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olasank [31]
3 years ago
5

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.]
Mathematics
1 answer:
Viktor [21]3 years ago
4 0

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 <u>c</u>ap z=\sqrt{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\times\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\times\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\times\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\times\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=\dfrac{\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

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