrotational, there exists a function f satisfying F = grad(f). Now, div(F ) = 0 implies divgrad(f) = ∆f = 0. 1 4 Find an example of a field which is both incompressible and irrotational. Solution. Find f which satisfies the Laplace equation ∆f = 0, like f(x; y) = x3 − 3xy2, then look at its gradient field F~ = rf. In that case, this gives F~ (x; y) = [3x2 − 3y2; −6xy] : We have now all the derivatives together. In dimension d, there are d fundamental derivatives. 1 grad 1 −! 1 grad curl 1 −! 2 −! 1 grad curl div 1 −! 3 −! 3 −! 1 If a surface S is parametrized as ~r(u; v) = [x(u; v); y(u; v); z(u; v)] over a domain R in the uv-plane and F~ is a vector field, then the flux integral of F~ through S is ZZ ~ F (~r(u; v)) · (~ru × ~rv) dudv : F 1 Compute the flux of F~ (x; y; z) = [0; 1; z2] through the upper half sphere S parametrized by ~r(u; v) = [cos(u) sin(v); sin(u) sin(v); cos(v)] : ~ 2 Solution. We have ~ru × ~rv = − sin(v)~r and F (~r(u; v)) = [0; 1; cos (v)] so that Z 2π Z π −[0; 1; cos2(v)] · [cos(u) sin2(v); sin(u) sin2(v); cos(v) sin(v)] dudv : 0 0 R 2π R π 2 3 R π 3 The flux integral is 0 π=2 − sin (v) sin(u)−cos (v) sin(v) dudv which is − π=2 cos v sin(v) dv = 4 π=2 cos (v)=4j0 = −1=4. 2 Calculate the flux of F~ (x; y; z) = [1; 2; 4z] through the paraboloid z = x2 + y2 lying above the region x2 + y2 ≤ 1. Solution: We can parametrize the surface as ~r(r; θ) = 2 2 2 ~ 2 [r cos(θ); r sin(θ); r ] where ~rr × ~rθ = [−2r cos(θ); −2r sin(θ); r] and F (~r(u; v)) = [1; 2; 4r ]. R ~ ~ R 2π R 1 2 2 3 We get S F · dS = 0 0 (−2r cos(v) − 4r sin(v) + 4r ) drdθ = 2π. RR ~ ~ 3 Evaluate the flux integral S curl(F ) · dS for F (x; y; z) = [xy; yz; zx], where S is the part of the paraboloid z = 4 − x2 − y2 that lies above the square [0; 1] × [0; 1] and has an upward orientation. Solution: curl(F ) = [−y; −z; −x]. The parametrization ~r(u; v) = 2 2 2 2 [u; v; 4 − u − v ] gives ru × rv = [2u; 2v; 1] and curl(F )(~r(u; v)) = [−v; u + v − 4; −u]. The R 1 R 1 2 2 flux integral is 0 0 [−2uv+2v(u +v −4)−u] dvdu = −1=2+1=3+1=2−4−1=2 = −25=6. 2.

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