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## 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?

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