Verify that the divergence theorem is true for the vector field f on the regione. give the flux. f(x, y, z) = x2i + xyj + zk, e is the solid bounded by the paraboloid z = 25 βˆ’ x2 βˆ’ y2 and the xy-plane.

Accepted Solution

Surface integral: Parameterize the closed surface by

[tex]\mathbf s_1(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+(25-u^2)\,\mathbf k[/tex]
[tex]\mathbf s_2(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j[/tex]

with [tex]0\le u\le5[/tex] and [tex]0\le v\le2\pi[/tex], where [tex]\mathbf s_1[/tex] defines the paraboloid part and [tex]\mathbf s_2[/tex] the planar part of the total surface [tex]\mathcal S[/tex].

We have

[tex]{\mathbf s_1}_u\times{\mathbf s_1}_v=2u^2\cos v\,\mathbf i+2u^2\sin v\,\mathbf j+u\,\mathbf k[/tex]
[tex]{\mathbf s_2}_u\times{\mathbf s_2}_v=u\,\mathbf k[/tex]

so we get

[tex]\displaystyle\iint_{\mathcal S}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=[/tex]
[tex]\displaystyle\iint_{\mathcal S_1}\mathbf f(\mathbf s_1(u,v))\cdot(2u^2\cos v\,\mathbf i+2u^2\sin v\,\mathbf j+u\,\mathbf k)\,\mathrm du\,\mathrm dv+\iint_{\mathcal S_2}\mathbf f(\mathbf s_2(u,v))\cdot(u\,\mathbf k)\,\mathrm du\,\mathrm dv[/tex]

The second integral vanishes when computing the dot product, so we're left with the first integral which reduces to

[tex]\displaystyle\int_{v=0}^{v=2\pi}\int_{u=0}^{u=5}(25u-u^3(2u\cos v-1))\,\mathrm du\,\mathrm dv=\frac{625\pi}2[/tex]

Volume integral (divergence theorem): We have divergence

[tex]\nabla\times\mathbf f(x,y,z)=\dfrac{\partial(x^2)}{\partial x}+\dfrac{\partial(xy)}{\partial y}+\dfrac{\partial z}{\partial z}=2x+x+1=3x+1[/tex]

By the divergence theorem, the flux is equivalent to the volume integral

[tex]\displaystyle\iiint_{\mathcal V}\nabla\times\mathbf f(x,y,z)\,\mathrm dV=\iiint_{\mathcal V}(3x+1)\,\mathrm dV[/tex]

where [tex]\mathcal V[/tex] denotes the space enclosed by the surface [tex]\mathcal S[/tex]. Converting to cylindrical coordinates lets us write the integral as

[tex]\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=5}\int_{z=0}^{z=25-r^2}(3r\cos\theta+1)r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=\frac{625\pi}2[/tex]

as desired.