Question

<0,0,z> outward flux through region bounded by paraboloid z = x^2+y^2 and z = 1 is double integral of <0,0,z> dot <-2x,-2y,1>, which is the double integral of z dydx on this region. Parameterizing z, this is the double integral of (x^2+y^2)dydx on the region. Using divergence theorem, this should equal the triple integral of 1 dV. However, if the vector field were <0,0,x^2+y^2>, the flux integral would be <0,0,x^2+y^2> dot <-2x,-2y,1>, which again yields the double integral of (x^2+y^2)dydx over this region. Divergence theorem gives triple integral 0 dV this time, a contradiction. Why?

Answer 1

∂/∂z {x^2+y^2} still has to be 1, because z = x^2+y^2

there is no contradiction

Answer 2

But what if it were <0,0,2x^2+3y^2>? Would you substitute in 2z + y^2? When do you substitute? For example <0,0,2x^2+6y^2> yields a different answer than <0,0,2x^2+3y^2> doing the flux integral out, but substituting 2z + 4y^2 for the z component of the vector field <0,0,2x^2+6y^2> would imply that both answers would be triple integral of 2 dV

Answer 3

you could even write <0,0,2x^2+6y^2> as <0,0,6z-4x^2> or <0,0,2z+4y^2>, which yields both the triple integral of 6 dV and the triple integral of 2 dV as answers, when they both are clearly different.

Answer 4

the surface is the paraboloid and the disc.

doing the surface integral on both in this is pretty OK, you don't need div theorem.

for the paraboloid, doing it either as <0,0,z> dot <2x,2y,-1> or <0,0,x^2+y^2> dot <2x,2y,-1> gets you the same result, -pi/2, because on that surface z = x^2+y^2. i have done them.

[NB vector points away from the surface, hence negative (inflow)].

but this changes when you factor in the cap on the paraboloid, ie the lid at z = 1. again doing it as a surface integral, the vector across the whole cap surface is 1 in z direction, applied to an area of pi(1)(1) = 1. the surface integral is pi.

so over all answer is pi - pi/2 = pi/2

if however, you applied a vector <0,0,x^2 + y^2> to cap, you would get a different result.

\[\hat n = <0,0,1> \\ \int\limits \int\limits <0,0,x^2 + y^2> \ \bullet \ <0,0,1> \ dA\]
quick switch to polar and you get pi/2, so overall integral is zero, which is what div theorem gets you.

does that move us forward any? it at least addresses the opening post.

Answer 5

Ok that makes sense thank you. For the other questions I found that you cannot substitute z in, but rather you do get 0 as expected from the divergence theorem if you include the cap.