Question

2. Consider the vector field F = (z v)a I zy (z + a)2. Consider also a frustum of cone defined as: (see figure). Let us call V the volume of this solid. Alio, let S be the closed surface enclosing the volume: S -S1 U S2 U S3, where Si is the flat bottom (z = 1), S2 is the curved surface and Ss is the flat top (z 4). (a) calculate the flux Ф-ISF ds, using the appropriate boundary theorem! (Gauss theorem or Stokes theorem? Think carefully). [Hint: use eylindrical coordinates) (b) Take a first step towards calculating Φ directly. compute the flux of F through S3 s, F dS. Do not use boundary theorems here. (o) Bonus: find the unit vector orthogonal to the surface Sz and pointing out of the volume V.

Answer 1

F = (z + y)x + z y + (z + x)z phivariant = integral_s F.ds = integral(div F)dv div(F) = partial differential F_x/partial differential x + partial differential F-y/partial differential y + partial differential F_z/partial differential z partial differential (x + y)/partial differential x + partial differential (z)/partial differential y + partial differential (z + x)/partial differential z integral div(F) dv = integral dv = integral^4_1 pi r^2 dr pi(4^3 - 1/3) = 63pi/3 21 pi integral_s_3F.ds f.ds_3 = (z + x)ds (4 + x)ds integral_s_3 F.ds = integral 4.ds + integral r ds 0 rightarrow as symmetric about x = 4 pi(4)^2 64 pi \r\n + 1/squareroot 2 cos theta I + 1/squareroot 2 sin theta j + - 1/squareroot 2 k