Q:

# 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 β x2 β y2 and the xy-plane.

Accepted Solution

A:
Surface integral: Parameterize the closed surface by

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

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

We have

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

so we get

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

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

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

Volume integral (divergence theorem): We have divergence

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

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

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

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

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

as desired.