Question

19. Show that the integral integration over s F. nds = 411, S where F= (x - z)i + (x3+ yz)j - 3xy²k and S is the surface of the cone z = 2 - root(x2+ y2) ? above the xy-plane.​

Answer 1

Answer:The shaddow in the xy plane is indeed a circle, of radius 2, so in cartesian:

V=∫2−2∫4−x2√−4−x2√∫4−x2+y2√x2+y2√1 dzdydx

or in polar (easier):

V=∫2π0∫20∫4−rr1 rdzdrdθ

By symmetry above/below the plane z=2 you can calculate the volume of the lower or upper portion and multiply by 2:

V=2∫2π0∫20∫2r1 rdzdrdθ      or       V=2∫2π0∫20∫4−r21 rdzdrdθ

Geometrically this is the volume of two cones each with base area B=π22 and height h=2. SO the volume should come out to be

V=2⋅13B⋅h=16π3

Step-by-step explanation: