Question

Use the Divergence Theorem to compute the net outward flux of the vector field F across the boundary of region D. D is the region between the spheres of radius 4 and 5 centered at the origin. F = <9z+4x, x-7y, y+9z>

Answer 1

By the divergence theorem,

$\displaystyle\iint_{\partial D}\vec F\cdot\mathrm d\vec S=\iiint_D(\nabla\cdot\vec F)\,\mathrm dV$

We have

$\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(9z+4x)}{\partial x}+\dfrac{\partial(x-7y)}{\partial y}+\dfrac{\partial(y+9z)}{\partial z}=6$

In the integral, convert to spherical coordinates, taking

$x=u\cos v\sin w$

$y=u\sin v\sin w$

$z=u\sin w$

so that

$\mathrm dV=u^2\sin w\,\mathrm du\,\mathrm dv\,\mathrm dw$

Then the flux is

$\displaystyle6\int_{w=0}^{w=\pi}\int_{v=0}^{v=2\pi}\int_{u=4}^{u=5}u^2\sin w\,\mathrm du\,\mathrm dv\,\mathrm dw=\boxed{488\pi}$

Answer 2

The net outward flux of the vector field F across the boundary of region D is 488$\pi$ and this can be determined by using the divergence theorem.

Given :

D is the region between the spheres of radius 4 and 5 centered at the origin. F = <9z+4x, x-7y, y+9z>

According to the divergence theorem:

$\rm \int\int_{\delta D} \bar{F}.d\bar{S} = \int\int\int_D(\bigtriangledown.\bar{F} )dV$

Now, the expression for $\rm \bigtriangledown .\bar{F}$ is given by:

$\rm \bigtriangledown .\bar{F}(x,y,z)=\dfrac{\delta(9z+4x)}{\delta x}+\dfrac{\delta(x-7y)}{\delta y}+\dfrac{\delta(y+9z)}{\delta z}$

Now, the spherical coordinates is given by:

x = u cosv sinw

y = u sinv sinw

z = u sinw

Therefore, the value of dV is given by:

$\rm dV = u^2sinw\;du\;dv\;dw$

Now, the net outward flux of the vector field F across the boundary of region D is given by:

$\rm \int\int_{\delta D} \bar{F}.d\bar{S} =\rm \int^{\pi}_0\int^{2\pi}_0\int^5_4 u^2sinw\;du\;dv\;dw$

Simplify the above integral.

$\rm \int\int_{\delta D} \bar{F}.d\bar{S} =488\pi$

For more information, refer to the link given below:

https://brainly.com/question/24308099