Tuesday, December 3 Solutions Surface integrals of vector fields and related theorems ∗ ∗ 1. Consider the region D in R3 bounded by the x y-plane and the surface x2 y2 z 1. + + = (a) Make a sketch of D.

Solution. The sketch of D is shown below.

(b) The boundary of D, denoted ∂D, has two parts: the curved top S1 and the flat bottom S2. Parameterize S and calculate the flux of F (0,0,z) through S with respect to the upward 1 = 1 pointing unit normal vector field. Check you answer with the instructor.

Solution. To parametrize S1, one has

r(u, v) u cosv, u sinv, 1 u2 , 0 u 1, 0 v 2π. = 〈 − 〉 ≤ ≤ ≤ ≤ In order to calculate the flux, first we have

r cosv, sinv, 2u , r u sinv, u cosv, 0 , u = 〈 − 〉 v = 〈− 〉 and so r r 2u2 cosv, 2u2 sinv, u . u × v = 〈 〉 Therefore, the flux of F (0,0,z) through S with respect to the upward pointing unit nor- = 1 mal vector field is

Ï Z 2π Z 1 F n dS F(u,v) (ru rv ) dudv S1 · = 0 0 · × Z 2π Z 1 (0,0,1 u2) 2u2 cosv, 2u2 sinv, u dudv = 0 0 − ·〈 〉 Z 2π Z 1 π (1 u2)u dudv . = 0 0 − = 2 (c) Without doing the full calculation, determine the flux of F through S2 with the downward pointing normals.

Solution. Since F 0 on S , we know the flux of F through S is = 2 2 Ï Ï F n dS 0 n dS 0. S2 · = S2 · =

(d) Determine the flux of F through ∂D with the outward pointing normals.

Solution. By adding up the result from (a) and (b), one gets Ï Ï Ï π F n dS F n dS F n dS . ∂D · = S1 · + S2 · = 2

(e) Apply the Divergence Theorem and your answer in (d) to find the volume of D. Check your answer with the instructor.

Solution. By the Divergence Theorem, one has Ï Ñ F n dS divF dV. ∂D · = D Since divF 1, one gets = Ñ Ñ Ï π Volume(D) 1 dV divF dV F n dS . = D = D = ∂D · = 2

2. Consider the vector field F ( y,x,z). = − (a) Compute curl F.

Solution.  i j k  curl F  ∂ ∂ ∂  (0,0,2). =  ∂x ∂y ∂z  = y x z −

Ï ¡ (b) For the surface S1 above, evaluate curl F) n dA. S1 ·

Solution. By applying the parametrization of S1 in 1(b), one gets

Ï Z 2π Z 1 ¡ curl F) n dA (curl F) (ru rv ) dudv S1 · = 0 0 · × Z 2π Z 1 2u dudv 2π. = 0 0 = (c) Check your answer in (b) using Stokes’ Theorem.

Solution. The boundary of S1 is a unit circle centered at the origin in the x y-plane. So we can parametrize it as

C : r(t) cost, sint, 0 , 0 t 2π. = 〈 〉 ≤ ≤ Thus, by Stokes’ Theorem, one has Ï Z Z 2π ¡ curl F) n dA F dr F(r(t)) r0(t)dt S1 · = C · = 0 · Z 2π Z 2π ( sint, cost, 0) sint, cost, 0 dt 1 dt 2π. = 0 − · 〈− 〉 = 0 =

3. If time remains:

(a) Check your answer in 1(e) by directly calculating the volume of D.

Solution. One can use the polar coordinate to calculate the volume of D in 1(e). Let

x r cosθ, y r sinθ, 0 r 1, 0 θ 2π. = = ≤ ≤ ≤ ≤ Then Ï Z 2π Z 1 π Volume(D) (1 x2 y2)d A (1 r 2)rdrdθ . = x2 y2 1 − − = 0 0 − = 2 + ≤ (b) Repeat 2 (b-c) for the surface S2 and also for the surface ∂D. What exactly does 2(c) mean for the surface ∂D?

Solution. The normal vector of S pointing downward is n k. Thus, 2 = − Ï Ï Ï ¡curl F) n dA (0,0,2) (0,0, 1)d A 2 d A 2π. S2 · = S2 · − = − S2 = − To check the above answer using Stokes’ Theorem, one needs the parametrization of the

boundary of S2. Notice that this boundary is the same as that of S1 except the orientation. The boundary of S2 is parametrized by

C 0 : r(t) cos(2π t), sin(2π t), 0 , 0 t 2π. = 〈 − − 〉 ≤ ≤ Thus, by Stokes’ Theorem, one has Ï Z Z 2π ¡ curl F) n dA F dr F(r(t)) r0(t)dt S2 · = C 0 · = 0 · Z 2π ( sin(2π t), cos(2π t), 0) sin(2π t), cos(2π t), 0 dt = 0 − − − ·〈 − − − 〉 Z 2π 1 dt 2π. = 0 − = − By adding up the two integrals one gets Ï Ï Ï ¡curl F) n dA ¡curl F) n dA ¡curl F) n dA 2π ( 2π) 0. ∂D · = S1 · + S2 · = + − =

Since ∂D is a surface without any curve boundary, then 2(c) shows that the integral of F along the curve boundary of the surface ∂D must be 0. (c) For the vector field F ( y,x,z) from the second problem, compute div(curlF). Now sup- = − pose F (F ,F ,F ) is an arbitrary vector field. Can you say anything about the function = 1 2 3 div(curlF)?

Solution. We already know, in 2(a), that curl F (0,0,2), so div(curlF) 0. Generally, sup- = = pose suppose F (F ,F ,F ) is an arbitrary vector field. Then = 1 2 3   i j k µ ¶ ∂F3 ∂F2 ∂F1 ∂F3 ∂F2 ∂F1 curlF det ∂ ∂ ∂  , , , =  ∂x ∂y ∂z  = ∂y − ∂z ∂z − ∂x ∂x − ∂y F1 F2 F3

and µ ¶ µ ¶ µ ¶ ∂F3 ∂F2 ∂F1 ∂F3 ∂F2 ∂F1 div(curlF) ∂x ∂y ∂z = ∂y − ∂z + ∂z − ∂x + ∂x − ∂y 2 2 2 2 2 2 ∂ F3 ∂ F2 ∂ F1 ∂ F3 ∂ F2 ∂ F1 0. = ∂x∂y − ∂x∂z + ∂y∂z − ∂y∂x + ∂z∂x − ∂z∂y =