Question

Problem 4: Let F = (2z + 2)k be the flow field. Answer the following to verify the divergence theorem: a) Use definition to find the upward flux (normal vector pointing upward in z direction) along the hemisphere given in Problem 1(b). b) Use definition to find the downward flux on the circle x 2 + y 2 ≤ 3 on the xy-plane. c) Using the results from parts (a) and (b), what is the net flux over the closed northern hemisphere? d) Use the divergence theorem to verify that this answer agrees. [Hint: dV = rho 2 sin ϕ drho dθ dϕ.]

Answer 1

Given that you mention the divergence theorem, and that part (b) is asking you to find the downward flux through the disk
`x^2+y^2\le3`, I think it's same to assume that the hemisphere referred to in part (a) is the upper half of the sphere
`x^2+y^2+z^2=3`.

a. Let
`C` denote the hemispherical cap
`z=√(3-x^2-y^2)`, parameterized by

`\vec r(u,v)=\sqrt3\cos u\sin v\,\vec\imath+\sqrt3\sin u\sin v\,\vec\jmath+\sqrt3\cos v\,\vec k`

with
`0\le u\le2\pi` and
`0\le v\le\frac\pi2`. Take the normal vector to
`C` to be

`\vec r_v*\vec r_u=3\cos u\sin^2v\,\vec\imath+3\sin u\sin^2v\,\vec\jmath+3\sin v\cos v\,\vec k`

Then the upward flux of
`\vec F=(2z+2)\,\vec k` through
`C` is

`\displaystyle\iint_C\vec F\cdot\mathrm d\vec S=\int_0^(2\pi)\int_0^(\pi/2)((2\sqrt3\cos v+2)\,\vec k)\cdot(\vec r_v*\vec r_u)\,\mathrm dv\,\mathrm du`

`\displaystyle=3\int_0^(2\pi)\int_0^(\pi/2)\sin2v(\sqrt3\cos v+1)\,\mathrm dv\,\mathrm du`

`=\boxed{2(3+2\sqrt3)\pi}`

b. Let
`D` be the disk that closes off the hemisphere
`C`, parameterized by

`\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath`

with
`0\le u\le\sqrt3` and
`0\le v\le2\pi`. Take the normal to
`D` to be

`\vec s_v*\vec s_u=-u\,\vec k`

Then the downward flux of
`\vec F` through
`D` is

`\displaystyle\int_0^(2\pi)\int_0^(\sqrt3)(2\,\vec k)\cdot(\vec s_v*\vec s_u)\,\mathrm du\,\mathrm dv=-2\int_0^(2\pi)\int_0^(\sqrt3)u\,\mathrm du\,\mathrm dv`

`=\boxed{-6\pi}`

c. The net flux is then
`\boxed{4\sqrt3\pi}`.

d. By the divergence theorem, the flux of
`\vec F` across the closed hemisphere
`H` with boundary
`C\cup D` is equal to the integral of
`\mathrm{div}\vec F` over its interior:

`\displaystyle\iint_(C\cup D)\vec F\cdot\mathrm d\vec S=\iiint_H\mathrm{div}\vec F\,\mathrm dV`

We have

`\mathrm{div}\vec F=(\partial(2z+2))/(\partial z)=2`

so the volume integral is

`2\displaystyle\iiint_H\mathrm dV`

which is 2 times the volume of the hemisphere
`H`, so that the net flux is
`\boxed{4\sqrt3\pi}`. Just to confirm, we could compute the integral in spherical coordinates:

`\displaystyle2\int_0^(\pi/2)\int_0^(2\pi)\int_0^(\sqrt3)\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=4\sqrt3\pi`