Surface Integral over a Hemisphere (Work check please I end up with zero)

[NastyAccident]

Homework Statement

Seawater has density 1025 kg/m^3 and flows in a velocity field v=yi+xj, where x, y, and z are measured in meters and the components of v in meters per second. Find the rate of flow outward through the hemisphere x^2+y^2+z^2=9, z≥0

Homework Equations

Surface integral of F over S is ∫∫ F • dS
In this case,

p * ∫∫_s F • n dS

Where n = the cross product between r_theta and r_phi.

The Attempt at a Solution

First, I parameterized the surface:

[tex]\vec r(\theta,\phi) = \langle 3\sin\phi\cos\theta,3\sin\phi\sin\theta,3\cos\phi \rangle[/tex]
Where 0 < theta < 2pi and 0 < phi < pi/2.

Partial with respect to theta:
[tex]\vec r_{\theta}(\theta,\phi) = \langle -3\sin\phi\sin\theta, 3\sin\phi\cos\theta, 0 \rangle[/tex]

Partial with respect to phi:
[tex]\vec r_{\phi}(\theta,\phi) = \langle 3\cos\phi\cos\theta, 3\cos\phi\sin\theta, -3\sin\phi \rangle[/tex]

Cross:
[tex]\vec r_{\phi}(\theta,\phi) \times r_{\theta}(\theta,\phi) = \langle 9\sin^{2}\phi\cos\theta, 9\sin^{2}\phi\sin\theta, 9\cos\phi\sin\phi \rangle[/tex]

Next, I look at the velocity field and grab the velocity vector:
[tex]\vec v = \langle 3\sin\phi\sin\theta,3\sin\phi\cos\theta,0 \rangle[/tex]

I am now set to integrate:
[tex]\int\int_S \delta\vec v \cdot d\vec S = \int\int_{(\phi,\theta)} \delta\vec v \cdot \vec r_\phi \times \vec r_\theta\ d\phi d\theta[/tex]

[tex]\int^{2\pi}_{0}\int^{\pi/2}_{0} (1025)* \langle 3\sin\phi\sin\theta,3\sin\phi\cos\theta,0 \rangle \cdot \langle 9\sin^{2}\phi\cos\theta, 9\sin^{2}\phi\sin\theta, 9\cos\phi\sin\phi \rangle d\phi d\theta[/tex]

After the dot product I end up with...
27sin^3(phi)cos(theta)sin(theta) + 27sin^3(phi)cos(theta)sin(theta)
which, I simpify to:
54sin^3(phi)cos(theta)sin(theta)

Split the integral in two.
[tex](1025)*54*(\int^{2\pi}_{0} \cos\theta\sin\theta d\theta) (\int^{pi/2}_{0} \sin_^{3}\phi d\phi[/tex]

Trig Identity Substitution:
[tex](1025)*27*(\int^{2\pi}_{0} -1/2\sin2\theta d\theta) (\int^{pi/2}_{0} (1-\cos\phi^{2})\sin\phi d\phi[/tex]

So, I end up with...

[tex]1025*27*((-1/2\cos\theta)^{2\pi}_{0}) ( \cos^{3}\theta/3-\cos\theta)^{\pi/2}_{0})[/tex]

Giving me...
[tex]1025*27*(0) ( \cos^{3}\theta/3-\cos\theta)^{\pi/2}_{0}) = 0[/tex]

Thoughts?

 

[HallsofIvy]

Yes, that is correct (and nice work!), the total flow through the hemisphere is 0. The flow is "anti- symmetric" about the z- axis. The flow through a given point, (x, y, z), on the hemisphere is canceled by the flow through (-x, -y, z).