Question

He integral i = z z ∂w f · ds when f(x, y, z) = y i − 4yz j + 3z 2 k and ∂w is the boundary of the solid w enclosed by the upper half of the sphere x 2 + y 2 + z 2 = 1

Answer 1

Use the divergence theorem:


`\displaystyle\iint_(\partial W)\mathbf f\cdot\mathrm d\mathbf S=\iiint_W\\abla\cdot\mathbf f\,\mathrm dV`

We have divergence


`\\abla\cdot\mathbf f=(\partial(y))/(\partial x)+(\partial(-4yz))/(\partial y)+(\partial(3z^2))/(\partial z)=-4z+6z=2z`

The volume integral is best computed by first converting to spherical coordinates:


`x=r\cos s\sin t`

`y=r\sin s\sin t`

`z=r\cos t`

Now,


`\displaystyle\iint_(\partial W)\mathbf f\cdot\mathrm d\mathbf S=\int_(t=0)^(t=\pi/2)\int_(s=0)^(s=2\pi)\int_(r=0)^(r=1)2r\cos t\,\mathrm dr\,\mathrm ds\,\mathrm dt=2\pi`