My attempt:
(a) flux 1 = double integral (u cos v, u sin v, 1) . (0, 0, u) du dv = pi
flux 2 = double integral ( u cos v, u sin v, u) . (-u cos v, -u sin v, u) du dv = 0
i think the answer for flux 1 makes sense because it is the area of unit circle but flux 2 should be pi because it is 3 times the volume of a unit cone.
(b) V = 1/3 triple integral dV , where V is the volume enclosed by a unit circle and a unit cone so V = pi/3
I just started to learn surface integrals and i am confused why i got 0 for flux 2 and i am not sure of my other answers. I really appreciate any feedbacks.
The first surface corresponds to a circle on the plane $z = 1$ with radius 1 and center in $(0,0,1)$. Although the equation for the vector field is the same, you have to imagine it as a collection of arrows transversal to the circle and pointing upwards.
On the other hand, the second surface is a cone, and the vector field is tangent to the surface; in particular, its component along the direction given by the unit normal is 0, so that the flux is null.