Problems: Del Notation; Flux 1. Verify the divergence theorem if F = xi + yj + zk and S is the surface of the unit cube with opposite vertices (0, 0, 0) and (1, 1, 1). Answer: To conﬁrm that S F·n dS = D divF dV we calculate each integral separately. The surface integral is calculated in six parts – one for each face of the cube. Flux through top: n = k ⇒ F · n dS = z dx dy = dx dy 1 ⇒ dx dy = 1. top bottom: right: 1 F · n dS = 0 0 n = −k ⇒ F · n dS = −z dx dy = 0 dx dy ⇒ n = j ⇒ F · n dS = y dx dz = dx dz 1 ⇒ left: front: 1 F · n dS = right F · n dS = 0. bottom dx dz = 1. 0 0 n = −j ⇒ F · n dS = −y dx dz = 0 dx dz ⇒ n = i ⇒ F · n dS = x dy dz = dy dz 1 ⇒ F · n dS = 0. left 1 F · n dS = dy dz = 1. 0 0 front n = −i ⇒ F · n dS = −x dy dz = 0 dy dz ⇒ F · n dS = 0. back The total ﬂux through the surface of the cube is 3. (We could have used geometric reasoning to see that the ﬂux through the back, left and bottom sides is 0; the vectors of F are parallel to the surface along those sides.) back: To calculate the divergence we start by noting divF = 1 + 1 + 1 = 3. Then 3 dV = 3 · (Volume) = 3. divF dV = D F · n dS = We have veriﬁed that S 2. Prove that ∂ operator P ∂x D divF dV in this example. D 1 V(F · F) = F × (V × F) + (F · V)F, where (P, Q, R) · V is the diﬀerential 2 ∂ ∂ + Q ∂y + R ∂z . Answer: We expand the left hand side, then the right hand side, then note that the expansions are equal. As usual, we assume F = (P, Q, R). LHS: 1 V (F · F) = 2 1 V(P 2 + Q2 + R2 ) 2 ∂P ∂Q ∂R ∂P ∂Q ∂R = P +Q +R i+ P +Q +R ∂x ∂x ∂x ∂y ∂y ∂y ∂P ∂Q ∂R + P +Q +R k. ∂z ∂z ∂z j RHS (in two parts): ∂R ∂Q ∂P ∂R ∂Q ∂P − i+ − j+ − k ∂y ∂z ∂z ∂x ∂x ∂y ∂Q ∂P ∂P ∂R = Q −Q −R +R i ∂x ∂y ∂z ∂x ∂R ∂Q ∂Q ∂P + R −R −P +P j ∂y ∂z ∂x ∂y ∂P ∂R ∂R ∂Q + P −P −Q +Q k ∂z ∂x ∂y ∂z ∂Q ∂R ∂R ∂P ∂P ∂Q = Q +R i+ R +P j+ P +Q k ∂x ∂x ∂y ∂y ∂z ∂z ∂P ∂P ∂Q ∂Q ∂R ∂R − Q +R i− R −P j− P +Q k. ∂y ∂z ∂z ∂x ∂x ∂y F × (V × F) = F × ∂ ∂ ∂ +Q +R F ∂x ∂y ∂z ∂P ∂P ∂P ∂Q ∂R ∂Q = P +Q +R i+ P +Q +R ∂x ∂y ∂z ∂x ∂y ∂z ∂R ∂Q ∂R + P +Q +R k. ∂x ∂y ∂z (F · V)F = P j Note that the negative terms in F × (V × F) are cancelled by positive terms in (F · V)F, leading to the desired result.